The density of a fluid in which an object is submerged can be determined using the original mass of the object, the apparent submerged mass of the object, and the density of the object, using the formula provided above. Applying this formula to a 4.00-kg aluminum ball submersed in a liquid and appearing to be 2.10 kg yields a liquid density of 1247 kg/m^3.
Explanation:According to Archimedes' Principle, the buoyant force on a submerged object is equal to the weight of the fluid displaced by the object. From this principle, if an object is fully submerged in a fluid then the volume of the fluid displaced is equal to the volume of the object.
To calculate the density of a liquid, using the mass of the object, the apparent mass, and the density of the object, the formula commonly used is:
ρliquid = (mobject - mapparent) / (mobject / ρobject - mapparent / ρobject).
Let's apply this formula to the situation you described. An aluminum ball of mass 4.00 kg is submerged in a liquid and has an apparent mass of 2.10 kg. The density of aluminum is approximately 2700 kg/m3. Therefore, the density of the liquid is (4.00kg - 2.10kg) / (4.00kg / 2700kg/m3 - 2.10kg / 2700kg/m3) = 1247 kg/m3.
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Archimedes' principle states that when a body is submerged in a liquid, it experiences an upward force equal to the weight of the liquid displaced by the object. The magnitude of the buoyant force is equal to the weight of the displaced liquid and is directed upward. The buoyant force can be used to determine the density of an object as well as the density of a liquid. The formula for Archimedes' principle is as follows:
Buoyant force = Weight of displaced fluid = density of fluid x volume of fluid displaced x g
where g is the acceleration due to gravity.
In this case, we can use the formula to determine the density of the liquid as follows:
Buoyant force = Weight of the object - Apparent weight of the object
Density of fluid x Volume of fluid displaced x g = m_object x g - m_apparent x g
Density of fluid = (m_object - m_apparent) / volume of fluid displaced
Now, substituting the given values, we get:
Density of fluid = (4.00 kg - 2.10 kg) / [(4.00 kg - 2.10 kg)/1000 kg/m³]
Density of fluid = 1900 kg/m³
Therefore, the density of the liquid is 1900 kg/m³.
Determine the density, specific gravity, and mass of the air in a room whose dimensions are 4 m×5 m× 6 m at 100 kPa and 25 C
Answer : The density, specific gravity, and mass of the air in a room is, 1.16825 g/L, 0.916 and 140.19 kg respectively.
Explanation :
First we have to calculate the volume of air.
[tex]Volume=Length\times Breadth\times Height[/tex]
[tex]Volume=4m\times 5m\times 6m[/tex]
[tex]Volume=120m^3=120000L[/tex] [tex](1m^3=1000L)[/tex]
Now we have to calculate the mole of air.
Using ideal gas equation:
[tex]PV=nRT[/tex]
where,
P = Pressure of air = 100 kPa = 0.987 atm (1 atm = 101.3 kPa)
V = Volume of air = 120000 L
n = number of moles of air = ?
R = Gas constant = [tex]0.0821L.atm/mol.K[/tex]
T = Temperature of air = [tex]25^oC=273+25=298K[/tex]
Putting values in above equation, we get:
[tex]0.987atm\times 120000L=n\times (0.0821L.atm/mol.K)\times 298K[/tex]
[tex]n=4841.04mol[/tex]
Now we have to calculate the mass of air.
[tex]\text{Mass of air}=\text{Moles of air}\times \text{Molar mass of air}[/tex]
As we know that the molar mass of air is, 28.96 g/mol
[tex]\text{Mass of air}=4841.04mol\times 28.96g/mol=140196.5184g=140.19kg[/tex]
Now we have to calculate the density of air.
[tex]\text{Density of air}=\frac{\text{Mass of air}}{\text{Volume of air}}[/tex]
[tex]\text{Density of air}=\frac{140.19kg}{120000L}=1.16825\times 10^{-3}kg/L=1.16825g/L[/tex]
Now we have to calculate the specific gravity of air.
[tex]\text{Specific gravity of air}=\frac{\text{Air density at given condition}}{\text{Air density at STP}}[/tex]
As we know that air density at STP is, 1.2754 g/L
[tex]\text{Specific gravity of air}=\frac{1.16825g/L}{1.2754g/L}=0.916[/tex]
Global warming will produce rising sea levels partly due to melting ice caps but also due to the expansion of water as average ocean temperatures rise. We will estimate this effect in this problem, assuming the thermal coefficient of volume expansion is 210×106/∘C for sea water. Calculate the change in length, in meters, of a column of water 1.00 km high for a temperature increase of 1.00∘C. Note that this calculation is only approximate because ocean warming is not uniform with depth.
Answer:
0.07 m
Explanation:
[tex]L_0[/tex] = Initial length = 1 km = 1000 m
[tex]\Delta T[/tex] = Change in temperature = 1.00°C
[tex]\alpha[/tex] = Coefficient of linear thermal expansion
Volumetric coefficient of expansion of water
[tex]\beta=210\times 10^{-6}^{\circ}C\\\Rightarrow \beta=3\alpha\\\Rightarrow \alpha=\dfrac{\beta}{3}\\\Rightarrow \alpha=\dfrac{210\times 10^{-6}}{3}\\\Rightarrow \alpha=70\times 10^{-6}\ ^{\circ}C[/tex]
Change in length is given by
[tex]\Delta L=L_0\alpha \Delta T\\\Rightarrow \Delta L=1000\times 1\times 70\times 10^{-6}\\\Rightarrow \Delta L=0.07\ m[/tex]
The change in length is 0.07 m
The change in length of a column of water for a temperature increase of 1.00°C is 210×10^6 meters.
Explanation:To calculate the change in length of a column of water for a temperature increase, we can use the thermal coefficient of volume expansion for sea water. The thermal coefficient of volume expansion is given as 210×106/°C. To find the change in length, we need to multiply the original length (1.00 km) by the thermal coefficient of volume expansion (210×106/°C) and the temperature change (1.00°C).
Change in length = Original length x Thermal coefficient x Temperature change
Change in length = (1.00 km) x (210×106/°C) x (1.00°C)
Change in length = 210×106 meters
The second-order rate constants for the reaction of oxygen atoms ·with aromatic hydrocarbons have been measured (R. Atkinson and J.N . Pitts, J. Phys. Chem. 79, 295 (1975)). In the reaction with benzene the rate constants are 1.44x 107dm' mor ' s ' at 300.3 K, 3.03 x 107 dm' mol"'s"' at 34 1.2K, and 6.9 x 107 dm' mor's ' at 392.2K. Find the frequency factor and activation energy of the reaction.
Answer: Frequency factor A = 8 × 10⁹
activation energy Ea = 15.5 KJ/Mol
Explanation: to begin, let us first define the parameters given;
K₁ = 1.44 × 10⁷dm³mol⁻¹s⁻¹
K₂ = 3.03 × 10⁷ dm³mol⁻¹s⁻¹
K₃ = 6.9 × 10 dm³mol⁻¹s⁻¹
also T₁ = 300.3 K
T₂ = 341.2 K
T₃ = 392.2 K
we know that;
㏑ K₂ / K₁ = Ea/R [1/T₁ -1/T₂]
where R is given as 8.314 J/mol-k
Ea = activation energy
K₁, K₂ = rate constant
T₁, T₂ = Temperature
therefore, ㏑ (3.03 × 10⁷/ 1.44 × 10⁷) = Ea / 8.314 [1/300.3 - 1/341.2]
this gives Ea = 15496.16 J/Mol ≈ 15.5 KJ /Mol
∴ Ea = 15.5 KJ/ Mol
also given that K = A e⁻∧Ea/RT
here A = frequency factor
∴ 6.9 × 10⁷ = A e⁻ ∧(15496.16/8.314 × 392.2)
A = 7.99 × 10⁹ = 8 × 10⁹
Final answer:
The Arrhenius equation can be used to determine the frequency factor and activation energy of a reaction.
Explanation:
The given question requires determining the frequency factor and activation energy for a reaction. The relationship between the rate constant and temperature is described by the Arrhenius equation:
k = Ae-Ea/RT
To find the frequency factor (A), we can rearrange the equation and use the rate constant (k) at a specific temperature and the activation energy (Ea) to solve for A. Similarly, to find the activation energy, we can rearrange the equation and use the rate constants at two different temperatures to solve for Ea.
An object consists of a rod (of length 3.0 m and negligible moment of inertia) to which four small 2.0-kg masses are attached, one at each end and one at each point on the rod 1.0 m from each end. (The masses are one meter apart.) The moment of inertia of this object about an axis perpendicular to the rod and through one of the inner masses:
Answer:
I = 12 Kg.m²
Explanation:
given,
mass of the small masses = 2 Kg
distance between the masses = 1 m
moment of inertia of object through one of the inner masses.
moment of inertia
taking second block from the left as the reference point
so,
I = m r₁² + m r₂² + m r₃² + m r₄²
r₁ = -1 m , r₂ = 0 m , r₃ = 1 m , r₄ = 2 m
I = m( r₁² + r₂² + r₃² + r₄² )
I = 2 x ( (-1)² + (0)² + (1)² + (2)² )
I = 2 x 6
I = 12 Kg.m²
Hence, the moment of inertia of the object is equal to 12 Kg.m²
The moment of inertia about an axis perpendicular to the rod and through one of the inner masses can be calculated using the parallel-axis theorem.
Explanation:The moment of inertia of this object about an axis perpendicular to the rod and through one of the inner masses can be calculated by using the parallel-axis theorem. By considering the rod and the four masses as separate point masses, the moment of inertia about the given axis can be written as the sum of the individual moments of inertia. The moment of inertia of a point mass is given by I = mr². Since we have four masses, we need to calculate the moment of inertia for each mass and then add them together.
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Here are the positions at three different times for a bee in flight(a bee's top speed is about 7 m/s).
time
4.8 s
5.3 s
5.8 s
position
< -3.2,7.7, 0 > m
< -1.0,6.4, 0 > m
< 0.7,4.6, 0 > m
(a) Between 4.8 s and 5.3 s, what was the bee's average velocity? Becareful with signs.
< 4.4correct check mark, -2.6correct check mark, 0 > m/s
(b) Between 4.8 s and 5.8 s, what was the bee's average velocity? Becareful with signs.
< 3.9correct check mark, -3.1correct check mark, 0 > m/s
(c) Of the two average velocities you calculated, which is the bestestimate of the bee's instantaneous velocity at time 4.8 s?
acorrect check mark
(d) Using the best information available, what was the displacementof the bee during the time interval from 4.8 s to 4.85 s?
< -2.98wrong check mark, 7.57wrong check mark, 0 > m
Answer:
(a) see attachment below
(b) see attachment below
(c) the average velocity in part (a) is a better estimate than that in part (b) because it was calculated for a shorter time interval compared to part (b). The basically definition of instantaneous velocity is that it is the limit of the average velocity as the time interview approaches zero (that is as the time interview becomes smaller and smaller). In other words the shorter the time interview the better the accuracy of the result.
(d) see attachment below.
Explanation:
Check the attachment below for the full solution of this problems.
Thank you for reading this post and I hope it is helpful to you. Thank you.
The average velocities between different time intervals are calculated using displacement and time intervals. The best estimate of the bee's instantaneous velocity is determined by one of the calculated average velocities. To find the displacement during a specific time interval, subtract the initial position from the final position.
Explanation:(a) The average velocity between 4.8 s and 5.3 s can be calculated by finding the displacement and dividing it by the time interval. The displacement is given by subtracting the initial position from the final position, which results in < -4.2, -1.3, 0 > m. Dividing this displacement by the time interval of 0.5 s gives an average velocity of < -8.4, -2.6, 0 > m/s.
(b) Similarly, between 4.8 s and 5.8 s, the displacement is < 3.7, -3.1, 0 > m and the time interval is 1 s. Dividing the displacement by the time interval gives an average velocity of < 3.7, -3.1, 0 > m/s.
(c) The best estimate of the bee's instantaneous velocity at time 4.8 s would be the average velocity between 4.8 s and 5.3 s, which was calculated in part (a) as < 4.4, -2.6, 0 > m/s.
(d) To find the displacement of the bee during the time interval from 4.8 s to 4.85 s, we can subtract the initial position at 4.8 s from the position at 4.85 s. The initial position is < -3.2, 7.7, 0 > m and the position at 4.85 s is < -1, 6.4, 0 > m. Subtracting these vectors gives a displacement of < 2.2, -1.3, 0 > m.
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A plastic sphere floats in water with 50.0% of its volume submerged. This same sphere floats in glycerin with 40.0% of its volume submerged. Determine the densities of:a. The glycerin b. The sphere.
Explanation:
weight of water displaced=weight of sphere,
Since, plastic sphere floats in water with 50.0% of its volume submerged
so, water=2 sphere,
so sphere's density = 1/2 of water's density
Now, weight of glycerin displaced=weight of sphere,
Also given This same sphere floats in glycerin with 40.0% of its volume submerged.
so glycerin =2.5 sphere,
so sphere's density = .4 of glycerin's = 1/2 of water's.
So glycerin = 1.25 water
Density of glycerin is 5/ 4 times of water density.
Density of sphere is 1/2 times of water density.
Density is defined as ratio of mass to volume , that is density is the amount of mass per unit of volume.
Since, plastic sphere floats in water with 50% of its volume submerged.
So, Density of sphere = [tex]\frac{1}{2} *water density[/tex]
Since, same sphere floats in glycerin with 40% of its volume submerged.
So, Density of sphere = [tex]\frac{40}{100}*Glycerin density=\frac{2}{5}*glycerindensity[/tex]
Density of Glycerin = 5/2 of sphere density
Density of glycerin = [tex]\frac{5}{2}*\frac{1}{2}*waterdensity[/tex]
Density of glycerin = [tex]\frac{5}{4} *waterdensity[/tex]
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You are given a sample of metal and asked to determine its specific heat. You weigh the sample and find that its weight is 27.2 N . You carefully add 1.45×104 J of heat energy to the sample and find that its temperature rises 16.0 ∘C
Answer:
c=326.5177 J/kg.K
Specific heat is c=326.5177 J/kg.K
Explanation:
In order ti find the specific heat, we will proceed as follow:
Formula we are going to use is:
[tex]Q=m*c*\Delta T[/tex]
Where:
Q is the heat energy added
m is the mass of sample
c is the specific heat
[tex]\Delta T[/tex] is the temperature Rise.
First we will find the mass:
Weight=m*g (g is gravitational acceleration=9.8 m/s^2)
[tex]m=\frac{weight}{g} \\m=\frac{27.2}{9.8} \\m=2.7755 \ kg[/tex]
Rearranging above formula:
[tex]c=\frac{Q}{m* \Delta T}[/tex]
[tex]c=\frac{1.45*10^4}{2.7755*16}[/tex]
c=326.5177 J/kg.K
Specific heat is c=326.5177 J/kg.K
An object of mass 0.67 kg is attached to a spring with spring constant 15 N/m. If the object is pulled 14 cm from the equilibrium position and released.
What is the maximum speed of the object?
Answer:
The maximum speed of the object is 0.662 m/s.
Explanation:
Given that,
Mass of the object, m = 0.67 kg
Spring constant of the spring, k = 15 N/m
The spring is pulled 14 cm or 0.14 m from the equilibrium position and released.
To find,
The maximum speed of the object.
Solution,
The maximum speed of the object is given by :
[tex]v=A\omega[/tex]........(1)
Where
[tex]\omega=\sqrt{\dfrac{k}{m}}[/tex]
[tex]\omega=\sqrt{\dfrac{15}{0.67}}[/tex]
[tex]\omega=4.73\ rad/s[/tex]
So,
[tex]v=0.14\times 4.73[/tex]
v = 0.662 m/s
So, the maximum speed of the object is 0.662 m/s.
A hydrogen atom emits a photon that has momentum 0.3059×10^(-27) kg·m/s. This photon is emitted because the electron in the atom falls from a higher energy level into the n = 4 level. What is the quantum number of the level from which the electron falls? Use values of h = 6.626×10^-34 J·s, c = 2.998×10^8 m/s, and e = 1.602×10^(-19) C.
Answer:
The quantum number of the higher energy level is 7
Explanation:
Given:
Momentum (p) = 0.3059×10⁻²⁷ kg·m/s
Planck's constant (h) = 6.626×10⁻³⁴ J·s
Speed of light (c) = 2.998×10⁸m/s
Charge of electron (e) = 1.602×10⁻¹⁹ C
Lower energy level (n₂) = 4
Higher energy level (n₁) = ?
From Bohr's model, change in energy of a photon is given as;
[tex]\delta E= (\frac{1}{(n_2)^2} -\frac{1}{(n_1)^2})*13.6eV[/tex]
ΔE = P*C
[tex]P*C = (\frac{1}{(n_2)^2} -\frac{1}{(n_1)^2})*13.6eV[/tex]
[tex]\frac{P*C}{13.6eV} =\frac{1}{(n_2)^2} -\frac{1}{(n_1)^2}[/tex]
[tex]\frac{1}{(n_1)^2}=\frac{1}{(n_2)^2} -\frac{P*C}{13.6eV}[/tex]
[tex]\frac{1}{(n_1)^2}=\frac{1}{(4)^2} -\frac{0.3059X10^{-27}*2.998X10^8}{13.6X1.602X10^{-19}}[/tex]
[tex]\frac{1}{(n_1)^2}=\frac{1}{(16)} -\frac{0.9177}{21.7812}}[/tex]
[tex]\frac{1}{(n_1)^2}=0.0625 -0.0421[/tex]
[tex]\frac{1}{(n_1)^2}=0.0204[/tex]
[tex](n_1)^2 = \frac{1}{0.0204}[/tex]
[tex](n_1)^2 = 49.0196[/tex]
[tex]n_1 =\sqrt{49.0196}[/tex]
n₁ = 7
Therefore, the quantum number of the higher energy level is 7
A vertical spring stores 0.962 J in spring potential energy whena 3.5-kg mass is suspended from it.(a)by what multiplicative factordoes the spring potential energy change if the mass atttached tothe spring is doubled?(b)verify your answer to part (a)bycalculating the spring potential energy when a 7.0-kg mass isattached to the spring.
Answer:
a) If the mass get double then the potential of the spring gets four times.
b) P'=3.848 J
Explanation:
Given that
P= 0.962 J
m = 3.5 kg
m'= 7 kg
Lets take extension in the spring is x when the mass 3.5 kg is attached to the spring.
m g = K x
K=Spring constant
[tex]x=\dfrac{mg}{K}[/tex]
We know that potential energy given as
[tex]P=\dfrac{1}{2}Kx^2[/tex]
[tex]P=\dfrac{1}{2}K\times \dfrac{m^2g^2}{K^2}[/tex]
[tex]P=\dfrac{1}{2}\times \dfrac{m^2g^2}{K}[/tex]
If the mass get double then the potential of the spring gets four times.
[tex]P'=\dfrac{1}{2}\times \dfrac{(2m)^2g^2}{K}[/tex]
[tex]P'=4\times \dfrac{1}{2}\times \dfrac{m^2g^2}{K}[/tex]
P'= 4 P
When mass ,m' = 7 kg
Then potential will be
[tex]0.962=\dfrac{1}{2}\times \dfrac{3.5^2\times g^2}{K} [/tex] -----1
[tex]P'=\dfrac{1}{2}\times \dfrac{7^2\times g^2}{K}[/tex] -------2
From equation 1 and 2
[tex]\dfrac{0.962}{P'}=\dfrac{\dfrac{1}{2}\times \dfrac{3.5^2\times g^2}{K} }{\dfrac{1}{2}\times \dfrac{7^2\times g^2}{K} }[/tex]
P'= 4 x 0.962 J
P'=3.848 J
Answer:
[tex]k=611.517\ N.m^{-1}[/tex]
[tex]U_7=3.8477\ J[/tex]
Explanation:
Given:
spring potential energy stored due to hanging mass, [tex]U=0.962\ J[/tex]mass attached to the spring, [tex]m=3.5\ kg[/tex]Now the force on the mass due to gravity:
[tex]F=m.g[/tex]
[tex]F=3.5 \times 9.8[/tex]
[tex]F=34.3\ N[/tex]
This force pulls the spring down, so:
[tex]F=k.\delta x[/tex]
[tex]34.3=k\times \delta x[/tex] ....................(1)
For the spring potential:
[tex]U=\frac{1}{2} k.\delta x^2[/tex]
[tex]0.962=0.5\times k\times \delta x^2[/tex]
[tex]1.924=k\times \delta x^2[/tex] .........................(2)
Using eq. (1) & (2)
[tex]\frac{1.924}{x^2} =\frac{34.3}{x}[/tex]
[tex]x=0.05609\ m[/tex]
a.
Now the spring factor:
using eq. (1)
[tex]34.3=k\times \delta 0.05609[/tex]
[tex]k=611.517\ N.m^{-1}[/tex]
b.
when mass attached is 7 kg.
The spring potential energy:
[tex]U_7=\frac{1}{2} \times k.\delta x'^2[/tex] ............(3)
Now the force on the mass due to gravity:
[tex]F=m.g[/tex]
[tex]F=7\times 9.8[/tex]
[tex]F=68.6\ N[/tex]
This force pulls the spring down, so:
[tex]F=k.\delta x[/tex]
[tex]68.6=611.517\times \delta x[/tex]
[tex]x=0.11218\ m[/tex]
Using eq. (3)
[tex]U_7=\frac{1}{2}\times 611.517\times 0.11218^2[/tex]
[tex]U_7=3.8478\ J[/tex]
A uniform brick of length 21 m is placed over
the edge of a horizontal surface with a maximum overhang of 10.5 m attained without
tipping. (PICTURE ONE)
Now two identical uniform bricks of length
21 m are stacked over the edge of a horizontal
surface. (PICTURE TWO)
What maximum overhang is possible for
the two bricks (without tipping)?
Answer in units of m.
Answer:
15.75 m
Explanation:
First, let's look at the top brick by itself. In order for it not to tip over the bottom brick, its center of gravity must be right at the edge of the bottom brick. So the edge of the top brick must be 10.5 m from the edge of the bottom brick.
Now let's look at both bricks as a combined mass. We know the total length of this combined brick is 10.5 m + 21 m = 31.5 m. And we know that for it to not tip over the edge of the surface, its center of gravity must be at the edge. So the edge of the combined brick must be 31.5 m / 2 = 15.75 m from the edge of the surface.
The distance of the combined bricks from the edge of the surface is 15.75 m.
How to calculate the maximum overhang possible?The maximum overhang is possible for the two bricks (without tipping) is calculated by applying the following method below;
From the diagram, the top brick's edge needs to be 10.5 meters away from the bottom brick's edge.
Also, if we consider both bricks as a single unit.
The total lengths of the bricks is calculated as follows;
L = 10.5 m + 21 m
L = 31.5 m
For this single unit, its center of gravity needs to be close to the edge in order for it to be balanced on the surface.
The distance of the combined bricks from the edge of the surface is calculated as;
x = L / 2
x = 31.5 / 2
x = 15.75 m
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The two spheres pictured above have equal densities and are subject only to their mutual gravitational attraction. Which of the following quantities must have the same magnitude for both spheres? (A) Acceleration (B) Velocity (C) Kinetic energy (D) Displacement from the center of mass (E) Gravitational force........ I understand the answer is E but dont know how to explain it
Answer:
Gravitational force
Explanation:
If two spheres have equal densities and they are subject only to their mutual gravitational attraction. We need to say that the quantities that must have the same magnitude for both spheres. So, the correct option is (E) i.e. gravitational force.
It is because of Newton's third law of motion. It states that the force due to object 1 to object 2 is same as force due to object 2 to object 1. The two forces act in opposite direction.
Hence, the correct option is (E) "Gravitational force".
The gravitational force has the same magnitude for both spheres.
By the Newton's law of gravitation and the Newton's third law we understand that gravitational force is directly proportional to the masses of the spheres and inversely proportional to the square of distance between centers, also that gravitational force experimented by one sphere has the same magnitude but opposite direction than the gravitational force experimented by the other sphere.
Therefore, we conclude that gravitational force has the same magnitude for both spheres. (Correct choice E)
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A spring operated dart gun fires 10 g darts. Arming the gun requires 185 N of force and results in the shortening of the spring by 10 cm. Find the spring constant. Find the energy stored in the spring. Find the muzzle velocity of the dart.
Answer:
(a) Spring constant = 1850 N/m.
(b) Energy stored in the spring = 9.25 J.
(c) Muzzle velocity of the dart = 43.01 m/s.
Explanation:
(a) Spring constant
From hook's law,
F = ke ...................... Equation 1
Where F = force on the gun, k = spring constant of the gun, e = extension of the gun's spring.
Making k subject of the equation,
k = F/e ............... Equation 2
Given: F = 185 N, e = 10 cm = 0.1 m.
Substitute into equation 2
k = 185/0.1
k = 185/0.1
k = 1850 N/m.
Hence the spring constant = 1850 N/m.
(b) Energy stored in the spring,
E = 1/2ke²............... Equation 3
Where E = Energy stored in the spring, k = spring constant, e = extension.
Given: k = 1850 N/m, e = 0.1 m.
Substitute into equation 3
E = 1/2(1850)(0.1)²
E = 925(0.01)
E = 9.25 J.
(c) Muzzle velocity of the dart.
Kinetic energy of the dart = 1/2mv²
Note: The kinetic energy of the dart is equal to the energy stored in the sprig.
E = 1/2mv²
Where m = mass of the dart, v = velocity of the dart.
Making v the subject of the equation,
v = √(2E/m)............... Equation 4
Given: E = 9.25 J, m = 10 g = 0.01 kg
Substituting these values into equation 4
v = √(2×9.25/0.01)
v = √(18.5/0.01)
v = √1850
v = 43.01 m/s.
Acceleration a velocity v and displacement s can be related to each other without the use of time as :
Answer:
The acceleration and velocity of a body can be related as
V² = U² + 2aS
Where V = the final velocity of the body in (m/s or ft/s)
U = initial velocity of the body (m/s or ft/s)
a = acceleration of the body in (m/s² or ft/s²)
Distance covered by the during that time interval of acceleration in (m or ft)
Explanation:
This equation is very useful in situations where the time interval of motion is not given.
The equation relates the quantities that are along the same axis (x or y).
That is the velocities (initial and final velocities), acceleration and the distance covered must be on the same axis for it to be used correctly. If some of the parameters are on different axis and used together it will lead to errors.
When dealing with multidimensional problems, care should be taken to treat parameters that are along the same axis together and a vector summation be done later to get the requested quantity. Thank you for reading.
A batter hits a softball over a third baseman's head with speed v0 and at an angle ?from the horizontal. Immediately after the ball is hit, the third baseman turns and runs at a constant velocity v=7.000m/s, for a time t=2.000s. He then catches the ball at the same height at which it left the bat. The third baseman was initiallyl=18.00m from home plate (the location where the ball was hit from).
a) Find v0. Use g=9.807m/s2 for the magnitude of the acceleration due to gravity. Assume that there is no air resistance.
b) Find the angle ? in degrees.
c)Find the components vxand vy of the ball’s velocity, v, 0.100 s before the ball is caught.
d)Find the vector components x and y of the ball’s position, r, 0.100 s before the ball is caught.
Answer:
a) The magnitude of the initial velocity is 18.77 m/s.
b) The launching angle is 31.51°.
c) The horizontal component of the velocity at t = 1.900 s is 16.00 m/s.
The vertical component of the velocity vector at t = 1.900 s is -8.823 m/s.
d) The horizontal component of the position vector at time t = 1.900 s is 30.40 m.
The vertical component of the position vector at time t = 1.900 s is 0.9375 m
Explanation:
Hi there!
The equations for the velocity and position vector of the ball are the following:
r = (x0 + v0 · t · cos α, y0 + v0 · t · sin α + 1/2 · g · t²)
v = (v0 · cos α, v0 · sin α + g · t)
Where:
r = position vector at time t
x0 = initial horizontal position.
v0 = initial velocity.
t = time.
α = launching angle.
y0 = initial vertical position.
g = acceleration due to gravity.
v = velocity vector at time t.
a and b) First, let´s find the range of the ball, i.e. the horizontal distance traveled by the ball.
The distance traveled by the baseman can be calculated with this equation:
x = v · t
Where:
x =traveled distance.
v = velocity.
t = time
Then:
x = 7.000 m/s · 2.000 s
x = 14.00 m
The baseman runs 14.00 m. Since he was located 18.00 from the home plate, the horizontal distance traveled by the ball is (14.00 m + 18.00 m) 32.00 m.
If we locate the origin of the frame of reference at the point where the ball is hit, the initial vertical and horizontal positions (x0 and y0) are zero. Since the ball is caught at the same height at which it left the bat, the vertical position of the ball when it is caught is 0.
So, the position vector of the ball at the time when it is caught (2 s after it is hit), is the following:
r = (32.00 m, 0 m)
Using the equations of the x- and y-components of the position vector, we can obtain the initial velocity and the angle:
rx = x0 + v0 · t · cos α (x0 = 0)
ry = y0 + v0 · t · sin α + 1/2 · g · t² (y0 = 0)
rx = 32.00 m = v0 · 2.000 s · cos α
ry = 0 m = v0 · 2.000 s · sin α - 1/2 · 9.807 m/s² · (2.000 s)²
Solving the first equation for v0:
16.00 m/s / cos α = v0
And replacing v0 in the second equation:
0 m = 32 m · sin α / cos α - 1/2 · 9.807 m/s² · (2.000 s)²
1/2 · 9.807 m/s² · (2.000 s)² = 32 m · tan α
1/2 · 9.807 m/s² · (2.000 s)² / 32 m = tan α
α = 31.51°
b) The launching angle is 31.51°
The initial velocity will be:
16.00 m/s / cos α = v0
16.00 m/s / cos (31.51°) = v0
v0 = 18.77 m/s
a) The magnitude of the initial velocity is 18.77 m/s.
c) Let´s use the equation of the velocity vector:
v = (v0 · cos α, v0 · sin α + g · t)
vx = v0 · cos α
vy = v0 · sin α + g · t
The horizontal component of the velocity does not depend on time (neglecting air resistance).
Then:
vx = 18.77 m/s · cos (31.51°)
vx = 16.00 m/s
0.100 s before the ball is caught, the horizontal component of the velocity is 16.00 m/s.
Now let´s calculate the vertical component of the velocity:
vy = 18.77 m/s · sin (31.51°) - 9.807 m/s² · 1.900 s
vy = -8.823 m/s
The vertical component of the velocity vector at t = 1.900 s is -8.823 m/s.
d) Let´s use the same equations we have used in part a).
x = x0 + v0 · t · cos α
x = 18.77 m/s · 1.900 s · cos (31.51°)
x = 30.40 m
The horizontal component of the position vector at time t = 1.900 s is 30.40 m
y = y0 + v0 · t · sin α + 1/2 · g · t²
y = 18.77 m/s · 1.900 s · sin (31.51°) - 1/2 · 9.807 m/s² · (1.900 s)²
y = 0.9375 m
The vertical component of the position vector at time t = 1.900 s is 0.9375 m
Water at room temperature is discharged from a pipe at a rate of 1000 gallons per minute (gpm). Express this flow rate in cubic meters per second, liters per minute and cubic feet per second.
Answer
given,
discharge rate from pipe = 1000 gallons/minutes
now,
flow rate in cubic meters per second
1 gallon = 0.00378541 m³
1 min = 60 s
Q = [tex]1000\times \dfrac{0.00378541\ m^3}{1\ gallon}\times \dfrac{1\ min}{60\ s}[/tex]
Q = 0.063 m³/s
flow rate in liters per minute
1 gallon = 3.78541 L
Q = [tex]1000\times \dfrac{3.78541\ m^3}{1\ gallon}[/tex]
Q = 3785.41 m³/min
flow rate in cubic feet per second
1 gallon = 0.133681 ft³
1 min = 60 s
Q = [tex]1000\times \dfrac{0.133681\ ft^3}{1\ gallon}\times \dfrac{1\ min}{60\ s}[/tex]
Q = 2.23 ft³/s
A spaceship far from all other objects uses its impulse power system to attain a speed of 104 m/s. The crew then shuts off the power. According to Newton's first law, what will happen to the motion of the spaceship from then on? (Select all that apply.)
Answer:
Velocity remains the same at 104 m/s
Explanation:
According to Newton's 1st law of motion, an object subjected to no force or net force equal 0 would maintain its velocity. In our case the crew shuts off the power, spaceship is in space and far from all other objects (so no gravity whatsoever) would have no force acting on it. Therefore its velocity would stay the same at 104 m/s
A regulation basketball has a 32 cm diameter
and may be approximated as a thin spherical
shell.
How long will it take a basketball starting
from rest to roll without slipping 4.8 m down
an incline that makes an angle of 39.4◦ with
the horizontal? The acceleration of gravity is
9.81 m/s^2
Answer in units of s
Answer:
1.8 s
Explanation:
Potential energy = kinetic energy + rotational energy
mgh = ½ mv² + ½ Iω²
For a thin spherical shell, I = ⅔ mr².
mgh = ½ mv² + ½ (⅔ mr²) ω²
mgh = ½ mv² + ⅓ mr²ω²
For rolling without slipping, v = ωr.
mgh = ½ mv² + ⅓ mv²
mgh = ⅚ mv²
gh = ⅚ v²
v = √(1.2gh)
v = √(1.2 × 9.81 m/s² × 4.8 m sin 39.4°)
v = 5.47 m/s
The acceleration down the incline is constant, so given:
Δx = 4.8 m
v₀ = 0 m/s
v = 5.47 m/s
Find: t
Δx = ½ (v + v₀) t
t = 2Δx / (v + v₀)
t = 2 (4.8 m) / (5.47 m/s + 0 m/s)
t = 1.76 s
Rounding to two significant figures, it takes 1.8 seconds.
To determine the time it takes for a basketball to roll without slipping down an incline, we use the motion formula as well as the formula for acceleration. With the given values, the time is approximately 1.243 seconds.
Explanation:SolutionTo solve this problem, first consider the formula that describes the motion of an object rolling down an incline without slipping. The formula is d = (1/2) * a * t^2, where d is the distance, a is the acceleration, and t is the time. We can rearrange this to solve for t, giving us t = sqrt(2 * d / a).
Next, we need to find the acceleration a. This is given by the formula a = g * sin(theta), where g is the acceleration due to gravity (9.81 m/s^2) and theta is the angle of the incline (39.4 degrees). Plugging these values in, we find that a = 9.81 * sin(39.4) = 6.24 m/s^2.
Finally, we can plug these values into our first formula to find the time it takes for the basketball to roll down the incline. This gives us t = sqrt(2 * 4.8 / 6.24) = 1.243 seconds.
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A bag containing originally 60 kg of flour is lifted through a vertical distance of 9 m. While it is being lifted, flour is leaking from the bag at such rate that the number of pounds lost is proportional to the square root of the distance traversed. If the total loss of flour is 12 kg find the amount of work done in lifting the bag.
Answer:
The amount of work done in lifting the bag is -20109.6 N-m
Explanation:
Given that,
Mass of bag = 60 kg
Distance = 9 m
Loss of mass = 12 kg
The number of pounds lost is proportional to the square root of the distance traversed
Mass of the bag containing flour at height is
[tex]m(y)=60-k\sqrt{y}[/tex]
Put the value into the formula
[tex]60-k\sqrt{y}=12[/tex]
[tex]k=144[/tex]
We need to calculate the work done
Using formula of work done
[tex]W=\int_{0}^{9}{m(y)gdy}[/tex]
Put the value into the formula
[tex]W=\int_{0}^{9}{(60-k\sqrt{y})gdy}[/tex]
[tex]W=((60y-\dfrac{2k}{3}\times y^{\frac{3}{2}}})_{0}^{9})\times9.8[/tex]
Put the value of y
[tex]W=(60\times9-\dfrac{2\times144}{3}\times 9^{\frac{3}{2}})\times9.8[/tex]
[tex]W=-20109.6\ N-m[/tex]
Hence, The amount of work done in lifting the bag is -20109.6 N-m
The sound source of a ship’s sonar system operates at a frequency of 22.0 kHz . The speed of sound in water (assumed to be at a uniform 20∘C) is 1482 m/s . What is the difference in frequency between the directly radiated waves and the waves reflected from a whale traveling straight toward the ship at 4.95 m/s ? Assume that the ship is at rest in the water.
Your target variable is Δf, the magnitude of the difference in frequency between the waves emitted from the sonar device and the waves received by the device after reflecting off the whale. Write an expression for Δf in terms of the relevant frequencies using the subscript notation introduced above.
Express your answer in terms of some or all of the variables fLe, fLr, fSe, and fSr.
Δf =_______________.
Answer:
147.45 Hz
[tex]\Delta f=f_{Lr}-f_{Se}[/tex]
Explanation:
v = Speed of sound in water = 1482 m/s
[tex]v_w[/tex] = Speed of whale = 4.95 m/s
The difference in frequency is given by
[tex]\Delta f=f\dfrac{v+v_w}{v-v_w}-f\dfrac{v}{v-v_w}\\\Rightarrow \Delta f=f_{Lr}-f_{Se}[/tex]
Frequency of the wave in stationary condition
[tex]f_{Lr}=f\dfrac{v+v_w}{v-v_w}[/tex]
Ship's frequency which is reflected back
[tex]f_{Se}=f\dfrac{v}{v-v_w}[/tex]
[tex]f_{Se}=22\ kHz[/tex]
[tex]\mathbf{\Delta f=f_{Lr}-f_{Se}}[/tex]
[tex]\Delta f=f_{Lr}-f_{Se}\\\Rightarrow \Delta f=22\times \dfrac{1482+4.95}{1482-4.95}-22\\\Rightarrow \Delta f=0.14745\ kHz\\\Rightarrow \Delta f=147.45\ Hz[/tex]
The difference in wavelength is 147.45 Hz
The Difference in frequency Δf = 147.4 Hz
Expression for Δf in terms of the relevant frequencies ; Δf = fL[tex]_{r}[/tex] - fL[tex]_{s}[/tex]
Given data :
Speed of sound in water ( V ) = 1482 m/s
Speed of whale ( Vw ) = 4.95 m/s
Frequency of ship sonar ( f ) = 22.0 kHz
Calculate the difference in frequencyΔf = [tex]f\frac{v + v_{w} }{v - v_{w} } - f \frac{v}{v-v_{w} }[/tex] ------ ( 1 )
where :
[tex]f \frac{v}{v-v_{w} }[/tex] = Frequency of ship ( F[tex]_{s}[/tex] ) = 22 kHz [tex]f\frac{v + v_{w} }{v - v_{w} }[/tex] = Frequency of wave when stationary ( FL[tex]_{r}[/tex] )= 22 * [ ( 1482 + 4.95) / ( 1482 - 4.95 ) ]
= 22.1474 kHz.
Back to equation ( 1 )
Δf = 22.1474 - 22
= 0.1474 kHz ≈ 147.4 Hz
Hence we can conclude that the Difference in frequency Δf = 147.4 Hz and
Expression for Δf in terms of the relevant frequencies ; Δf = fL[tex]_{r}[/tex] - fL[tex]_{s}[/tex]
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2.09 J>g °C, and that of steam is 2.01 J>g °C. 72. How much heat (in kJ) is evolved in converting 1.00 mol of at - 10.0 °C, to steam at 110.0 °C? The heat capacity of ice is ## 2.01 J>g °C, and that of ice is 2.09 J>g °C. Phase Diagrams steam at 145 °C to ice at - 50 °C? The heat capacity of steam is ##
Answer:
1,54.95KJ
2. phase diagram can be found as attached
Explanation:
Definition of terms
Heat is the degree of hotness or coldness in a body
Specific heat capacity is the amount of heat to raise one kg mass of a substance by 1 degree rise in temperature
How much heat (in kJ) is evolved in converting 1.00 mol of iceat - 10.0 °C, to steam at 110.0 °C?
we take it step by step
the heat needed to raise ice from -10 to 0C
Q=mcdT
mass=mole*relative molecular mass of water
mass (g)=1*18g/mol
mass=18g
Q=18*2.1*(0-(-10)=378J
2. the heat of fusion of ice
Qf=mlf
Qf=18*334J/g
Qf=6012J
3. heat to take water from 0c to 100c
Q=18*4.18*(100)
7524J
4. heat of vapourization
Qv=mLv
Qv=18*2260=40680J
5. heat to raise the steam from 0c to 110c
Q=mCsteam*dT
Q=18*2.01*(110-100)
Q=361.8J
add up all the heat evolved
378+6012+7524+40680+361.8
=54955.8
54.95KJ
A round pipe of varying diameter carries petroleum from a wellhead to a refinery. At the wellhead, the pipe's diameter is 50.9 cm ( 0.509 m) and the flow speed of the petroleum is 11.5 m/s. At the refinery, the petroleum flows at 5.25 m/s. What is the volume flow rate of the petroleum along the pipe and what is the pipe's diameter at the refinery
Answer:
Flow rate 2.34 m3/s
Diameter 0.754 m
Explanation:
Assuming steady flow, the volume flow rate along the pipe will always be constant, and equals to the product of flow speed and cross-section area.
The area at the well head is
[tex]A = \pi r_w^2 = \pi (0.509/2)^2 = 0.203 m^2[/tex]
So the volume flow rate along the pipe is
[tex]\dot{V} = Av = 0.203 * 11.5 = 2.34 m^3/s[/tex]
We can use the similar logic to find the cross-section area at the refinery
[tex]A_r = \dot{V}/v_r = 2.34 / 5.25 = 0.446 m^2[/tex]
The radius of the pipe at the refinery is:
[tex]A_r = \pi r^2[/tex]
[tex]r^2 =A_r/\pi = 0.446/\pi = 0.141[/tex]
[tex]r = \sqrt{0.141} = 0.377m[/tex]
So the diameter is twice the radius = 0.38*2 = 0.754m
Assume that the motor has accelerated the wheel up to an angular velocity Ï1 with angular acceleration α in time t1. At this point, the motor is turned off and a brake is applied that decelerates the wheel with a constant angular acceleration of â5α. Find t2, the time it will take the wheel to stop after the brake is applied (that is, the time for the wheel to reach zero angular velocity). Express your answer in terms of some or all of the following: Ï1, α, and t1.
Answer:
t₂ = t₁ / 5
Explanation:
Rotational kinematics using: ωf = ωi + αt
Starting from rest and speeding up:
ω₁ = 0 + αt₁ .. Eq1
Starting from ω₁ and slowing to a stop:
0 = ω₁ - 5αt₂
Substituting for ω₁ from Eq 1
0 = αt₁ - 5αt₂
5αt₂ = αt₁
5t₂ = t₁
t₂ = t₁ / 5
Find the value of 15.0 N in pounds. Use the conversions 1slug=14.59kg and 1ft=0.3048m. Express your answer in pounds to three significant figures.
Given the equivalent value to convert the units from kilograms to slug and feet to meters, we will proceed to define the equivalence of the 'Newton' in simplified units, that is,
[tex]1N=1kg\cdot m\cdot s^{-2}[/tex]
Then,
[tex]15.0N=15.0kg\cdot m\cdot s^{-2}[/tex]
Converting this value to British units we have that
[tex]15N = 15.0kg\cdot m\cdot s^{-2} (\frac{1slug}{14.59kg})(\frac{1ft}{0.3048m})[/tex]
[tex]15N = 3.37 slug \cdot ft \cdot s^{-2} (\frac{1 lb \cdot ft^{-1}}{1slug})[/tex]
[tex]15N =3.37lb[/tex]
Therefore the value of 15.0 N in pounds is 3.37 lb.
To convert 15 N into pounds, one must first convert Newton to slugs, using the conversion 1slug=14.59kg and then convert slugs to pounds using the conversion 1ft=0.3048m. Using these conversions, we find that 15.0 N is approximately 3.37 pounds.
Explanation:The Newton is a measure of force in the International System of Units (SI), while the pound is a measure of force in the imperial system. To find the equivalent pounds, you'll need to use the given conversions 1slug=14.59kg and 1ft=0.3048m.
Firstly, we need to convert Newtons to slugs. This is because the pound is a unit of force in the American engineering system, where the basic mass unit is the slug, not the kilogram. 1 N is the force required to give a 1 kg mass an acceleration of 1 m/[tex]s^{2}[/tex]. Therefore, we need to convert kilograms to slugs. One slug is equivalent to 14.59 kg. So, we convert the kilograms resulting from the Newton's definition (1 N = 1 kg*m/[tex]s^{2}[/tex]) into slugs with 1slug/14.59kg.
Also, since a pound is .454 kg under the force of gravity, and using 1ft=0.3048m, we can express 1 pound as 1 slug*ft/[tex]s^{2}[/tex]. Therefore, we find that 1N = 0.2248 pounds approximately. Therefore, 15.0 Newtons is equal to 15*0.2248 = 3.37 pounds, when rounded to three significant figures.
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Styrofoam has a density of 300kg/m3. What is the maximum mass that can hang without sinking from a 20.0 cm -diameter Styrofoam sphere in water?
Answer:
1.257 kilograms is the maximum mass that can hang without sinking.
Explanation:
Mass of styrofoam = m
Volume of the sphere = V
V = [tex]\frac{4}{3}\pir^3[/tex]
Diameter of the sphere = d = 20.0 cm
Radius of the of sphere = r = 0.5 d = 10.0 cm = 0.1 m ( 1 cm = 0.01 m)
[tex]V=\frac{4}{3}\times 3.14\times (0.1 m)^3=0.00419 m^3[/tex]
Density of the sphere = [tex]\rho =300 kg/m^3[/tex]
[tex]Density=\frac{Mass}{Volume }[/tex]
Weight of the styrofoam sphere , [tex]W= \rho Vg[/tex]
[tex]m \times g=\rho Vg[/tex]
[tex]m=\rho V[/tex]
[tex]m=\rho \times V = 300 kg/m^3\times 0.00419 m^3=1.257 kg[/tex]
1.257 kilograms is the maximum mass that can hang without sinking.
The question involves finding the density of a polystyrene cube that is partially submerged in water, the effect of adding mass to the cube, and the behavior of the cube in a different fluid. By applying the principles of density, buoyancy, and Archimedes' Principle, one can calculate the necessary parameters involved.
Explanation:The subject of the question relates to the concepts of density, buoyancy, and Archimedes' Principle in Physics. To address the various parts of the question, we will need to apply these principles to calculate the density of the polystyrene, determine the effect of adding mass to the block in water, and examine the block's behavior in a fluid with different density.
Part a: Density of PolystyreneThe question states that 90% of the polystyrene floats above the water's surface. Since the polystyrene is floating, the weight of the displaced water is equal to the weight of the polystyrene. We can use the formula for buoyancy Fb = ρfluidVdisplacedg, where ρfluid is the density of the fluid, Vdisplaced is the volume of the displaced fluid, and g is the acceleration due to gravity.
For 90% of the polystyrene to float, it implies 10% is submerged, thus displacing 10% of its volume in water. Given the density of water as 1000 kg/m³, we find that the density of the polystyrene must be the same as the density of the water multiplied by the submerged volume percentage, yielding a density of 100 kg/m³ for the polystyrene.
Part b: Percentage of Block above Water after Adding MassWhen an additional 0.5 kg mass is placed on the block, the block must displace an additional amount of water equivalent to the weight of the added mass to remain afloat. This results in more of the block submerging to increase the displaced water volume. The exact new percentage of the block that remains above water can be calculated using the same principle, where the new weight of the system (polystyrene + added mass) equals the weight of the displaced water.
Considering Different FluidsWhen the container's fluid changes to ethyl alcohol with a lower density, the buoyancy effect is reduced due to the altered density-weight relationship. For the block to float, the weight of the block plus any mass on top must be less than or equal to the buoyancy force provided by the ethyl alcohol.
Determining Density of a Spherical PolystyreneTo determine the density of a spherical polystyrene without using water, one could weigh the sphere to find its mass (m) and then measure its volume (V) using the formula for the volume of a sphere V = 4/3πr³, where r is the radius. The density ρ is then calculated using the formula ρ = m/V.
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An egg drops from a second-story window, taking 1.13 s to fall and reaching a speed of 11.1 m/s just before hitting the ground. On contact with the ground, the egg stops completely in 0.140 s. Calculate the average magnitudes of its acceleration while falling and while stopping.
m/s2 (acceleration while falling)
m/s2 (deceleration while stopping)
Answer:
While falling, the magnitude of the acceleration of the egg is 9.82 m/s²
While stopping, the magnitude of the deceleration of the egg is 79.3 m/s²
Explanation:
Hi there!
The equation of velocity of the falling egg is the following:
v = v0 + a · t
Where:
v = velocity at time t.
v0 = initial velocity.
a = acceleration.
t = time
Let´s calculate the acceleration of the egg while falling. Notice that the result should be the acceleration of gravity, ≅ 9.8 m/s².
v = v0 + a · t
11.1 m/s = 0 m/s + a · 1.13 s (since the egg is dropped, the initial velocity is zero). Solving for "a":
11.1 m/s / 1.13 s = a
a = 9.82 m/s²
While falling, the magnitude of the acceleration of the egg is 9.82 m/s²
Now, using the same equation, let´s find the acceleration of the egg while stopping. We know that at t = 0.140 s after touching the ground, the velocity of the egg is zero. We also know that the velocity of the egg before hiiting the ground is 11.1 m/s, then, v0 = 11.1 m/s:
v = v0 + a · t
0 = 11.1 m/s + a · 0.140 s
-11.1 m/s / 0.140 s = a
a = -79.3 m/s²
While stopping, the magnitude of the deceleration of the egg is 79.3 m/s²
The average acceleration while the egg was falling is 9.8 m/s² and the average deceleration while the egg was stopping is -79.3 m/s².
Explanation:The subject of the question is Physics, namely kinematics in the field of Mechanics. The main concepts involved are acceleration, velocity, and time. Acceleration is the rate of change of velocity, and it can be calculated as the change in velocity divided by the change in time.
Acceleration while falling: Since the egg is falling, it's subject to gravity. On Earth the acceleration due to gravity is approximately 9.8 m/s². However if this value wasn't given, it could have been calculated using the formula: final velocity divided by the time taken. Hence, 11.1 / 1.13 = 9.8 m/s² (rounded).
Deceleration while stopping: This is calculated in the same way as acceleration, the only difference being that its direction is opposite to the direction of motion hence it's often referred as deceleration. The initial velocity, in this case the speed it was falling just before hitting the ground, will be used as the initial velocity while calculating acceleration after hitting the ground. Therefore (0 - 11.1) / 0.140 = -79.3 m/s², the negative sign indicates deceleration.
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An ambulance is driving towards the hospital at a velocity 108 km/h and emitting a steady 798-Hz sound from its siren. The sound reflects off the front of the hospital and is received by the same ambulance. In addition to it's own siren, the ambulance hears a shifted tone from the reflection at what frequency? The speed of sound on this day is 343 m/s. a. 1.06e+3Hz b. 855Hz c. 850Hz d. 1.08e+3Hz e. 950 Hz
The ambulance will hear a reflected sound with a frequency of 855Hz, which is calculated using the Doppler Effect.
Explanation:The subject of this question is the Doppler Effect, which states that the observed frequency of a wave depends on the relative speed of the source and the observer. In this case, the source of the sound is the ambulance and the observer is also the ambulance (after reflection of the sound from the hospital). We can use the formula for Doppler Effect in this case:
F' = F *(v+v0) / (v-vS)
Where:
F' is the apparent frequency (the frequency heard by the observer), F is the source frequency (the frequency of the emitted sound), v is the speed of sound on this day, v0 is the speed of the observer (which is zero since the sound after reflection is observed at the same spot), vS is the speed of the source (which is the speed of the ambulance converted to m/s).
Putting the given values into this formula, we can calculate:
F' = 798Hz * (343m/s + 0) / (343m/s - 108km/h converted to m/s) = 855Hz
Therefore, the frequency of the reflected sound heard by the ambulance is 855Hz. The correct option is b. 855Hz.
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When a rigid object rotates about a fixed axis, what is true about all the points in the object? (There could be more than one correct choice.)
A) they all have the same tangential speed
B) they all have the same tangential acceleration
C) they all have the same angular speed
D) they all have the same angular acceleration
Answer:
C) they all have the same angular speed
D) they all have the same angular acceleration
Explanation:
Wrong --> they all have the same tangential speed. The points close to the axis will have less speed than the points away from the axis.
Wrong --> they all have the same tangential acceleration. Similarly, the points close to the axis will have smaller acceleration than the points away from the axis.
Correct --> they all have the same angular speed. Angular speed is the same for all the particles in the rotating object.
Correct --> they all have the same angular acceleration. Angular acceleration is the same for all the particles in the rotating object.
This all comes from the following relations:
v = ωR
a = αR
where ω is the angular velocity and α is the angular acceleration.
As can be seen from above, tangential velocity and acceleration depends on the distance from the axis, whereas the angular velocity and acceleration is the same for all the points on the rotating body.
The true statements about all points in the object rotating about a fixed point are;
C) they all have the same angular speed
D) they all have the same angular acceleration
For a circular motion about a given point, the angular speed is same for all points on the circular path and it is calculated as;
[tex]\omega = \frac{2\pi N}{T}[/tex]
Where;
N is the number of turns about the fixed pointT is the time of motionThus, angular speed is independent of the position of an object rotating about a fixed point.
The angular acceleration is given as;
[tex]\alpha = \frac{\omega}{T}[/tex]
Tangential speed and acceleration depends on the position of each object along the circular path.
Thus, we can conclude that the true statements about all points in the object rotating about a fixed point are, they all have the same angular speed and they all have the same angular acceleration.
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food product with 10 kg mass is being transported to thesurface of the moon,where the acceleration due to gravity is1.624 m/s2; about 1/6 of the value on earth. Compute thefollowing:a.The force exerted by the product on the earth’s surface; inSI units and English units.b.The product force exerted on the surface of the moon; inSI and English units
Answer:
In SI units 98.1 N, 16.24 N
English units 22.053861 lbf, 3.6509144 lbf
Explanation:
g = Acceleration due to gravity
m = Mass = 10 kg
Weight on Earth
[tex]W=mg\\\Rightarrow W=10\times 9.81\\\Rightarrow W=98.1\ N[/tex]
Converting to lbf
[tex]98.1\times 0.22481=22.053861\ lbf[/tex]
On Moon
[tex]W=10\times 1.624\\\Rightarrow W=16.24\ N[/tex]
Converting to lbf
[tex]16.24\times 0.22481=3.6509144\ lbf[/tex]
In SI units 98.1 N, 16.24 N
English units 22.053861 lbf, 3.6509144 lbf
An electrical motor spins at a constant 2695.0 rpm. If the armature radius is 7.165 cm, what is the acceleration of the edge of the rotor?
A) 28.20 m/s^2
B) 572,400 m/s^2
C) 281.6 m/s^2
D) 5707 m/s^2
Answer:
Acceleration will be [tex]5706.77rad/sec^2[/tex]
So option (D) will be correct answer
Explanation:
We have given angular speed of the electrical motor [tex]\omega =2695rpm[/tex]
We have to change this angular speed in rad/sec for further calculation
So [tex]\omega =2695rpm=2695\times \frac{2\pi }{60}=282.2197rad/sec[/tex]
Armature radius is given r = 7.165 cm = 0.07165 m
We have to find the acceleration of edge of motor
Acceleration is given by [tex]a=\omega ^2r=282.2197^2\times 0.07165=5706.77rad/sec^2[/tex]
So acceleration will be [tex]5706.77rad/sec^2[/tex]
So option (D) will be correct answer
Final answer:
The acceleration of the edge of the rotor is approximately 572400 m/s^2.
Explanation:
The acceleration of the edge of the rotor can be determined using the formula:
acceleration = radius x (angular velocity)²
In this case, the radius of the armature is given as 7.165 cm, which is equal to 0.07165 m. The angular velocity can be calculated by converting the given rpm (2695.0) to radians per second. One revolution is equal to 2π radians, and one minute is equal to 60 seconds, so:
angular velocity = (2695.0 rpm) x (2π rad/1 rev) x (1 min/60 s)
Once the angular velocity is determined, it can be substituted into the formula for acceleration along with the radius:
acceleration = (0.07165 m) x [(2695.0 rpm) x (2π rad/1 rev) x (1 min/60 s)]²
Mathematically solving this equation will give the value of the acceleration in m/s². After calculating the expression, we find that the acceleration of the edge of the rotor is approximately 572400 m/s².