A simple harmonic oscillator completes 1550 cycles in 30 min. (a) Calculate the period. s (b) Calculate the frequency of the motion. Hz

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

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]

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

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.

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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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.

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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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.

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.

The 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.

When 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.

When 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.

To 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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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.

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.

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

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.

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

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

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.

(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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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

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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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

Δf =  [tex]f\frac{v + v_{w} }{v - v_{w} } - f \frac{v}{v-v_{w} }[/tex]  ------ ( 1 )

where :

                   = 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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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

Answer:

14062.5 J

Explanation:

From the law of conservation of momentum,

Total momentum before collision  = Total momentum after collision.

V = (m₁u₁ + m₂u₂)/(m₁+m₂).................1

Where V = common velocity after collision

Given: m₁ = m₂ = 25000 kg, u₁ = 2.5 m/s, u₂ = 1 m/s

Substitute into equation 1

V = [25000(2.5) + 25000(1)]/(25000+25000)

V = (62500+25000)/50000

V = 87500/50000

V = 1.75 m/s.

Note: The collision is an inelastic collision as such there is lost in kinetic energy of the system.

Total Kinetic energy before collision = kinetic energy of the first train car + kinetic energy of the second train car

E₁ = 1/2m₁u₁² + 1/2m₂u₂²........................ Equation 2

Where E₁ = Total kinetic energy of the body before collision, m₁ and m₂ = mass of the first train car and second train car respectively. u₁ and u₂ = initial velocity of the first train car and second train car respectively.

Given: m₁ = m₂ = 25000 kg, u₁ = 2.5 m/s, u₂ = 1 m/s

Substitute into equation 2

E₁ = 1/2(25000)(2.5)² + 1/2(25000)(1.0)²

E₁ = 12500(6.25) + 12500

E₁ = 78125+12500

E₁ = 90625 J.

Also

E₂ = 1/2V²(m₁+m₂)....................... Equation 3

Where E₂ = total kinetic energy of the system after collision, V = common velocity, m₁ and m₂ = mass of the first and second train car respectively.

Given: V = 1.75 m/s, m₁ = m₂ = 25000 kg

Substitute into equation 3

E₂ = 1/2(1.75)²(25000+25000)

E₂ = 1/2(3.0625)(50000)

E₂ = (3.0625)(25000)

E₂ = 76562.5 J.

Lost in kinetic Energy of the system = E₁ - E₂ = 90625 - 76562.5

Lost in kinetic energy of the system = 14062.5 J

The decrease in  kinetic energy of the system during the collision is 14062.5 kg m/s².

The question involves a collision between two train cars of equal mass moving in the same direction, which falls under the principles of momentum conservation and kinetic energy assessment in physics. First, we calculate the final velocity of the combined train cars using the law of conservation of momentum. Then, we compare the initial and final kinetic energies to find the decrease in kinetic energy due to the collision.

pi = m1v1 + m2v2 = (25000×2.50) + (25000×1) = 87500 kg m/s.

Using momentum conservation, vf = pi / (m1 + m2) = 87500/(25000 + 25000) = 1.75 m/s.

KEi = ½m1v1² + ½m2v2² = ½(25000)(2.5)² + ½(25000)(1)² = 90625 kg m/s²

KEf = ½(m1+m2)vf² = ½(25000+25000)(1.75)² = 76562.5 kg m/s².

The difference in kinetic energy: ΔKE = KEi - KEf = 90625 - 76562.5 = 14062.5 kg m/s².

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

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

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.

Answer:

3.7

Explanation:

[tex]\epsilon_0[/tex] = Permittivity of free space = [tex]8.85\times 10^{-12}\ F/m[/tex]

r = Radius = [tex]5.1\times 10^{-2}\ m[/tex]

A = Area = [tex]\pi r^2[/tex]

V = Voltage = 1120 V

d = Distance of plate seperation = 0.23 mm

Charge is given by

[tex]Q=CV\\\Rightarrow Q=\dfrac{k\epsilon_0 A}{d}\times V\\\Rightarrow k=\dfrac{Qd}{\epsilon_0 AV}\\\Rightarrow k=\dfrac{1.3\times 10^{-6}\times 0.23\times 10^{-3}}{8.85\times 10^{-12}\times \pi (5.1\times 10^{-2})^2\times 1120}\\\Rightarrow k=3.69164\\\Rightarrow k=3.7[/tex]

The value of dielectric constant is 3.7

Final answer:

To find the dielectric constant (κ) for a parallel-plate capacitor, given its physical dimensions and charge-potential difference, we calculate the capacitance with and without the dielectric and then solve for κ using the relation between charge, capacitance, and potential difference.

Explanation:

The question involves finding the dielectric constant (κ) of the material placed between the plates of a parallel-plate capacitor given the radius of the plates, the separation between them, the charge on the capacitor, and the potential difference across it. The capacitance of a parallel-plate capacitor is given by C = ε_0 κ A / d, where ε_0 is the permittivity of free space, κ is the dielectric constant, A is the area of the plates, and d is the separation between them. The charge (Q) on a capacitor is related to the capacitance (C) and the potential difference (V) across it by Q = CV. From the given information, we can calculate the area of the plates from their radius, and using the given separation, charge, and potential difference, solve for the dielectric constant, κ.

Given:

Radius of the plates, r = 5.10×10−2 m

Separation between the plates, d = 0.23 mm = 0.23×10−3 m

Charge on the capacitor, Q = 1.3 μC = 1.3×10−6 C

Potential difference, V = 1120 V

We can first calculate the area (A) of the plates using the formula for the area of a circle, A = πr2, substitute the value of A and the given values into the formula for C, and then use the relationship Q = CV to solve for κ.

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.

To 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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Answer:

a)  e% = 49.0%

, b)  e% = 19.7%

Explanation:

The percentage difference is the absolute uncertainty between the value accepted by 100

               e% = Δx / x_value

a) for this case the correct accepted value is x_value = 0.482 m / s²

               Δx = 0.718 - 0.482

               Δx = 0.236 m / s²

               e% = 0.236 / 0.482 100

               e% = 48.96%

b) In this case the results of the two experiments are real and the correct value is the average

              x_value = (0.482 + 0.718) / 2

              x_value = 0.600 m / s²

              e% = (0.718 - 0.600) / 0.600 100

              e% = 19.7%

The ambulance will hear a reflected sound with a frequency of 855Hz, which is calculated using the Doppler Effect.

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:

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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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.

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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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;

Thus, 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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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:

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]

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.

Answer:

(a) the acceleration a = 6.23 x 10^10 m/s²

(b) the time interview required to reach that velocity is t = 22.5microseconds

(c) the distance traveled is 15.7m

(d) K.E = 1.64 x 10^-15 J

Explanation:

Answer:

(a) the acceleration a = 6.23 x 10^10 m/s²

(b) the time interview required to reach that velocity is t = 22.5microseconds

(c) the distance traveled is 15.7m

(d) K.E = 1.64 x 10^-15 J

Explanation:

The detailed step by step solution to this problem can be forced below. The electric force on the charge is equal in magnitude to (Eq) front columb's law. The electric force on the charge gives it an acceleration of magnitude a. The force is also equal in magnitude to (ma) from newton's second law.

The acceleration of the charge in the electric field is constant and as a result the equations for constant acceleration motion applies to its motion.

KE = 1/2(MV²)

The complete solution can be found in the attachment below.

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