Answer:
at zero point : GPE = 0 J
at max height : GPE = 633.7 J
Explanation:
the gravitational potential energy at the lowest point is zero
maximum height relative to the lowest point = h =1.70 m
G potential energy at max height = mgh = (38kg)(9.81m/s^2)(1.7)
= 633.7 J
If the magnitude of a charge is twice as much as another charge, but the force experienced is the same, then the electric field strength of this charge is _____ the strength of the other charge.
Answer:
The new electric field strength of this charge is half of the strength of the other charge.
Explanation:
The electric force acting on the charge particle is given by :
F = q E
[tex]E=\dfrac{F}{q}[/tex]
Where
q is the charged particle
E is the electric field
If the magnitude of a charge is twice as much as another charge, q' = 2q, but the force experienced is the same, then the new electric field is given by :
[tex]E'=\dfrac{F}{q'}[/tex]
[tex]E'=\dfrac{F}{(2q)}[/tex]
[tex]E'=\dfrac{1}{2}\times \dfrac{F}{q}[/tex]
[tex]E'=\dfrac{E}{2}[/tex]
So, the new electric field strength of this charge is half of the strength of the other charge. Hence, this is the required solution.
Final answer:
The electric field strength of the charge with twice the magnitude is half the strength of the electric field of the other charge because the force experienced by both is equal.
Explanation:
If the magnitude of a charge is twice as much as another charge, but the force experienced by each is the same, then the electric field strength that the larger charge is in is half the strength of the other charge’s electric field. The electric field strength (E) at a point is defined as the force (F) experienced by a positive test charge (q) placed at that point divided by the magnitude of the charge itself: E = F/q.
If we have two charges, q1 and q2, where q2 is twice q1 (“q2 = 2 × q1”), and the forces on both charges are equal (“F1 = F2”), we can set up the following equations: E1 = F1/q1 and E2 = F2/q2. Combining our known relationships, we get E1 = F/q1 and E2 = F/(2 × q1), which simplifies to E1 = 2 × E2. Therefore, E2 is half the magnitude of E1, meaning that the electric field strength of the charge with twice the magnitude is half the strength of the electric field of the other charge.
An engineer claims that solar cells for the generation of electrical power can never meet our high demand for electrical energy. The engineer points out that such cells____________________.
Answer:
The sentence is following with "are made from mostly silicon material".
Explanation:
Silicon is extracted and refined by the use of electrical energy. The amount of energy required to produce the silicon in a solar cell is greater than the energy that cell will produce during its useful working life.
Calculate the change in entropy as 0.3071 kg of ice at 273.15 K melts. (The latent heat of fusion of water is 333000 J / kg)
Answer:
374.39 J/K
Explanation:
Entropy: This can be defined as the degree of disorder or randomness of a substance.
The S.I unit of entropy is J/K
ΔS = ΔH/T ..................................... Equation 1
Where ΔS = entropy change, ΔH = Heat change, T = temperature.
ΔH = cm................................... Equation 2
Where,
c = specific latent heat of fusion of water = 333000 J/kg, m = mass of ice = 0.3071 kg.
Substitute into equation 2
ΔH = 333000×0.3071
ΔH = 102264.3 J.
Also, T = 273.15 K
Substitute into equation 1
ΔS = 102264.3/273.15
ΔS = 374.39 J/K
Thus, The change in entropy = 374.39 J/K
The change in entropy of the gas is 374.39 J/K.
The given parameters:
Mass of the ice, m = 0.3071 kgTemperature of the gas, T = 273.15 KLatent heat of fusion of water, L = 333,000 J/kgThe heat of fusion of the ice is calculated as follows;
[tex]\Delta H = mL\\\\\Delta H = 0.3071 \times 333,000\\\\\Delta H = 102,264.3 \ J[/tex]
The change in entropy of the gas is calculated as follows;
[tex]\Delta S = \frac{\Delta H}{T} \\\\\Delta S = \frac{102,264.3}{273.15} \\\\\Delta S = 374.39 \ J/K[/tex]
Thus, the change in entropy of the gas is 374.39 J/K.
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When you swim in a pool, Select one: a. rolling friction occurs. b. fluid friction occurs. c. sliding friction occurs. d. static friction occurs.
Answer:
b. fluid friction occurs
Explanation:
Water is a liquid and both going through air gas and liquid cause fluid friction.
A woman is reported to have fallen 161 ft from a building, landing on a metal ventilator box, which she crushed to a depth of 29 in. She suffered only minor injuries. Ignoring air resistance, calculate:(a) the speed of the woman just before she collided with the ventilator and (b) her average acceleration while in contact with the box. (c) Modeling her acceleration as constant, calculate the time interval it took to crush the box.
Answer:
a) 31.02 m/s
b) - 653.24 m/s²
c) 50 ms
Explanation:
a) For a vertical movement, let's suppose that the woman fell by the rest, thus she's initial velocity v0 = 0 m/s. If she makes a distance (S) of 161 ft = 49.1 m, thus, her final velocity at the box, v is:
v² = v0² +2aS
The acceleration is the gravity acceleration, 9.8 m/s².
v² = 2*9.8*49.1
v² = 962.36
v =√962.36
v = 31.02 m/s
b) She crushes the box until she stops, so her final velocity will be 0 m/s. The initial velocity is now 31.02 m/s, and S = 29 in = 0.7366 m
0 = 31.02² + 2a*0.7366
-1.4732a = 962.36
a = - 653.24 m/s² (the negative signal indicates that she's dessacelaranting)
c) The time can be calculate by:
v = v0 + at
0 = 31.02 -653.24*t
653.24t = 31.02
t = 0.05 s = 50 ms
John weighs 90 lbs and Jane weighs 60 lbs. They are both sitting on a seesaw. If John is seated 10 feet away from Jane, how far should each be from the fulcrum of the seesaw?
Answer: Jane 6ft from fulcrum
John 4 ft from fulcrum
Explanation: You use the law of moment of force.
That is Clockwise moment equals anti clockwise moment.
Here is the attachment involving the diagram and procedure of calculations and the answers.
The small child should sit 1.30 m away from the pivot point on the seesaw to maintain balance.
The small child should sit 1.30 m away from the pivot point on the seesaw to maintain balance.
A 130 g ball and a 230 g ball are connected by a 34-cm-long, massless, rigid rod. The balls rotate about their center of mass at 120 rpm .
What is the speed of the 100 g ball?
The linear velocity of the 100 g ball in the rotating system with a 130 g ball and a 230 g ball, connected by a rigid rod and rotating at 120 RPM, is approximately [tex]\(4.24 \, \text{m/s}\).[/tex]
Given values:
- Mass of the 130 g ball [tex](\(m_1\))[/tex]: 130 g
- Mass of the 230 g ball [tex](\(m_2\))[/tex]: 230 g
- Length of the rod [tex](\(r\)):[/tex] 34 cm = 0.34 m
- Initial angular velocity [tex](\(ω_{\text{initial}}\))[/tex]: [tex]\( ω_{\text{initial}} = \frac{2π \times 120}{60} \)[/tex]
Now, let's calculate the initial angular velocity:
[tex]\[ ω_{\text{initial}} = \frac{2π \times 120}{60} = 4π \, \text{rad/s} \][/tex]
Next, calculate the moment of inertia for each ball:
[tex]\[ I_1 = \frac{2}{5}m_1r^2 = \frac{2}{5} \times 0.13 \times (0.34)^2 \][/tex]
[tex]\[ I_2 = \frac{2}{5}m_2r^2 = \frac{2}{5} \times 0.23 \times (0.34)^2 \][/tex]
Now, calculate the total moment of inertia:
[tex]\[ I_{\text{total}} = I_1 + I_2 \][/tex]
Substitute the values into the conservation of angular momentum equation:
[tex]\[ I_{\text{total}} \times ω_{\text{initial}} = I_{\text{total}} \times ω_{\text{final}} \][/tex]
Solve for [tex]\(ω_{\text{final}}\)[/tex] and then use it to calculate the linear velocity [tex](\(v\))[/tex] of the 100 g ball:
[tex]\[ v = ω_{\text{final}} \times r \][/tex]
After performing these calculations, we can determine the speed of the 100 g ball. Let me provide you with the numerical results in the next response.
After performing the calculations, the final angular velocity [tex](\(ω_{\text{final}}\))[/tex] is determined to be the same as the initial angular velocity [tex](\(4π \, \text{rad/s}\))[/tex]. Now, we can calculate the linear velocity [tex](\(v\))[/tex] of the 100 g ball using the formula [tex]\(v = ω_{\text{final}} \times r\).[/tex]
[tex]\[ v = 4π \times 0.34 \][/tex]
[tex]\[ v \approx 4 \times 3.14 \times 0.34 \][/tex]
[tex]\[ v \approx 4.24 \, \text{m/s} \][/tex]
Therefore, the speed of the 100 g ball in this rotating system is approximately [tex]\(4.24 \, \text{m/s}\).[/tex]
A zero-order reaction has a constant rate of 2.30×10−4 M/s. If after 80.0 seconds the concentration has dropped to 1.50×10−2 M, what was the initial concentration?
Answer:
Initial concentration of the reactant = 3.34 × 10^(-2)M
Explanation:
Rate of reaction = 2.30×10−4 M/s,
Time of reaction = 80s
Final concentration = 1.50×10−2 M
Initial concentration = Rate of reaction × Time of reaction + Final concentration
= 2.30×10−4 M/s × 80s + 1.50×10−2 M = 3.34 × 10^(-2)M
Initial concentration = 3.34 × 10^(-2)M
Life of a Star Cluster. Imagine you could watch a star cluster from the time of its birth to an age of 13 billion years. Describe in one or two paragraphs what you would see happening during that time?
Explanation:
The life span of a star is quite as many more than 10 billion years. During these many years it would pass through different stages of its life, from nebula to ( perhaps, the current stage) supernova to neutron star or Black Hole.
The stages of the life of star over a span of 13 billion year can be summarized as
the transition stages are
Nebula → Blue star → Blue-white super-giant star → Supernova → Neutron star →Black Hole.
The ease with which the charge distribution in a molecule can be distorted by an external electrical field is called the __________.
A) electronegativity
B) hydrogen bonding
C) polarizability
D) volatility
E) viscosity
Answer:
C
Explanation:
The answer is polarization. In a normal molecule the positive and negative charges are distributed in a way that it makes a continuous distribution with no net force. but when you apply an external electric force, the opposite charges go to the opposite ends of the molecule making them a polarized molecule.
This is called polarization.
I hope it helps you.
Thank you.
projectile motion of a particle of mass M with charge Q is projected with an initial speed V in a driection opposite to a uniform electric fiedl of magnitude E.
Answer:
Range, [tex]R = MV²/2QE[/tex]
Explanation:
The question deals with the projectile motion of a particle mass M with charge Q, having an initial speed V in a direction opposite to that of a uniform electric field.
Since we are dealing with projectile motion in an electric field, the unknown variable here, would be the range, R of the projectile. We note that the electric field opposes the motion of the particle thereby reducing its kinetic energy. The particle stops when it loses all its kinetic energy due to the work done on it in opposing its motion by the electric field. From work-kinetic energy principles, work done on charge by electric field = loss in kinetic energy of mass.
So, [tex]QER = MV²/2{/tex} where R is the distance (range) the mass moves before it stops
Therefore {tex}R = MV²/2QE{/tex}
An electron moving parallel to a uniform electric field increases its speed from 2.0 ×× 1077 m/sm/s to 4.0 ×× 1077 m/sm/s over a distance of 1.3 cmcm.
What is the electric field strength?
Answer:
262 kN/C
Explanation:
If the electrons is moving parallel, thus it has a retiline movement, and because the velocity is varing, it's a retiline variated movement. Thus, the acceleration can be calculated by:
v² = v0² + 2aΔS
Where v0 is the initial velocity (2.0x10⁷ m/s), v is the final velocity (4.0x10⁷ m/s), and ΔS is the distance (1.3 cm = 0.013 m), so:
(4.0x10⁷)² = (2.0x10⁷)² + 2*a*0.013
16x10¹⁴ = 4x10¹⁴ + 0.026a
0.026a = 12x10¹⁴
a = 4.61x10¹⁶ m/s²
The electric force due to the electric field (E) is:
F = Eq
Where q is the charge of the electron (-1.602x10⁻¹⁹C). By Newton's second law:
F = m*a
Where m is the mass, so:
E*q = m*a
The mass of one electrons is 9.1x10⁻³¹ kg, thus, the module of electric field strenght (without the minus signal of the electron charge) is:
E*(1.602x10⁻¹⁹) = 9.1x10⁻³¹ * 4.61x10¹⁶
E = 261,866.42 N/C
E = 262 kN/C
A two stage rocket leaves its launch pad moving vertically with an average acceleration of 4 m/s2. at 10 s after launch the first stage of the rocket (now without fuel) is released. the second stage now had an acceleration of 6 m/s2
a) how high is the rocket when the first stage seperates?
b)how fast is the rocket moving upon first stage seperation?
c) what will be the maximum height attained by the first stage after seperation?
d) what will be the distance between the first and second stages 2 s after separation
Answer:
a) 200m
b) 40 m/s
c) 81.55m
d) 31.62m
Explanation:
Solution
a)
y = y0 + u×t+⅟2×a×t2 =
y0 = 0
u = 0
y = unknown
a = 4m/s2
t = time = 10 seconds
y = 0.5×4×102 = 200m
b) v = u + at
v = 0 + 4×10 = 40 m/s
c) v2 = u2 - 2×g×y
at maximum height v = 0
we have
402 = 2×9.81×y
Y =81.55m
d)
for the stage 2 we haace
y = y0 + u×t+⅟2×a×t2 =
y = 0 + 4×2+0.5×6×22 = 92m
for the stage one we have
y = 0+40×2-0.5×9.81×4= 60.38m
distance between the first and second stage 2s aftee separation = 92-60.38 = 31.62m
A force of 14 N acts on a 5 kg object for 3 seconds.
a. What is the object’s change in momentum?
b. What is the object’s change in velocity?
Answer: a) 42Nm b) 8.4m/s
Explanation:
Impulse is defined as object change in momentum.
Since Force = mass × acceleration
F = ma
Acceleration is the rate of change in velocity.
F = m(v-u)/t
Cross multiply
Ft = m(v-u)
Since impulse = Ft
and Ft = m(v-u)... (1)
The object change in velocity (v-u) = Ft/m from eqn 1
Going to the question;
a) Impulse = Force (F) × time(t)
Given force = 14N and time = 3seconds
Impulse = 14×3
Impulse = 42Nm
b) The object change in velocity (v-u) = Ft/m where mass = 5kg
v-u = 14×3/5
Change in velocity = 42/5 = 8.4m/s
A 4.0 kg model rocket is launched, shooting 50.0 g of burned fuel from its exhaust at an average velocity of 625 m/s. What is the velocity of the rocket after the fuel has burned?
Answer:
Velocity of rocket will be equal to 7.81 m/sec
Explanation:
We have given mass of the rocket [tex]m_1=4kg[/tex]
Mass of fuel is given [tex]m_2=50gram=0.05kg[/tex] ( As 1 kg is equal to 1000 gram )
Velocity of fuel [tex]v_2[/tex] = 625 m/sec
We have to find the velocity of rocket [tex]v_1[/tex]
From conservation of momentum we know that initial momentum is equal to final momentum '
So [tex]m_1v_1=m_2v_2[/tex], here [tex]m_1[/tex] is mass of rocket [tex]v_1[/tex] is velocity of rocket [tex]m_2[/tex] is mass of fuel and [tex]v_2[/tex] is velocity of fuel
So [tex]4\times v_1=625\times 0.05[/tex]
[tex]v_1=7.81m/sec[/tex]
So velocity of rocket will be equal to 7.81 m/sec
You are given the masses, in grams (g), and the volumes, in cubic centimeters (cm3), for a series of substances. Arrange them in order of increasing density. Rank from lowest density to highest density. To rank items as equivalent, overlap them.
The question requires arranging substances by density, a fundamental physics concept, calculated as mass divided by volume. Understanding densities, from light gases to dense neutron stars, is essential in both everyday applications and advanced scientific fields.
Explanation:The question involves arranging substances in order of increasing density based on given masses and volumes. Density is the measure of how much mass is contained in a given volume, and this concept is fundamental in physics. By calculating the density of each substance using the formula density = mass/volume (expressed in g/cm³ for solids and liquids or g/L for gases), we can compare the densities of the substances and arrange them accordingly. It is crucial to recognize that the density of a material is a characteristic property that can help in identifying the material and understanding its behavior in different environments.
Density plays a significant role in various applications, such as determining whether an object will float or sink in a fluid, in material selection for construction, and in understanding the structure of the universe, from the density of air we breathe (1.2 g/L) to the dense core of neutron stars reaching up to 10¹⁵ g/cm³.
To arrange the substances in order of increasing density, we need to calculate the density of each substance using the formula [tex]\[ \text{Density} = \frac{\text{Mass}}{\text{Volume}} \][/tex]
Then, we compare the calculated densities to determine their relative order.
Here's how the substances could be ranked:
1. Helium gas: Since helium gas is the least dense substance, it will have the lowest density.
2. Cork: Cork typically has a low density compared to other materials.
3. Water: Water has a moderate density, higher than helium gas and cork.
4. Aluminum: Aluminum generally has a higher density than water.
5. Iron: Iron is denser than aluminum, making it the densest substance among the listed ones.
By comparing the calculated densities, we can arrange the substances in increasing order of density from lowest to highest: helium gas, cork, water, aluminum, and iron.
Tribes of the Sioux Nation (among other Plains Indians) maintained historical calendars composed of winter counts. Tribe historians would_____________________.
Answer:
Tribe historians would: Depict a significant event for each year on a buffalo or deer skin
Tribes of the Sioux Nation maintained historical calendars composed of winter counts, using pictorial representations to record significant events. Winter counts served as a visual record of the tribe's history and were passed down through generations.
Explanation:Tribes of the Sioux Nation (among other Plains Indians) maintained historical calendars composed of winter counts. Tribe historians would use pictorial representations to record significant events that occurred during each year. Each year, a new pictorial symbol would be added to the count. For example, if a significant battle occurred during a year, the historian would draw a symbol representing that battle. Winter counts served as a visual record of the tribe's history and were often passed down from generation to generation.
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Complete the following statement:When work is done on a positive test charge by an external force to move it from one location to another, electric potential _________.( increase or decrease).
Answer:
increase
Explanation:
Electric potential is defined as the work done in bringing a unit positive charge from infinity to that point against the electrical forces of the field. Taking a test positive charge from one point to another means that work is done against the field hence electric potential increases.
To push a 25.0 kg crate up a frictionless incline, angled at 25.0° to the horizontal, a worker exerts a force of 209 N parallel to the incline. As the crate slides 1.50 m, how much work is done on the crate by:_________
(a) the worker’s applied force,
(b) the gravitational force on the crate, and
(c) the normal force exerted by the incline on the crate?
(d) What is the total work done on the crate?
Answer:
a. [tex]W_w=313.5\ J[/tex]
b. [tex]W_g=-155.312\ J[/tex]
c. [tex]F_N=222.045\ N[/tex]
d. [tex]W_t=313.5\ J[/tex]
Explanation:
Given:
angle of inclination of the surface,[tex]\theta=25^{\circ}[/tex]mass of the crate, [tex]m=25\ kg[/tex]Force applied along the surface, [tex]F=209\ N[/tex]distance the crate slides after the application of force, [tex]s=1.5\ m[/tex]a.
Work done by the worker who applied the force:
[tex]W_w=F.s\ cos 0^{\circ}[/tex] since the direction of force and the displacement are the same.
[tex]W_w=209\times 1.5[/tex]
[tex]W_w=313.5\ J[/tex]
b.
Work done by the gravitational force:
[tex]W_g=m.g\times h[/tex]
where:
g = acceleration due to gravity
h = the vertically downward displacement
Now, we find the height:
[tex]h=s\times sin\ \theta[/tex]
[tex]h=1.5\times sin\ 25^{\circ}[/tex]
[tex]h=0.634\ m[/tex]
So, the work done by the gravity:
[tex]W_g=25\times 9.8\times (-0.634)[/tex] ∵direction of force and displacement are opposite.
[tex]W_g=-155.312\ J[/tex]
c.
The normal reaction force on the crate by the inclined surface:
[tex]F_N=m.g.cos\ \theta[/tex]
[tex]F_N=25\times 9.8\times cos\ 25[/tex]
[tex]F_N=222.045\ N[/tex]
d.
Total work done on crate is with respect to the worker: [tex]W_t=313.5\ J[/tex]
In order to understand the full scope of a disease, we take its occurrence into account. The __________ of a disease is the number of people in a population who develop a disease at a specified time.
a. incidence
b. PREVALENCE
c. endemic infection
d. sporadic infection
Answer: b. PREVALENCE
Explanation:
Prevalence can be defined as the proportion of individuals in a population having a disease or common characteristic. Prevalence of a particular disease is number of people in a population who have develop a disease at a particular time.
It can be expressed in form of proportion Which equals the number of developed cases at a particular time divided by the total number of population.
A potential energy function for a system in which a two-dimensional force acts is of the form U = 3x5y − 3x. Find the force that acts at the point (x, y).
Answer:
[tex]\overrightarrow{F} = -\left ( 15x^{4}y-3 \right )\widehat{i}-3x^{5}\widehat{j}[/tex]
Explanation:
[tex]U = 3x^{5}y-3x[/tex]
Force is defined as the
[tex]\overrightarrow{F} = -\frac{dU}{dx}\widehat{i}-\frac{dU}{dy}\widehat{j}[/tex]
[tex]\frac{dU}{dx}=15x^{4}y-3[/tex]
[tex]\frac{dU}{dy}=3x^{5}[/tex]
So, force is given by
[tex]\overrightarrow{F} = -\left ( 15x^{4}y-3 \right )\widehat{i}-3x^{5}\widehat{j}[/tex]
Assume that all four H-R diagrams below represent a star in different stages of its life, after it starts to fuse hydrogen in its core. Rank the HR diagrams based on when each stage occurs, from first to last. star in the sequencea little outsidemore outsidethen below the line
A star goes through multiple stages of life represented on the H-R Diagram, from the main sequence stage to a red giant, then a brief stable period after a helium flash, back to a giant, and finally, death. The 'main-sequence turnoff' on the H-R diagram marks where stars begin to leave the main sequence. These stages occur faster for more massive stars.
Explanation:Stars progress through various stages on the H-R diagram as they age. Initially, in the main sequence stage, stars start by fusing hydrogen in their cores. The most massive stars evolve into red giants and supergiants less than a million years after they reach the main sequence, moving up and to the right on the diagram. After, lower mass stars also begin their departure from the main sequence.
During the next phase of evolution, a star experiences a helium flash which causes a change in its internal structure, leading to a brief period of stability. Following the exhaust of the central helium, the star turns into a giant again and its position on the H-R diagram shifts to high luminosity and low temperature. During the final stage of the star's life, it depletes its inner resources and begins to die.
This process of evolution is faster for more massive stars, as the rate at which they go through each stage of life is higher. The main-sequence turnoff point on the H-R diagram denotes the location where stars begin to leave the main sequence and transform towards the red giant region. It is lower on the H-R diagram and the mass of the stars is lower in older clusters.
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A circuit has a current of 2.4 A. The voltage is increased to 4 times its original value, while the resistance stays the same. How should the resistance change to return the current to its original value if the voltage remains at its increased amount?
Answer:
Resistance will become 4 times the previous value
Explanation:
We have given current in the circuit i = 2.4 A
According to ohm's law current in the circuit is given by [tex]I=\frac{V}{R}[/tex]
So [tex]2.4=\frac{V}{R}[/tex]............eqn 1
Now voltage is increased to 4 times so new voltage = 4 V
And current in the circuit is same as 2.4 A
We have to fond the resistance so that after increasing voltage current will be same
So [tex]2.4=\frac{4V}{R_{unknown}}[/tex]..........eqn 2
Dividing eqn 1 and 2
[tex]1=\frac{V}{R}\times \frac{R_{unkown}}{4V}[/tex]
[tex]R_{unknown}=4R[/tex]
So resistance will become 4 times the previous value
Answer:
It should increase to four times its value.
Explanation:
A projectile of mass 0.607 kg is shot straight up with an initial speed of 20.1 m/s. (a) How high would it go if there were no air resistance?
Answer:
20.6m
Explanation:
The mass is useless in this case.
Kinetic Energy = Potential Energy
g is acceleration due to gravity.
(1/2)mv² = mgh (m's drop out)
(1/2)v² = gh
(1/2)(20.1²) = 9.81(h)
h = (20.1 x 20.1)/9.81 x 2
20.61 meters
A calorimeter is used to determine the specific heat capacity of a test metal. If the specific heat capacity of water is known, what quantities must be measured?
Answer:
initial and final temperatures of both the water and metal, mass of the metal, and mass of the water
Explanation:
Heat lost by the metal, [tex]Q = mc(t_{2} - t_{1})[/tex]
Heat gained by the water in the calorimeter, [tex]Q_{w} = m_{w}c_{w}(t_{2w} - t_{1w})[/tex]
For energy to be conserved in the system, the heat lost by the metal will equal the heat gain by the water in the calorimeter.
[tex]mc(t_{2} - t_{1}) = m_{w}c_{w}(t_{2w} - t_{1w})[/tex]
Where,
m is the mass of the metal
c is specific heat capacity of the metal
t₂ is the final temperature of the metal
t₁ is the initial temperature of the metal
[tex]m_{w} [/tex] is the mass of the water
[tex]c_{w} [/tex] is specific heat capacity of water
[tex]t_{2w} [/tex] is the final temperature of water
[tex]t_{1w} [/tex] is the initial temperature of water
From the question given, specific heat capacity of the water is known, the quantities to be measured are;
Initial and final temperatures of both the water and metal,
Mass of the metal, and mass of the water
What is the mechanical advantage of the lever if you hold on to the end of the long side and lift something that's placed on the short side?
Answer:
[tex]MA=\frac{F_l}{F_e}>1[/tex]
Explanation:
The mechanical advantage with a shorter lever will always be greater than 1.
It is so because with a longer effort arm we need to apply lesser force to lift a unit mass which is at a shorter distance from the fulcrum. This is in accordance with the conservation of moments.
[tex]F_e\times d_e=F_l\times d_l[/tex]
where:
[tex]F_e=[/tex] force on the effort arm
[tex]F_l=[/tex] force on the load arm
[tex]d_e\ \&\ d_l[/tex] are the lengths of load effort arm and load arm respectively.
So, one factor balances the other keeping the product of the two constant.
And we know that mechanical advantage is :
[tex]MA=\frac{F_l}{F_e}[/tex]
Does air pressure increase or decrease with an increase in altitude?
Answer:
increase
Explanation:
think of climbing a mountain. the higher you go the harder it is to breathe. its because air pressure is increasing
Answer:
decrease.
Explanation:
The air pressure is given by
P = h x d x g
where, h is the height of air column above the surface and g be the acceleration due to gravity and d be the density of air.
As we go up, the height of air column decreases, density of air decreases and acceleration due to gravity also decreases, so the value of pressure decreases at altitudes.
1A soccer ball is kicked horizontally off a 22.0-meter high hill and lands a distance of 35.0 meters from the edge of the hill. Determine the initial horizontal velocity of the soccer ball.
The initial horizontal velocity of the soccer ball is 16.5 m/s
Explanation:
When we throw a ball, there is a constant velocity horizontal motion and there is an accelerated vertical motion. These components act independently of each other. Horizontal motion is constant velocity motion.
[tex]v_{x f}=v_{x i}=v_{x}[/tex]
[tex]a_{x}=0[/tex], so [tex]x=v_{i x} t+\left(\frac{1}{2}\right) a_{x} t^{2}[/tex] for horizontal motion
[tex]y=v_{i y} t+\left(\frac{1}{2}\right) a_{y} t^{2}[/tex] for vertical motion
Given:
x = 35 m
[tex]a_{x}=0 \mathrm{m} / \mathrm{s}^{2}[/tex]
Need to find [tex]v_{i x}[/tex]
y = - 22 m
[tex]v_{i y}=0 \mathrm{m} / \mathrm{s}[/tex]
[tex]a_{y}=-9.8 \mathrm{m} / \mathrm{s}^{2}[/tex] (negative sign indicates downward motion)
By substituting all known values, we can solve for 't' value as below
[tex]y=v_{i y} t+\left(\frac{1}{2}\right) a_{y} t^{2}[/tex]
[tex]-22=0(t)+\left(0.5 \times-9.8 \times t^{2}\right)[/tex]
[tex]t^{2}=\frac{-22}{0.5 \times-9.8}=\frac{-22}{-4.9}=4.4897[/tex]
Taking square root, we get t = 2.12 seconds
Now, substitute these to find initial horizontal velocity
[tex]x=v_{i x} t+\left(\frac{1}{2}\right) a_{x} t^{2}[/tex]
[tex]35=v_{i x}(2.12)+\left(0.5 \times 0 \times(2.12)^{2}\right)[/tex]
[tex]35=v_{i x}(2.12)+0[/tex]
[tex]v_{i x}=\frac{35}{2.12}=16.5 \mathrm{m} / \mathrm{s}[/tex]
Where did Earth’s water come from?, select the two competing scientific theories supported by evidence that explain the formation of the hydrosphere.
Answer:
Explanation:
There are two prevailing theories: One is that the Earth held onto some water when it formed, as there would have been ice in the nebula of gas and dust (called the proto-solar nebula) that eventually formed the sun and the planets about 4.5 billion years ago. Some of that water has remained with the Earth, and might be recycled through the planet's mantle layer, according to one theory.
The second theory holds that the Earth, Venus, Mars and Mercury would have been close enough to that proto-solar nebula that most of their water would have been vaporized by heat; these planets would have formed with little water in their rocks. In Earth's case, even more water would have been vaporized when the collision that formed the moon happened. In this scenario, instead of being home-grown, the oceans would have been delivered by ice-rich asteroids, called carbonaceous chondrites.
Two theories explain the formation of Earth's hydrosphere: water from interstellar grains and water from comets and asteroids.
Explanation:There are two competing scientific theories supported by evidence that explain the formation of Earth's hydrosphere:
Water from comets and asteroids: Another theory suggests that the water may have been brought to Earth when comets and asteroids impacted it. Scientists estimate that comet impacts during Earth's early years could have contributed enough water to account for what we see today.
Obliquity describes:
a. the roundness of an object
b. the circularity of the orbit
c. the tilt of the axis of rotation with respect to the Plane of the Ecliptic
d. none of the above
Answer:
c. the tilt of the axis of rotation with respect to the Plane of the Ecliptic
Explanation:
The inclination of the ecliptic (or known only as obliqueness) refers to the angle of the axis of rotation with respect to a perpendicular to the plane of the eclipse. He is responsible for the seasons of the year that the planet Earth lends. It is not constant but changes through the movement of nutation. The terrestrial plane of Ecuador and the ecliptic intersect in a line that has an end at the point of Aries and at the diametrically opposite point of Libra.
When the Sun crosses the Aries, the spring equation occurs (between March 20 and 21, the beginning of spring in the northern hemisphere and the early autumn of the southern hemisphere), and from which the Sun is in the North Hemisphere; Pound until you reach the point of the autumn equinox (around September 22-23, beginning fall in the northern hemisphere and spring in the southern hemisphere).