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s344n2d4d5 [400]
3 years ago
5

Five metal samples, with equal masses, are heated to 200oC. Each solid is dropped into a beaker containing 200 ml 15oC water. Wh

ich metal will cool the fastest? A) aluminum B) copper C) gold D) platinu Which sample of copper would demonstrate the GREATEST increase in temperature if 500 J of energy were added to the sample? A) 0.5 kg B) 1.0 kg C) 1.5 kg D) 2.0 kg Eliminate
Physics
1 answer:
Ksju [112]3 years ago
4 0
Part 1) Which metal will cool the fastest?
To answer this question, we should have a look at the formula of the heat flow rate, which says "how fast" a material is able to heat/cool:
\frac{\Delta Q}{\Delta t}  = -k  \frac{A \Delta T}{x}
where:
\Delta Q is the heat exchanged
\Delta t is the time interval
k is thermal conductivity of the material
A the  surface where the exchange of heat occurs
\Delta T the variation of temperature
x is the thickness of the material
We see that the heat flow rate \frac{\Delta Q}{\Delta t} is linearly proportional to k, the thermal conductivity of the material. So, the larger k, the fastest the metal will cool. 
If we have a look at the thermal conductivity of each metal, we find:
- Aluminium: 237 W/(mK)
- Copper: 401 W/(mK)
- Gold: 314 W/(mK)
- Platinum: 69 W/(mK)
Therefore, copper is the material with highest heat flow rate, so the metal which cools fastest.

Part 2) Which sample of copper demonstrates the greatest increase in temperature
To solve this part, we can have a look at how the amount of heat exchanged Q is related to the increase in temperature \Delta T:
Q=m C_S \Delta T
where m is the mass and Cs the specific heat of the material. Re-arranging the formula, we get
\Delta T= \frac{Q}{m C_s}
therefore, we see that the increase in temperature is inversely proportional to the mass m. This means that the block that will show the largest increase in temperature is the block with the smallest mass, so the correct answer is A) 0.5 kg.
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A spherical bowling ball with mass m = 3.6 kg and radius R = 0.101 m is thrown down the lane with an initial speed of v = 8.7 m/
luda_lava [24]

Answer:

1)  The magnitude of the angular acceleration = 67.92 rad/s^{2}

2) Magnitude of the linear acceleration = 2.744 m/s^{2}

3) How long does it take the bowling ball to begin rolling without slipping = 0.906 s

4) How long does it take the bowling ball to begin rolling without slipping = 6.75 m

5) the final velocity is 6.21 m/s

Explanation:

the given information :

Bowling mass m = 3.6 kg

Radius = 0.101 m

Initial speed v_{0} = 8.7 m/s

Coefficient of kinetic friction μ = 0.28

1) he magnitude of the angular acceleration of the bowling ball is

F = m a

F_{g}  = μ N  ,   N = m g

F_{g}  = μ m g

1) The magnitude of the angular acceleration of the bowling ball as it slides down the lane:

momen inersia of Bowling ball I = (2/5) m R^{2}

torque τ = I α

τ = F R

I α = F R

(2/5) m R^{2}  α = μ m g R

α = (5 μ g / 2R) μ g R

  = (5 x 0.28 x 9.8/ 2 x 0.101)

  = 67.92 rad/s^{2}

2) Magnitude of the linear acceleration of the bowling ball as it slides down the lane

F = - F_{g} , F_{k} is the force of kinetic friction

m a = - μ m g, remove m

the magnitude of linear accelaration is

a = μ g

  = (0.28) (9.8)

  = 2.744 m/s^{2}

3) The bowling ball takes time to begin rolling without slipping:

The linear speed, v_{t} = v_{0} - a t

                            v_{t}  =  v_{0} - μ g t

the angular speed, ω = ω0 + α t

                                ω = ω0 + (5  μ g/2R ) t

v_{t} = ω R

v_{0} - μ g t = ω0 R + (5  μ g/2R ) t R

7 μ g t/2 = v_{0} + ω0 R

hence,

t = (2 v_{0} + ω0 R)/  7 μ g

ω0 = 0 (no initial spin), therefore

t = 2 v_{0} / 7 μ g

 = 2 x 8.7 / 7 (0.28) (9.8)

 = 0.906 s

4) How long it takes for the bowling ball to begin rolling without slipping, S

S = v_{0}  t - (1/2) a t^{2}

  = (8.7) (0.906) - (1/2) (2.744) 0.906^{2}

  = 6.75 m

5) The final velocity

v_{t} = v_{0} - a t

v_{t} = 8.7 - (2.744) (0.906)

v_{t} = 6.21 m/s

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Escape velocity of an object from the surface of a planet depends upon:
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Answer:

Escape velocity: Measuring the gravitational strength of an object

The escape velocity is the exact amount of energy you would need to escape the gravitational clutches of an object with mass. Since all objects have mass, they all have a measureable gravitational strength. A good way to think about escape velocity is to think about a deep well (physicists like to think of this as an energy well). If you are at the bottom of the well and want to get out (to escape), you need enough energy to climb out. The deeper the well, the more energy you will have to expend in order to climb to

the top. If you have only enough energy to get half way out, you will eventually fall back to the bottom. The escape velocity is a way of measuring the exact amount of energy needed to reach the lip of the well -- and have no energy left over for walking away.

When a ball is thrown up into the air from the surface of the Earth, it does not have enough energy to escape. So it falls back down. How might we enable the ball to escape? Throw it harder, give it more energy. How hard must we throw it? Just hard enough to get over the top, over the edge of the well.

We can find this energy directly by saying that the kinetic energy of the thrown ball must exactly equal the 'potential energy' of the well. From basic physics we know that the potential energy for an object at a height above a surface is:

Epotential= GMm/R

where

G = Newton's universal constant of gravity = 6.67 x 10-11 N-m2/kg3

M = the mass of the 'attracting object' [the planet] [in units of kg]

m = the mass of the object trying to escape [e.g., me or a ball or a rocket or a molecule] [in kg]

R = the distance between the centers of objects M and m [in units of m]

note: provided we do everything in the same units, we don't have to worry about units

while the kinetic energy we know from above:

Ekinetic=0.5 m v2

where

m = mass of the moving object [in kg]

v = the velocity of object m [in m/sec]

If we set these two energies equal to each other, and solve for v, we find the exact velocity needed to escape from the energy well:

0.5 m v2= GMm/R

v= (2GM/R)0.5

and since this velocity is exactly what is needed to 'escape,' it is called the escape velocity:

vescape= (2GM/R)0.5

Explanation:

that's my all i know

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