All categories

3 years ago

Why does heat flow from a body at higher temperature to body at lower temperature?

1 answer:

5 0

Because, body at higher temperature has higher average KE of molecules and body at lower temperature has lower average KE of molecules. When two bodies having higher temperature and lower temperature are keep in contact, the body at higher temperature shares their energy to molecules at lower temperature.

please mark me as brainliest

You might be interested in

A 4.67-g bullet is moving horizontally with a velocity of +357 m/s, where the sign + indicates that it is moving to the right (s

Leni [432]

Answer:

(a)0.531m/s

(b)0.00169

Explanation:

We are given that

Mass of bullet, m=4.67 g= $4.67\times 10^{-3} kg$

1 kg =1000 g

Speed of bullet, v=357m/s

Mass of block 1, $m_1=1177g=1.177kg$

Mass of block 2, $m_2=1626 g=1.626 kg$

Velocity of block 1, $v_1=0.681m/s$

(a)

Let velocity of the second block after the bullet imbeds itself=v2

Using conservation of momentum

Initial momentum=Final momentum

$mv=m_1v_1+(m+m_2)v_2$

$4.67\times 10^{-3}\times 357+1.177(0)+1.626(0)=1.177\times 0.681+(4.67\times 10^{-3}+1.626)v_2$

$1.66719=0.801537+1.63067v_2$

$1.66719-0.801537=1.63067v_2$

$0.865653=1.63067v_2$

$v_2=\frac{0.865653}{1.63067}$

$v_2=0.531m/s$

Hence, the velocity of the second block after the bullet imbeds itself=0.531m/s

(b)Initial kinetic energy before collision

$K_i=\frac{1}{2}mv^2$

$k_i=\frac{1}{2}(4.67\times 10^{-3}\times (357)^2)$

$k_i=297.59 J$

Final kinetic energy after collision

$K_f=\frac{1}{2}m_1v^2_1+\frac{1}{2}(m+m_2)v^2_2$

$K_f=\frac{1}{2}(1.177)(0.681)^2+\frac{1}{2}(4.67\times 10^{-3}+1.626)(0.531)^2$

$K_f=0.5028 J$

Now, he ratio of the total kinetic energy after the collision to that before the collision

= $\frac{k_f}{k_i}=\frac{0.5028}{297.59}$

=0.00169

5 0

3 years ago

Why is it cold in the mountains if heat rises?

hjlf

Answer:

look at explanation

Explanation:

If heat rises, then why is it so cold at the top of a mountain? Heat does indeed rise. More specifically, a mass of air that is warmer than the air around it expands, becomes less dense, and will therefore float atop the cooler air. ... So when warm air rises, it cools off.

6 0

3 years ago

An student in a swimming pool weighs 450 N and in a regular balance out of the swimming pool weighs 700 N. What is the value of

rewona [7]

The value of buoyant force is 700-450=300N

7 0

4 years ago

20 cubic inches of a gas with an absolute pressure of 5 psi is compressed until its pressure reaches 10 psi. What's the new volu

Anna71 [15]

Answer:

B. $V_{f}= 10\,cubic\,inches$

Explanation:

Assuming we are dealing with a perfect gas, we should use the perfect gas equation:

$PV=nRT$

With T the temperature, V the volume, P the pressure, R the perfect gas constant and n the number of mol, we are going to use the subscripts i for the initial state when the gas has 20 cubic inches of volume and absolute pressure of 5 psi, and final state when the gas reaches 10 psi, so we have two equations:

$P_{i}V_{i}=n_{i}RT_{i}$ (1)

$P_{f}V_{f}=n_{f}RT_{f}$ (2)

Assuming the temperature and the number of moles remain constant (number of moles remain constant if we don't have a leak of gas) we should equate equations (1) and (2) because $T_{i}=T_{f}$ , $n_{i}=n_{f}$ and R is an universal constant:

$P_{i}V_{i}= P_{f}V_{f}$ , solving for $V_{f}$

$V_{f} =\frac{P_{i}V_{i}}{P_{f}} =\frac{(5)(20)}{10}$

$V_{f}= 10 cubic\,inches$

6 0

4 years ago

After polishing his 2-kg wrestling trophy, Mike sets it down on the ground and walks away to find more polish. Meanwhile, Julie

klio [65]

1) The initial momentum of the trophy is zero

2) The initial momentum of the bowling ball is 160 kg m/s

3) The total momentum before the collision is 160 kg m/s

4) The total momentum of the system after the collision is 160 kg m/s

5) The final velocity of the trophy is 32 m/s

Explanation:

The momentum of an object is given by

$p=mv$

where

m is the mass of the object

v is its velocity

In this problem, the data for the trophy before the collision are:

m = 2 kg is the mass

v = 0 is its initial velocity

Therefore, the initial momentum of the trophy is

$p_1=(2)(0)=0$

Using the same equation used in part 1), the initial momentum of the bowling ball is

$p=mv$

where

m is the mass of the bowling ball

v is its initial velocity

The data of the problem are

m = 8 kg is the mass

v = 20 m/s is the velocity

Substituting,

$p_2=(8)(20)=160 kg m/s$

The total momentum of the system before the collision is given by the sum between the initial momentum of the trophy and the initial momentum of the bowling ball:

$p_i = p_1 + p_2$

where

$p_1$ is the initial momentum of the trophy

$p_2$ is the initial momentum of the ball

Here we have

$p_1 = 0$

$p_2 = 160 kg m/s$

Therefore, the total momentum is

$p_i = 0 + 160 = 160 kg m/s$

According to the law of conservation of momentum, for an isolated system (=no external unbalanced forces acting on the system), the total momentum of the system is conserved before and after the collision:

$p_i = p_f$

where

$p_i$ is the total momentum before the collision

$p_f$ is the total momentum after the collision

If we consider the system in the problem to be isolated (i.e. no frictional forces acting on the ball or the trophy), we can therefore say that the total momentum after the collision must be equal to the total momentum before the collision: therefore,

$p_f = 160 kg m/s$