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3 years ago

Select the correct answer

Physics

1 answer:

Sergeu [11.5K]3 years ago

5 0

Answer:

I may be a little late, but it's A i got it right on the exam

Explanation:

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A 2.64-kg copper part, initially at 400 K, is plunged into a tank containing 4 kg of liquid water, initially at 300 K. The coppe

marin [14]

Answer:

a) $T_f=305.7049\ K$

b) $\Delta S=313.51\ J.K^{-1}$

Explanation:

Given:

mass of copper, $m_c=2.64\ kg$

initial temperature of copper, $T_{ic}=400\ K$

specific heat capacity of copper, $c_c=385\ J.kg^{-1}.K^{-1}$

mass of water, $m_w=4\ kg$

initial temperature of water, $T_{iw}=300\ K$

specific heat capacity of water, $c_w=4200\ J.kg^{-1}.K^{-1}$

<u>∵No heat is lost in the environment and the heat is transferred only between the two bodies:</u>

Heat rejected by the copper = heat absorbed by the water

$2.64\times 385\times (400-T_f)= 4\times 4200\times (T_f-300)$

$T_f=305.7049\ K$

<u>Now the amount of heat transfer:</u>

$Q=m_c.c_c.(T_{ic}-T_{f})$

$Q=2.64\times 385\times (400-305.7049)$

$Q=95841.5841\ J$

∴Entropy change

$\Delta S=\frac{dQ}{T}$

$\Delta S=\frac{95841.5841}{305.7049}$

$\Delta S=313.51\ J.K^{-1}$

5 0

4 years ago

Which list places different units of matter in the correct sequence from largest to smallest

weqwewe [10]

Answer:

What do you mean?

6 0

3 years ago

Read 2 more answers

A man pulls his dog (m=20kg) on a sled with a force of 100N at a 60° angle from the horizontal. What is the horizontal component

sertanlavr [38]

Answer:

the horizontal component of the force is 50 N

Explanation:

Given;

force applied by the man, F = 100 N

angle of inclination of the force, θ = 60⁰

mass of the dog, m = 20 kg

The horizontal component of the force is calculated as;

$F_x = F\times cos(\theta)\\\\F_x = 100 \ N \times cos(60^0)\\\\F_x = 100\ N \times 0.5\\\\F_x = 50 \ N$

Therefore, the horizontal component of the force is 50 N

4 0

4 years ago

Convert 1.5 radian into milliradians.

Alex

Answer:

1500 milliradians

Explanation:

Data provided in the question:

1.5 radians

Now,

1 radians consists of 1000 milliradians

1 milli = 1000

thus for the 1.5 radians, we have

1.5 radians = 1.5 multiplied by 1000 milliradians

1.5 radians = 1500 milliradians

Hence, after the conversion

1.5 radians equals to the value 1500 milliradians

4 0

3 years ago

A plastic block of mass 0.20.2kg is initially at rest on a horizontal frictionless surface. A bullet of mass 1212 g is fired wit

Juli2301 [7.4K]

Answer:

94.29619%

Explanation:

$m_1$ = Mass of bullet= 0.012 kg

$m_2$ = Mass of block = 0.2 kg

$u_1$ = Initial Velocity of bullet= 550 m/s

$u_2$ = Initial Velocity of block = 0 m/s

$v_1$ = Final Velocity of bullet = 20 m/s

$v_2$ = Final Velocity of block

As the linear momentum of the system is conserved

$m_1u_1+m_2u_2=m_1v_1+m_2v_2\\\Rightarrow v_2=\dfrac{m_1u_1+m_2u_2-m_1v_1}{m_2}\\\Rightarrow v_2=\dfrac{0.012\times 550+0.2\times 0-0.012\times 20}{0.2}\\\Rightarrow v_2=31.8\ m/s$

Change in kinetic energy is given by

$\Delta K=\dfrac{1}{2}(m_1u_1^2-m_1v_1^2-m_2v_2^2)\\\Rightarrow \Delta K=\dfrac{1}{2}(0.012\times 550^2-0.012\times 20^2-0.2\times 31.8^2)\\\Rightarrow \Delta K=1711.476\ J$

Percentage change is given by

$\dfrac{\Delta K}{\dfrac{1}{2}m_1u_1^2}\times 100=\dfrac{1711.476}{\dfrac{1}{2}0.012\times 550^2}\times 100=94.29619\ \%$

The percent of the initial kinetic energy of the bullet block system is lost during the collision is 94.29619%

4 0

3 years ago