· A hot, just-minted copper coin is placed in 101 g of water to cool. The water

A 2.5 g bullet traveling at 350 m/s hits a tree and slows uniformly to a stop while penetrating a distance of 12 cm into the tre

Cloud [144]

Answer: Work done = 153.125Joules, Work done = 0.003Nm

Explanation:

Kinetic energy of a body is the energy possessed by a body by virtue of its motion.

Mathematically,

K.E = 1/2MV²

Where;

M = mass of the body = 2.5g = 0.0025kg

V = velocity of the body = 350m/s

Substituting this values in the formula, we have;

K.E = 1/2× 0.0025×350²

K.E = 153.125Joules

Work done is the force applied to body to cause it to move through a distance.

Work = Force × distance

Force = ma = 0.0025 × 10

Force = 0.025N

Distance = 12cm = 0.12m

Work = 0.025×0.12

Work = 0.003Nm

work done by the tree in stopping the bullet is 0.003N

4 0

3 years ago

Read 2 more answers

A doctor pushes the plunger of syringe down and then pulls it up to a draw liquid into a syringe? give reason please help me wit

ruslelena [56]

Answer:

When the doctor has the syringe, it is full of air. So, now after the doctor pushes the plunger, hence the air gets released into the medicine container. After this the doctor then takes the plunger and pull it back. Now , the air gets pulled up back into the syringe, but not only does the air come in but also the medicine because of the pressure build up.

If my answer helped, please mark me as the brainliest!!

Thank You!

4 0

3 years ago

Q 1: Calculate the pressure.

Oksana_A [137]

Answer:

1: 300pa

2: 10cm2

3: 430pa

4: a: 1.6667pa

b:2.5m2

c:20pa

Explanation:

7 0

3 years ago

ANSWER FOR BRAINLIEST

Vedmedyk [2.9K]

I think the answer is A

8 0

3 years ago

What would be the coefficient of performance if the refrigerator (operating between the same temperatures) was instead used as a

viktelen [127]

Complete Question

A certain refrigerator, operating between temperatures of -8.00°C and +23.2°C, can be approximated as a Carnot refrigerator.

What is the refrigerator's coefficient of performance? COP

(b) What If? What would be the coefficient of performance if the refrigerator (operating between the same temperatures) was instead used as a heat pump? COP

Answer:

a

$COP = 8.49$

b

$COP_1 = 9.49$

Explanation:

From the question we are told that

The lower operation temperature of refrigerator is $T_1 = -8.00^oC = 265 \ K$

The upper operation temperature of the refrigerator is $T_2 = 23.2 ^oC = 296.2 \ K$

Generally the refrigerators coefficient of performance is mathematically represented as

$COP = \frac{T_1}{T_2 - T_1 }$

=> $COP = \frac{265}{296.2 - 265 }$

=> $COP = 8.49$

Generally if a refrigerator (operating between the same temperatures) was instead used as a heat pump , the coefficient of performance is mathematically represented as

$COP_1 = \frac{T_2}{ T_2 - T_1}$

=> $COP_1 = \frac{296.2}{ 296.2 - 265 }$

=> $COP_1 = 9.49$

8 0

3 years ago