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2 years ago

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A copper coin resting on a piece of cardboard is placed on a beaker as shown in the diagram below. When the cardboard is rapidly

removed, the coin drops into the beaker. The two properties of the coin which best explain its fall are its weight and its-
A: Temperature
B: Density
C: Volume
D: Inertia

1 answer:

Gelneren [198K]2 years ago

6 0

The answer should be d) inertia

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A real heat engine operates between temperatures TcTcT_c and ThThT_h. During a certain time, an amount QcQcQ_c of heat is releas

Nookie1986 [14]

The maximum amount of work performed is

$W_{max}=\frac{T_H-T_C}{T_C}Q_C$

Explanation:

The efficiency of a real heat engine is given by the equation:

$\eta = 1-\frac{T_C}{T_H}$ (1)

where

$T_C$ is the temperature of the cold reservoir

$T_H$ is the temperature of the hot reservoir

However, the efficiency of a real heat engine can be also written as:

$\eta = \frac{W_{max}}{Q_H}$

where

$W_{max}$ is the maximum work done

$Q_H$ is the heat absorbed from the hot reservoir

$Q_H$ can be written as

$Q_H=W_{max}+Q_C$

where

$Q_C$ is the heat released to the cold reservoir

So the previous equation can be also written as

$\eta=\frac{W_{max}}{W_{max}+Q_C}$ (2)

By combining eq.(1) and (2) we get

$1-\frac{T_C}{T_H}=\frac{W_{max}}{W_{max}+Q}$

And re-arranging the equation and solving for $W_{max}$ , we find

$W_{max}=\frac{T_H-T_C}{T_C}Q_C$

Learn more about work and heat:

brainly.com/question/4759369

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3 years ago

Is our body in thermal equilibrium with the environment ?

LenKa [72]

Answer:

The human body runs at a constant 37 ºC but the air around you at room temperature is about 20-25 ºC which means heat is constantly leaving your body to warm your surroundings and maintain thermal equilibrium. You don't lose much energy doing this however as air heats reasonably quickly

Explanation:

4 0

3 years ago

Two hockey pucks, labeled A and B, are initially at rest on a smooth ice surface and are separated by a distance of 18.0 m . Sim

Nonamiya [84]

Answer:

The distance covered by puck A before collision is $z = 8.56 \ m$

Explanation:

From the question we are told that

The label on the two hockey pucks is A and B

The distance between the two hockey pucks is D 18.0 m

The speed of puck A is $v_A = 3.90 \ m/s$

The speed of puck B is $v_B = 4.30 \ m/s$

The distance covered by puck A is mathematically represented as

$z = v_A * t$

=> $t = \frac{z}{v_A}$

The distance covered by puck B is mathematically represented as

$18 - z = v_B * t$

=> $t = \frac{18 - z}{v_B}$

Since the time take before collision is the same

$\frac{18 - z}{V_B} = \frac{z}{v_A}$

substituting values

$\frac{18 -z }{4.3} = \frac{z}{3.90}$

=> $70.2 - 3.90 z = 4.3 z$

=> $z = 8.56 \ m$

8 0

3 years ago

Which topic are you most likely to find in a physics textbook?

Darina [25.2K]

I think the answer would be physics.

5 0

3 years ago

Read 2 more answers

A train pulls the boxcars. What are the vertical forces present? (1 point)

Greeley [361]

The vertical forces present in a boxcars being pulled by a train are weight and pull of air on the boxcars.

The forces exerted on an object can be resolved into <em>horizontal</em> and <em>vertical components.</em>

<em />

The horizontal components of the forces on an object being pulled include the following;

The frictional force on the object; which tends to reduce the motion,

force dragging the object forward

if the object is inclined to an angle, the horizontal component of the weight

This horizontal force is written as;

$\Sigma F_x = 0\\\\F_x - F_k = ma$

The vertical component of the force on the object include the following;

the weight of the object acting downwards

the pull of air on the object acting upwards

The vertical force on the object is written as;

$\Sigma F_n = 0\\\\F_n - W = 0$

Thus, the vertical forces present in a boxcars being pulled by a train are weight and pull of air on the boxcars.

Learn more here: brainly.com/question/2000189

5 0

3 years ago