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3 years ago

Explain Why must a cooling system do work to transfer thermal energy?

1 answer:

8 0

Although internal energy will not spontaneously flow from a cold region to a hot region, it can be forced to do so by doing work on the system.Refrigerators and heat pumps are examples of heat engines which cause energy to be transferred from a cold area to a hot area. Usually this is done with the aid of a phase change, i.e., a refrigerant liquid is forced to evaporate and extract energy from the cold area. Then it is compressed and forced to condense in the hot area, dumping its heat of vaporization into the hot area. ARE YOU FROM K12?

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You push your couch a distance of 3.9 m across the living room floor with a horizontal force of 220.0 n. the force of friction i

Mrac [35]

The general formula to calculate the work is:

$W=Fd \cos \theta$

where F is the force, d is the displacement of the couch, and $\theta$ is the angle between the direction of the force and the displacement. Let's apply this formula to the different parts of the problem.

(a) Work done by you: in this case, the force applied is parallel to the displacement of the couch, so $\theta=0^{\circ}$ and $\cos \theta=1$ , therefore the work is just equal to the product between the horizontal force you apply to push the couch and the distance the couch has been moved:

$W=Fd=(220.0 N)(3.9 m)=858 J$

(b) work done by the frictional force: the frictional force has opposite direction to the displacement, therefore $\theta=180^{\circ}$ and $\cos \theta=-1$ . Therefore, we must include a negative sign when we calculate the work done by the frictional force:

$W=-Fd=-(144.0 N)(3.9 m)=-561.6 J$

(c) The work done by gravity is zero. In fact, gravity (which points downwards) is perpendicular to the displacement of the couch (which is horizontal), therefore $\theta=90^{\circ}$ and $\cos \theta=0$ : this means

$W=0$ .

(d) Work done by the net force:

The net force is the difference between the horizontal force applied by you and the frictional force:

$F=220 N-144 N=76 N$

And the net force is in the same direction of the displacement, so $\theta=0^{\circ}$ and $\cos \theta=1$ and the work done is

$W=Fd=(76 N)(3.9 m)=296.4 J$

4 0

3 years ago

A 40 kg girl and an 8.4 kg sled are on the surface of a frozen lake, 15 m apart. By means of a rope, the girl exerts a 5.2 N for

stealth61 [152]

Answer:

(a) $a_s=0.62\frac{m}{s^2}$

(b) $a_s=0.13\frac{m}{s^2}$

Explanation:

(a) According to Newton's second law, the acceleration of a body is directly proportional to the force exerted on it and inversely proportional to it's mass.

$a_s=\frac{F}{m_s}\\a_s=\frac{5.2N}{8.4kg}\\a_s=0.62\frac{m}{s^2}$

(b) According to Newton's third law, the force that the sled exerts on the girl is equal in magnitude but opposite in the direction of the force that the girl exerts on the sled:

$a_g=\frac{F}{m_g}\\a_g=\frac{5.2N}{40kg}\\a_g=0.13\frac{m}{s^2}$

$x_f=x_0+v_0t \pm \frac{at^2}{2}$

For the girl, we have $x_0=0$ and $v_0=0$ . So:

$x_f_g=\frac{a_gt^2}{2}(1)$

For the sled, we have $v_0=0$ . So:

$x_f_s=x_0_s-\frac{a_st^2}{2}(2)$

When they meet, the final positions are the same. So, equaling (1) and (2) and solving for t:

$x_0_s-\frac{a_st^2}{2}=\frac{a_st^2}{2}\\t^2(a_g+a_s)=2x_0_s\\t=\sqrt{\frac{2x_s_0}{a_g+a_s}}\\t=\sqrt{\frac{2(15m)}{0.13\frac{m}{s^2}+0.62\frac{m}{s^2}}}\\t=6.32s$

Now, we solve (1) for $x_f_g$

$x_f_g=\frac{0.13\frac{m}{s^2}(6.32s)^2}{2}\\x_f_g=2.6m\\x_f=2.6m$

5 0

3 years ago

The intensity of electromagnetic waves from the sun is 1.4kW/m2 just above the earth's atmosphere. Eighty percent of this reache

Arlecino [84]

Answer:

$E=8.02*10^{5}J$

Explanation:

Given data

Electromagnetic waves from the sun is I=1.4kW/m² at 80%

Area a=(0.30×0.51)m²

Time t=1.30 hr

To find

Energy E

Solution

The energy received by your back is calculated as:

$E=Pt=Iat\\E=(0.80)(1400W/m^{2} )(0.30*0.51m^{2} )(1.30*3600s)\\E=8.02*10^{5}J$

3 0

3 years ago

Read 2 more answers

!!HELP HELP!!

borishaifa [10]

Answer:

7.5s

Explanation:

Given parameters:

Velocity = 30m/s

Deceleration = 4m/s²

Unknown:

Time it takes for the car to come to complete rest = ?

Solution:

To solve this problem, we use the kinematics expression below:

v = u + at

Since this is a deceleration

v = u - at

v is the final velocity

u is the initial velocity

a is the acceleration

t is the time taken

v - u = -at

0 - 30 = -4 x t

-30 = -4t

t = 7.5s

6 0

3 years ago

A mass on a string is swung in a circle of radius 0.75m at 7.0m/s.what its rate of acceleration.

Ulleksa [173]

Explanation:

ac = v^2/r

= (7.0 m/s)^2/(0.75 m)

= 65 m/s^2

3 0

2 years ago