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gtnhenbr [62]
2 years ago
14

An athlete is working out in the weight room. He steadily holds 50 kilograms above his head for 10 seconds. Which statement is t

rue about this situation?
The athlete isn’t doing any work because he doesn’t move the weight.
The athlete isn’t doing any work because he doesn’t hold the weight long enough.
The athlete is doing work because he prevents the weight from falling downward.
The athlete is doing work because 50 kilograms is a significant load to lift.
Physics
2 answers:
Kamila [148]2 years ago
7 0

Answer: The true statement form the given statements is "the athlete is not doing any work because he does not move the weight.".

Explanation:

The formula for work done is as follows,

W = Fscos\theta

Here, F is the force, s is the displacement and theta is the angle between force and displacement.

In the given problem, an athlete holds 50 kg on his head for 10 seconds. The angle between force and displacement is 90 degree.

In this case, the work done is zero.

Therefore, the athlete is not doing any work because he does not move the weight.

Nataly_w [17]2 years ago
4 0
<span>The athlete is sweating, straining, sweating, trembling and groaning. 
But if he passed high school physics, then he may remember that
'work' in physics means (force) x (distance).  If there's no distance,
then there's no work. 

He realizes that he isn’t doing any work because he's not moving the weight.</span>
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5 0
2 years ago
A solid cylindrical bar conducts heat at a rate of 25 W from a hot to a cold reservoir under steady state conditions. If both th
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Answer:

Using the new cylinder the heat rate between the reservoirs would be 50 W

Explanation:

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  2. For the original cylinder the Fourier's law would be: kA_1\frac{T_1-T_2}{L_1}=25W, and if A_1=\frac{\pi D_{1}^{2}}{4}, then the expression would be:k\frac{\pi D_1^{2}}{4} \frac{T_1-T_2}{L_1}=25W where D_1 is the diameter of the original cylinder, and {L_1} is the length of the original cylinder.
  3. For the new cylinder, in the same fashion that for the first, Fourier's Law would be: Q_2=k\frac{\pi D_2^2}{4}\frac{T_1-T_2}{L_2},where Q_2 is the heat rate in the second case, D_2 and {L_2 are the new diameter and length.
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  5. Rearranging: Q_2=\frac{2^2}{2}(k\frac{\pi D_1^2}{4}\frac{T_1-T_2}{L_1}).
  6. In the last declaration of  Q_2, it could be noted that the expressión inside the parenthesis is actually  Q_1, then:  Q_2=\frac{2^2}{2}(25W)=50W.
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7 0
2 years ago
Two resistors of 5.0 and 9.0 ohm are connected in parallel. A 4.0 ohm resistor is then connected in series with the parallel com
rewona [7]

Answer:

I1 = 0.772 A

Explanation:

<u>Given</u>: R1 = 5.0 ohm, R2 = 9.0 ohm, R3 = 4.0 ohm, V = 6.0 Volts

<u>To find</u>:  current I = ? A

<u>Solution: </u>

Ohm's law  V= I R

⇒   I = V / R

In order to find R (total) we first find R (p) fro parallel combination. so

1 / R (p) = 1 / R1  + 1/ R2          ∴(P) stand for parallel

R (p) = R1R2 / ( R1 + R2)

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R (p) = 3.214 ohm

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R (total) = 3.214 ohm + 4.0 Ohm

R (total) = 7.214 ohm

now I (total) = 7.214 ohm / 6.0 Volts

I (total) = 1.202 A

This the total current supplied by 6 volts battery.

as voltage drop across R (p) = V = R (p) × I (total)

V (p) = 3.214 ohm × 1.202 A  = 3.864 volts

Now current through 5 ohms resister  is I1 = V (P) / R1

I1 = 3.864 volts / 5 ohm

I1 = 0.772 A

3 0
2 years ago
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