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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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Answer: d

Explanation:

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2 years ago
Inna Hurry is traveling at 6.8 m/s, when she realizes she is late for an appointment. She accelerates at 4.5 m/s^2 for 3.2 s. Wh
Alborosie

Answer:

1) v = 21.2 m/s

2) S = 63.33 m

3) s = 61.257 m

4) Deceleration, a = -4.32 m/s²

Explanation:

1) Given,

The initial velocity of Inna, u = 6.8 m/s

The acceleration of Inna, a = 4.5 m/s²

The time of travel, t = 3.2 s

Using the first equation of motion, the final velocity is

                v = u + at

                   = 6.8 + 4.5 x 3.2

                   = 21.2 m/s

The final velocity of Inna is, v = 21.2 m/s

2) Given,

The initial velocity of Lisa, u = 12 m/s

The final velocity of Lisa, v = 26 m/s

The acceleration of Lisa, a = 4.2 m/s²

Using the III equations of motion, the displacement is

                          v² = u² +2aS

                         S = (v² - u²) / 2a

                            = (26² -12²) / 2 x 4.2

                            = 63.33 m

The distance Lisa traveled, S = 63.33 m

3) Given,

The initial velocity of Ed, u = 38.2 m/s

The deceleration of Ed, d = - 8.6 m/s²

The time of travel, t = 2.1 s

Using the II equations of motion, the displacement is

                        s = ut + 1/2 at²

                           =38.2 x 2.1 + 0.5 x(-8.6) x 2.1²

                           = 61.257 m

Therefore, the distance traveled by Ed, s = 61.257 m

4) Given,

The initial velocity of the car, u = 24.2 m/s

The final velocity of the car, v = 11.9 m/s

The time taken by the car is, t = 2.85 s

Using the first equations of motion,

                         v = u + at

∴                        a = (v - u) / t

                            = (11.9 - 24.2) / 2.85

                            = -4.32 m/s²

Hence, the deceleration of the car, a = = -4.32 m/s²

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