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Rama09 [41]
3 years ago
7

A man in a strength competition pulls an 18-wheel truck 3.10 m in 20.5 s. There is a cable that is attached to his body that exe

rts a horizontal force of 756 N. What is the power expended by the man?
Physics
1 answer:
larisa [96]3 years ago
4 0

Answer:

114.32195122 but Round your answer to three significant figures.) is 114

Explanation:

Just took the test

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During a baseball game, a batter hits a high pop-up. If the ball remains in the air for 6.82 s, how high does it rise? The accel
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The answer is 56.98 m

The dislocation (d) is:
d = v1 * t + 1/2 * a * t²
v1 - initial velocity
t - time
a - <span>acceleration of gravity

We know:
d = ?
v1 = ?
t = 6.82 s / 2 = 3.41  (it reaches the peak at half time)
a = - 9.8 m/s</span>²

Let's first calculate v1:
v2 = v1 + at
v2 - final velocity (v2 = 0 when it reaches peak)

0 = v1 + -9.8 * 3.41
0 = v1 - 33.418

v1 = 33.418 m/s

d = v1 * t + 1/2 * a * t²
d = 33.418 * 3.41 +  1/2 * -9.8 * 3.41²
d = 113.96 - 56.98
d = 56.98 m
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The result of a wave generator traveling faster than the speed of a wave is:
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The result of a wave generator traveling faster than the speed of a wave is called as a boom. If the wave is a sound wave, it is called a sonic boom. However, if the wave is light, it is called as a luminal boom.  Luminal bloom happens in some industries and is commonly called as the Cherenkov radiation.
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8. Contrast. The energy that can be released during a nuclear fission reaction with the energy that can be released during a nuc
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7 0
3 years ago
B. Projectile on cliff (range)
dimulka [17.4K]

Answer:

x = 41.28 m

Explanation:

This is a projectile launching exercise, let's find the time it takes to get to the base of the cliff.

Let's start by using trigonometry to find the initial velocity

         cos 25 = v₀ₓ / v₀

         sin 25 = Iv_{oy} / v₀

         v₀ₓ = v₀ cos 25

         v_{oy} = v₀ sin 25

         v₀ₓ = 22 cos 25 = 19.94 m / s

         v_{oy} = 22 sin 25 = 0.0192 m / s

let's use movement on the vertical axis

         y = y₀ + v_{oy} t - ½ g t²

     

when reaching the base of the cliff y = 0 and the initial height is y₀ = 21 m

         0 = 21 + 0.0192 t - ½ 9.81 t²

         4.905 t² - 0.0192 t - 21 = 0

         t² - 0.003914 t - 4.2813 =0

we solve the quadratic equation

        t = \frac{ 0.003914\  \pm \sqrt{0.003914^2 + 4 \ 4.2813 }   }{2}

        t = \frac{0.003914 \ \pm 4.13828}{2}

        t₁ = 2.07 s

        t₂ = -2.067 s

since time must be a positive scalar quantity, the correct result is

        t = 2.07 s

now we can look up the distance traveled

         x = v₀ₓ t

         x = 19.94  2.07

         x = 41.28 m

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