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Snowcat [4.5K]
3 years ago
8

The 75.0 kg hero of a movie is pulled upward with a constant acceleration of 2.00 m/s2 by a rope. What is the tension on the rop

e?
585N


75.0N


885N


11.8N
Physics
1 answer:
lutik1710 [3]3 years ago
7 0

Use Newton's second law. The net force on the hero is

∑ <em>F</em> = <em>T</em> - <em>m g</em> = <em>m</em> (2.00 m/s²)

where

• <em>T</em> = tension in the rope

• <em>m</em> = 75.0 kg = mass of the hero

• <em>g</em> = 9.80 m/s² = acceleration due to gravity

Solve for <em>T</em> :

<em>T</em> = (75.0 kg)<em> </em>(<em>g</em> + 2.00 m/s²) = 885 N

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A car is moving with a constant acceleration. At t = 5.0 s its velocity is 8.0 m/s and at t = 8.0 s its
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A car traveling at 50 km/h hits a bridge abutment. A passenger in the car moves forward a distance of 61 cm (with respect to the
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Answer:

6957.04N

Explanation:

Using

vf2=vi2+2ad

But vf = 0 .

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3 0
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A simple harmonic wave of wavelength 18.7 cm and amplitude 2.34 cm is propagating along a string in the negative x-direction at
storchak [24]

Answer

given,

wavelength = λ = 18.7 cm

                    = 0.187 m

amplitude , A = 2.34 cm

v = 0.38 m/s

A)  angular frequency = ?

     f = \dfrac{v}{\lambda}

     f = \dfrac{0.38}{0.187}

     f =2.03\ Hz

angular frequency ,

ω = 2π f

ω = 2π x 2.03

ω = 12.75 rad/s

B) the wave number ,

      K = \dfrac{2\pi}{\lambda}

     K= \dfrac{2\pi}{0.187}

    K =33.59\ m^{-1}

C)

as the wave is propagating in -x direction, the sign is positive between x and t

y ( x ,t) = A sin(k  x - ω t)

y ( x ,t) = 2.34  x  sin(33.59 x - 12.75 t)

4 0
3 years ago
14
Minchanka [31]

Answer:

C

Explanation:

8 0
2 years ago
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