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Nikolay [14]
2 years ago
11

A simple pendulum, 1.00 m in length, is released from rest when the support string is at an angle of 35.0 from the vertical. Wh

at is the speed of the suspended mass at the bottom of the swing? (g = 9.80 m/s2 and ignore air resistance)
Physics
1 answer:
taurus [48]2 years ago
7 0

Answer:

Explanation:

Length = 1.00 m

If the length is 1.0, the vertical distance pivot to bob is cos 35 = 0.819

At the lowest point, vertical distance is 1.0, so the change is the difference, 0.181 meter

The potential energy of that height is converted to kinetic energy of motion, which determines the speed.

PE = KE

mgh = ½mV²

V = √(2gh) = 1.88 m/s

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A mars prototype carrying scientific instruments has a mass of 1060kg. a net force of 52000 n is applied to this roverat a test
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8 0
3 years ago
MathPhys i need your help please helpppo
Lena [83]

Answer:

6.77 m/s

Explanation:

First, in the x direction:

Given:

Δx = 3.17 m

v₀ = v cos 30.8° = 0.859 v

a = 0 m/s²

Δx = v₀ t + ½ at²

(3.17 m) = (0.859 v) t + ½ (0 m/s²) t²

3.17 = 0.859 v t

3.69 = v t

Next, in the y direction:

Given:

Δy = 0.432 m

v₀ = v sin 30.8° = 0.512 v

a = -9.81 m/s²

Δy = v₀ t + ½ at²

(0.432 m) = (0.512 v) t + ½ (-9.81 m/s²) t²

0.432 = 0.512 v t − 4.905 t²

Two equations, two variables.  Solve for t in the first equation and substitute into the second equation:

t = 3.69 / v

0.432 = 0.512 v (3.69 / v) − 4.905 (3.69 / v)²

0.432 = 1.89 − 66.8 / v²

66.8 / v² = 1.458

v² = 45.8

v = 6.77

7 0
2 years ago
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