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NeTakaya
2 years ago
9

Find the acceleration of a box of mass 3.0 kg, if applied force is 20.0 N at 37° above the x axis, and the coefficient of kineti

c friction is 0.20.​
Physics
1 answer:
ExtremeBDS [4]2 years ago
4 0

Hello:

First, <u>the normal force</u> in y axis:

∑Fy = 0

∑Fy = N - Fy

∑Fy = N - 20 N * sen 37°

∑Fy = N = 12,87 N

Formula:

R = ma

F - F' = ma

F - (μN) = ma

20 N - (0.2 * 12,87 N) = 3 kg * a

20 N - 2.574 N = 3 kg * a

17,426 N / 3 kg = a

a = 5,8 m/s²

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A 2.2-kg block slides on a horizontal surface with a speed of v=0.80m/s and an
mina [271]

Answer:

μ = 0.33

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5 0
3 years ago
Classify the animal hydra and plant mustard with two characters of each​
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if my answer helps you than mark me as brainliest

4 0
3 years ago
The stopping sight distance for a car traveling at 50 mph is 461 ft (including both perception-reaction distance and braking dis
Ivan

Answer:

2.08 s

Explanation:

We are given that

Speed,v=50mph=73.3ft/s

1 mile=5280 feet

1 hour=3600 s

Distance,d=461 ft

t=2.5 s

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4 0
3 years ago
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