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Otrada [13]
3 years ago
10

A(n) 93 kg clock initially at rest on a horizontal floor requires a(n) 617 N horizontal force to set it in motion. After the clo

ck is in motion, a horizontal force of 528 N keeps it moving with a constant velocity. The acceleration of gravity is 9.81 m/s 2 . a) Find µs between the clock and the floor
Physics
1 answer:
Viefleur [7K]3 years ago
5 0

Answer:

\mu_s = 0.676

Explanation:

As we know that the force required to move the clock from rest position must be equal to the maximum limiting friction

So we will have

F = F_f

now we know that

F_f = \mu_s Mg

here we will have

F = 617 N

m = 93 kg

now from above formula we will have

617 = \mu_s (93)(9.81)

\mu_s = 0.676

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Suppose you hit a steel nail with a 0.500-kg hammer, initially moving at 15.0 m/s and brought to rest in 2.80 mm. How much is th
katrin [286]

Complete Question

Suppose you hit a steel nail with a 0.500-kg hammer, initially moving at 15.0 m/s and brought to rest in 2.80 mm. How much is the nail compressed if it is 2.50 mm in diameter and 6.00-cm long.What Average force is excreted on the Nail

Answer:

F=2*10^{4}N

Explanation:

From the question we are told that:

Mass m=0.500kg

Initial Velocity V=15.0m/s

Distance x=2.80mm=>0.00280m

Diameter d=2.50mm=>0.00250m

Length l=6.00cm=>0.6m

Generally the equation for Force is mathematically given by

 F=\frac{mv^2}{2d}

 F=\frac{0.500*15^2}{2.80*10^{-3}}

 F=2*10^{4}N

6 0
3 years ago
If you can answer my last post ill give you 75 points pls its very important and please make sure it correct!!!!!!!
Hoochie [10]

Answer:

I can't see the post :/

Explanation:

5 0
3 years ago
Read 2 more answers
Faiyez wrote the problem below
Ilya [14]

445/100 - 5/4 = 445/100 - 125/100 = 320/100 = 16/5 = 3 1/5.

6 0
4 years ago
A pendulum oscillates 12 times in 4 seconds. what is the length of the pendulum?
seraphim [82]

Answer:

L = 2.8 cm

Explanation:

Period T = 4 / 12 = 1/3 s

T = 2π√(L/g)

L = (T/2π)²g

L = ((1/3)/2π)²9.8 = 0.02758... ≈ 2.8 cm

6 0
3 years ago
A space station in the shape of a 100 m-diameter (50m radius) wheel is spinning so as to impart a linear tangential speed of 22.
zalisa [80]

Answer:

correct option is b. 31.3 m/s

Explanation:

given data

artificial gravity a1 = 1 g

artificial gravity a2 = 2 g

diameter = 100 m

radius  r= 50 m

speed v1 = 22.1 m/s

solution

As acceleration is  ∝ v²

so we can say

\frac{a2}{a1} = \frac{v2}{v1}    .....................1

put here value

\frac{2}{1} = \frac{v2}{22.1}  

solve it

v2 = \sqrt{2 } × 22.1

v2 = 31.25 m/s

so correct option is b. 31.3 m/s

4 0
3 years ago
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