Please help me long answer plz

A 25kg chair initially at rest on a horizontal floor requires 165 N force to set it in motion. Once the chair is in motion, a 12

bazaltina [42]

The coefficient of static friction between the chair and the floor is 0.67

Explanation:

Given:

Weight of the chair = 25kg

Force = 165 N (F_applied)

Force = 127 N (F_max)

To find: Coefficient of static friction

The “coefficient of static friction” between a chair and the floor is defined as the ration of maximum force to the normal force acting on the chair

μ_s= $F_{max}/F_{n}$

The F_n is equal to the weight multiplied by its gravity

∴ $F_{n}$ =mg

Thus the coefficient of static friction changes as

μ_s= $F_{max}/mg$

μ_{s} = $=165N/((25kg)\times(9.80 m/s^2 ) )$

= 0.67

3 0

2 years ago

Inez uses hairspray on her hair each morning before going to school. The spray spreads out before reaching her hair partly becau

Tamiku [17]

Answer:

$7.0\cdot 10^{-13}C$

Explanation:

The magnitude of the electrostatic force between two charged objects is

$F=k\frac{q_1 q_2}{r^2}$

where

k is the Coulomb's constant

q1 and q2 are the two charges

r is the separation between the two charges

The force is attractive if the charges have opposite sign and repulsive if the charges have same sign.

In this problem, we have:

$r=0.070 cm =7\cdot 10^{-4} m$ is the distance between the charges

$q_1=q_2=q$ since the charges are identical

$F=9.0\cdot 10^{-9}N$ is the force between the charges

Re-arranging the equation and solving for q, we find the charge on each drop:

$F=\frac{kq^2}{r^2}\\q=\sqrt{\frac{Fr^2}{k}}=\sqrt{\frac{(9.0\cdot 10^{-9})(7\cdot 10^{-4})^2}{8.99\cdot 10^9}}=7.0\cdot 10^{-13}C$

8 0

2 years ago

If μs is greater than some critical value, the woman cannot start the crate moving no matter how hard she pushes. calculate this

fiasKO [112]

weight = mg acts downwards 
normal force = N acts upwards.
and force F acts at an angle θ below the horizontal.
(Let us assume that the woman pushes from the left, so F is acted towards the right, which is below the horizontal)
so that, Frictional force, f=us*N acts towards the left

Now we balance the forces along x and y directions:
y direction: N = mg + F sinΘ
x direction: us * N = F cosΘ

We let the value of µs be equal to a value such that any F will not be able to move the crate. Then, if we increase F by an amount F', then the force pushing the crate towards the right also increases by F' cosΘ. Additionally, the frictional force f must raise by exactly this amount.
Since f can’t exceed us*N, so the normal force must increase by F' cosΘ/us.
Also, from the y direction equation, the normal force exceeds by F' sin Θ.

These two values must be the same, therefore:
us = cot θ

4 0

2 years ago

How far away is the moon?

Fed [463]

The moon is 230,100 miles from planet earth.

5 0

2 years ago

Read 2 more answers

If a 60 kg woman, runs up a flight of stairs having a total rise of 4.0m in a time of 4.2s. What average power did she supply?

LenKa [72]

PE = m * g * h
PE = 60 * 9.8 * 4
PE = 2352 J

Power = E/t

Power = PE/t
Power = 2352/4.2
Power = 560 Watt

6 0

3 years ago