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nikdorinn [45]
3 years ago
7

Find the force necessary to start the crate moving, given that the mass of the crate is 32 kg and the coefficient of static fric

tion between the crate and the floor is 0.57. Express your answer using two significant figures.
Physics
1 answer:
fiasKO [112]3 years ago
3 0

Answer:

178.75 N

Explanation:

The force necessary to start moving the crate must be equal to or more than the frictional force (resistive force) acting on the crate but moving in an opposite direction to the frictional force.

So, we find the frictional force, Fr:

Fr = -μmg

Where μ = coefficient of friction

m = mass

g = acceleration due to gravity

The frictional force is negative because it acts against the direction of motion of the crate.

Fr = -0.57 * 32 * 9.8

Fr = - 178.75 N

Hence, the force necessary to move the crate must be at least equal to but opposite in direction to this frictional force.

Therefore, this force is 178.75 N

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Find the magnitude and direction of an electric field that exerts a 4.80×10−17N westward force on an electron. (b) What magnitud
jok3333 [9.3K]

Explanation:

(a) E = F/q

E = 4.8×10^-17/1.6×10^-19

E = 300 N/C

(b) same magnitude of electric field is exerted on proton

4 0
3 years ago
A 56 kg astronaut stands on a bathroom scale inside a rotating circular space station. The radius of the space station is 250 m.
Zielflug [23.3K]

Answer:

The speed of space station floor is 49.49 m/s.

Explanation:

Given that,

Mass of astronaut = 56 kg

Radius = 250 m

We need to calculate the speed of space station floor

Using centripetal force and newton's second law

F=mg

\dfrac{mv^2}{r}=mg

\dfrac{v^2}{r}=g

v=\sqrt{rg}

Where, v = speed of space station floor

r = radius

g = acceleration due to gravity

Put the value into the formula

v=\sqrt{250\times9.8}

v=49.49\ m/s

Hence, The speed of space station floor is 49.49 m/s.

6 0
3 years ago
Can a goalkeeper at his goal kick a soccer ball into the opponent’s goal without the ball touching the ground? The distance will
zmey [24]

The goalkeeper at his goal cannot kick a soccer ball into the opponent’s goal without the ball touching the ground

Explanation:

Consider the vertical motion of ball,

We have equation of motion v = u + at

     Initial velocity, u  = u sin θ

     Final velocity, v =  0 m/s    

     Acceleration = -g

     Substituting

                      v = u + at  

                      0 = u sin θ - g t

                      t=\frac{usin\theta }{g}

This is the time of flight.

Consider the horizontal motion of ball,

        Initial velocity, u =  u cos θ

        Acceleration, a =0 m/s²  

        Time, t=\frac{usin\theta }{g}  

     Substituting

                      s = ut + 0.5 at²

                      s=ucos\theta \times \frac{usin\theta }{g}+0.5\times 0\times (\frac{usin\theta }{g})^2\\\\s=\frac{u^2sin\theta cos\theta}{g}\\\\s=\frac{u^2sin2\theta}{2g}

This is the range.

In this problem

              u = 30 m/s

              g = 9.81 m/s²

              θ = 45° - For maximum range

Substituting

               s=\frac{30^2\times sin(2\times 45)}{2\times 9.81}=45.87m

Maximum horizontal distance traveled by ball without touching ground is 45.87 m, which is less than 95 m.

So the goalkeeper at his goal cannot kick a soccer ball into the opponent’s goal without the ball touching the ground

6 0
3 years ago
A baseball with a mass of 0.3kg is thrown into the air with a speed of 50 m/s and is at a height of 10m above. What is the speed
makkiz [27]

a una velocidad de

22 m/s, quien lo golpea y devuelve en la misma

dirección con una velocidad de 14 m/s. Si el

tiempo de contacto del balón con la jugadora es

de 0,03 s, ¿con qué fuerza golpeó la jugadora el

balón?

21 Una bala de 0,8 g, está en la recámara de un rifl e

cuando se g

6 0
3 years ago
What happens to the coefficient of friction when the weight is increased? Why is this?
Crazy boy [7]

Answer:

Usually the coefficient of friction remains unchanged

Explanation:

The coefficient of friction should in the majority of cases, remain constant no matter what your normal force is. When you apply a greater normal force, the frictional force increases, and your coefficient of friction stays the same. Here's another way to think about it: because the force of friction is equal to the normal force times the coefficient of friction, friction is increased when normal force is increased.

Plus, the coefficient of friction is a property of the materials being "rubbed", and this property usually does not depend on the normal force.

6 0
3 years ago
Read 2 more answers
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