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3 years ago

What are 3 types of frictions and when does each apply/9029244/29edc63f?utm_source=registration

1 answer:

5 0

- Static friction: when an object is not moving
- Kinetic friction: if an object is moving
- Rolling friction: when there is rolling (wheel,..)

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As a marble rolls across the floor, it gradually slows to a stop due to friction. Which statement best describes the change in m

Olegator [25]

Answer:

I belive it would be "C"

Explanation:

If it was any of the other answers "B" it would instantly stop. "A" it would roll forever.

3 0

2 years ago

Consider two wheels with fixed hubs. The hub cannot move, but the wheel can rotate about it. The hubs are fixed to a stationary

kykrilka [37]

Answer:

Magnitude of force on wheel B is 4 N

Explanation:

Given that

$I=mr^2$

For wheel A

m= 1 kg

d= 1 m,r= 0.5 m

F=1 N

We know that

T= F x r

T=1 x 0.5 N.m

T= 0.5 N.m

T= I α

Where I is the moment of inertia and α is the angular acceleration

$I_A=1 \times 0.5^2\ kg.m^2$

$I_A=0.25\ kg.m^2$

T= I α

0.5= 0.25 α

$\alpha = 2\ rad/s^2$

For Wheel B

m= 1 kg

d= 2 m,r=1 m

$I_B=1 \times 1^2\ kg.m^2$

$I_B=1 \ kg.m^2$

Given that angular acceleration is same for both the wheel

$\alpha = 2\ rad/s^2$

T= I α

T= 1 x 2

T= 2 N.m

Lets force on wheel is F then

T = F x r

2 = F x 1

So F= 2 N

Magnitude of force on wheel B is 2 N

3 0

3 years ago

A box weighing 43.2 N is pulled horizontally until it slides uniformly lat a constant

GREYUIT [131]

Your diagram should include four forces:

• the box's weight, pointing down (magnitude w = 43.2 N)

• the normal force, pointing up (mag. n)

• the applied force, pointing the direction in which the box is sliding (mag. p = 6.30 N, with p for "pull")

• the frictional force, pointing oppoiste the applied force (mag. f )

The box is moving at a constant speed, so it is inequilibrium and the net forces in both the vertical and horizontal directions sum to 0. By Newton's second law, we have

n + (-w) = 0

and

p + (-f ) = 0

So then the forces have magnitudes

w = 43.2 N

n = w = 43.2 N

p = 6.30 N

f = p = 6.30 N

5 0

2 years ago

Suppose you have a 136-kg wooden crate resting on a wood floor. The coefficient of static friction is 0.4 and coefficient of kin

stealth61 [152]

Answer:

The the maximum force acting on the crate is 533.12 newtons.

Explanation:

It is given that,

Mass of the wooden crate, m = 136 kg

The coefficient of static friction, $\mu_s=0.4$

The coefficient of kinetic friction, $\mu_k=0.2$

We need to find the maximum force exerted horizontally on the crate without moving it. As the crate is not moving than the coefficient of static friction will act and the force is given by :

$F=\mu_s mg$

$F=0.4\times 136\ kg\times 9.8\ m/s^2$

F = 533.12 N

So, the maximum force acting on the crate is 533.12 newtons. Hence, this is the required solution.

4 0

3 years ago

Why don't asteroids have atmosphere

EleoNora [17]

Asteroids are too small to have atmospheres

4 0

2 years ago

Read 2 more answers