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ivolga24 [154]
1 year ago
12

a 150 N force is used to pull a wooden box across a wooden surface at a constant velocity. what is the mass of the box?

Physics
1 answer:
IRINA_888 [86]1 year ago
7 0

Answer:

The mass of the box:

m =  60 kg

Explanation:

Given:

F = 150 N

g = 10 m/s²

_________

m - ?

Coefficient of friction wood on wood:

μ = 0.25

Friction force:

F₁ = μ*m*g

Newton's Third Law:

F = F₁

F = μ*m*g

The mass of the box:

m = F / ( μ*g) = 150 / (0.25*10) =  60 kg

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Sonja [21]

When thermal energy of a substance increases, it's entropy(randomness) & Kinetic energy increases.

For more appropriate answer, you should put the options 'cause there could be more than one answer for this question.

4 0
3 years ago
Please help! I'm not sure what equation or the process to do this question.
lions [1.4K]

Answer:

The momentum is 1.94 kg m/s.

Explanation:

To solve this problem we equate the potential energy of the spring with the kinetic energy of the ball.

The potential energy U of the compressed spring is given by

U = \dfrac{1}{2} kx^2,

where x is the length of compression and k is the spring constant.

And the kinetic energy of the ball is

K.E = \dfrac{1}{2}mv^2.

When the spring is released all of the potential energy of the spring goes into the kinetic energy of the ball; therefore,

\dfrac{1}{2}mv^2 = \dfrac{1}{2}kx^2,

solving for v we get:

v = x \sqrt{\dfrac{k}{m} }.

And since momentum of the ball is p=mv,

p =mx \sqrt{\dfrac{k}{m} }.

Putting in numbers we get:

p =(0.5kg)(0.25m) \sqrt{\dfrac{(120N/m)}{0.5kg} }.

\boxed{p=1.94kg\: m/s}

5 0
3 years ago
At t = 0 a grinding wheel has an angular velocity of 24.0 rad/s. It has a constant angular acceleration of 30.0rad/s2 until a ci
kozerog [31]

Answer:

θ=108rad

t =10.29seconds

α=-8.17rad/s²

Explanation:

Given that

At t=0, Wo=24rad/sec

Constant angular acceleration =30rad/s²

At t=2, θ=432rad as it try to stop because the circuit break

Angular motion

W=Wo+αt

θ=Wot+1/2αt²

W²=Wo²+2αθ

We need to find θ between 0sec to 2sec when the wheel stop

a. θ=Wot+1/2αt²

θ=24×2+1/2×30×2²

θ=48+60

θ=108rad.

b. W=Wo+αt

W=24+30×2

W=84rad/s

This is the final angular velocity which is the initial angular velocity when the wheel starts to decelerate.

Wo=84rad/sec

W=0rad/s, because the wheel stop at θ=432rad

Using W²=Wo²+2αθ

0²=84²+2×α×432

-84²=864α

α=-8.17rad/s²

It is negative because it is decelerating

Now, time taken for the wheel to stop

W=Wo+αt

0=84-8.17t

-84=-8.17t

Then t =10.29seconds.

a. θ=108rad

b. t =10.29seconds

c. α=-8.17rad/s²

3 0
3 years ago
28. Ken and Musa shared a cake such that Ken got twice the size
ratelena [41]

Answer:

Musa = \frac{1}{3}

Ken = \frac{2}{3}

Explanation:

Given

Ken = 2 * Musa --- Ken's share

Required

The fraction each got

Since they both shared a cake, we have:

Ken + Musa = 1

Substitute: Ken = 2 * Musa

2 * Musa+ Musa = 1

Factorize

Musa(2+ 1)= 1

Musa(3)= 1

Divide both sides by 3

Musa = \frac{1}{3}

Recall that: Ken = 2 * Musa

Ken = 2 * \frac{1}{3}

Ken = \frac{2}{3}

3 0
3 years ago
If i go to the moon what happens to my mass and weight
mixas84 [53]

Your mass stays the same because mass is the matter of something so the only way it could be changed is if you cut off something

Your weight changes because the gravitational field strength of different

6 0
3 years ago
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