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Alekssandra [29.7K]
2 years ago
10

I need help with nine and 10 I will really appreciate it

Physics
1 answer:
horsena [70]2 years ago
7 0

Answer:

9] V = D ÷ T

Take any distance value from the graph and its relevant time.

V = 4 ÷ 2

V = 2 m/s

[You will notice that any distance values with its time will give you 2 m/s as its speed. This means that speed is constant throughout.]

10] Take the distance value and its time for the highest peak of B.

V = 20 ÷ 2

V = 10 m/s

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What effect would this have on the momentum?
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PLEASE HELP ME
pogonyaev

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The answer is Insulator, Conductor

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Write the difference between convert and<br>Concave minores​
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3 years ago
A force of 10. N toward the right is exerted on a wooden crate initially moving to the right on a horizontal wooden floor. the c
Artyom0805 [142]

1) 5 N

The crate is initially moving, so we must calculate the force of kinetic friction, which is given by:

F_f = \mu_k mg

where

\mu_k=0.2 is the coefficient of friction between the crate (made of wood) and the floor (made of wood). The coefficient of kinetic friction between wood and wood is about 0.2.

mg=25 N is the weight of the crate

Substituting the numbers into the formula, we find

F_f=(0.2)(25 N)=5 N


2) 5 N

There are two forces acting on the crate along the horizontal direction:

- The force that pushes the crate toward the right, of magnitude F=10 N

- The force of friction, which acts in the opposite direction (so, towards the left), of magnitude F_f = 5 N

Since the two forces are in opposite directions, the net force is given by their difference:

F_{net}=F-F_f = 10 N-5 N=5 N


3) Yes

The crate is accelerating. In fact, according to Second Newton's Law:

F_{net}=ma (1)

where Fnet is the net force on the crate, m is its mass, a is its acceleration. We can immediately see that since Fnet is not zero, the acceleration is also non-zero, so the crate is accelerating.

We can even calculate the magnitude of the acceleration. In fact, the mass of the crate is given by:

m=\frac{Weight}{g}=\frac{25 N}{9.8 m/s^2}=2.55 kg

And by using (1) we find

a=\frac{F_{net}}{m}=\frac{5 N}{2.55 kg}=1.96 m/s^2


3 0
3 years ago
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