I’m sorry this is a quiz and we’re correcting it and I still don’t get it anyone know ?

Physics

1 answer:

kotegsom [21]3 years ago

5 0

Answer:

3.0M

Explanation:

Thats two wavelengths,not one.

Pretty honest mistake I would've made the same if I was rushing

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The velocity of 200kg object changes from 10m/s to 22m/s in 12 sec.what is the force on the accelerated object?

Vikentia [17]

Initial velocity (u) = 10 m/s
Final velocity (v) = 22 m/s
Time (t) = 12 s
Mass (m) = 200 Kg
Let the acceleration be a.
By using the equation of motion,

v = u + at, we have

22 m/s = 10 m/s + 12 s × a
or, 22m/s - 10 m/s = 12 s × a
or, 12 m/s = 12 s × a
or, a = 1 m/s^2
Let the force be F.
We know, F = ma
Therefore, the force on the accelerated object (F)
= ma
= (200 × 1) N
= 200 N

Answer:

b) 200 N

Hope you could understand.

If you have any query, feel free to ask.

4 0

3 years ago

suppose the ball has the smallest possible frequency that allows it to go all the way around the circle. what tension in the str

tatyana61 [14]

The complete question is missing, so i have attached the complete question.

Answer:

A) FBD is attached.

B) The condition that must be satisfied is for ω_min = √(g/r)

C) The tension in the string would be zero. This is because at the smallest frequency, the only radially inward force at that point is the weight(force of gravity).

Explanation:

A) I've attached the image of the free body diagram.

B) The formula for the net force is given as;

F_net = mv²/r

We know that angular velocity;ω = v/r

Thus;

F_net = mω²r

Now, the minimum downward force is the weight and so;

mg = m(ω_min)²r

m will cancel out to give;

g = (ω_min)²r

(ω_min)² = g/r

ω_min = √(g/r)

The condition that must be satisfied is for ω_min = √(g/r)

C) The tension in the string would be zero. This is because at the smallest frequency, the only radially inward force at that point is the weight(force of gravity).