All categories

2 years ago

Explain the centripetal force or Newton's second law...

2 answers:

8 0

Answer: a centripetal force is toward the center of curvature, the same as the direction of centripetal acceleration. According to Newton's second law of motion, net force is mass times acceleration: net . For uniform circular motion, the acceleration is the centripetal acceleration

Explanation:

stepan [7]2 years ago

6 0

Newton’s second law states that the acceleration of an object is directly related to the net force and inversely related to its mass. Acceleration of an object depends on two things, force and mass.

You might be interested in

A 5 N force pushes on the right side of a box. At the same time, a 10 N force pushes on the left side of the box. What happens t

notka56 [123]

The resultant force is 5N.So the box moves to right with constant acceleration.

8 0

3 years ago

Read 2 more answers

please help me guys please please please please please please please please please please please please please

seropon [69]

Answer:

1.F = 256 N

2.a = 8 m/s/s

By looking at the given information, you know force, and an acceleration. Therefore you have enough information to use the first formula.

F = ma

256 N = m * 8 m/s/s

m = 256 N/8 m/s/s

m = 32 Kg

6 0

2 years ago

A constant unbalanced force is applied to an object for a period of time. Which graph best represents the acceleration of the ob

balu736 [363]

Answer:

Here is an image attached with similar questions.

The correct answer is D where acceleration is function of time.

Explanation:

The force acting on the object is constant.

There is no change in the application of forces.

And we know that Force is the product of acceleration and masses.

Newtons second law: $F=m\times a$

Regarding mass we know that it can neither be created nor destroyed.

So in $F=m \times a$ we have two constant terms, constant divided by constant will give the same result as $a=\frac{Force}{mass}$

There is no change in acceleration $(a)$ with respect to time $(t)$ .

So the most appropriate graph where time (t) is changing on $x-axis$ but acceleration doesn't changes is D.

The graph will be similar to $y=any\ constant$ and will be horizontal to the $x-axis$ .

Option D depicts the same.

5 0

2 years ago

Imagine that I have a ping-pong ball and a bowling ball resting on the floor of our classroom. I go up to the bowling ball and g

MatroZZZ [7]

Answer:

the speed and aceleration of the ping pong ball is greater than that of the bowling ball.

Explanation:

We can analyze this exercise from several points of view, if we use Newton's second law

Bowling ball

F = M a₁

pingpongg ball

F = m a₂

as the forces the same

M a₁ = m a₂

a₂ = $\frac{M}{m}$ a₁

Since the mass of the bowling ball is much greater than the ping pong ball,

a₂ »a₁

so the acceleration of the ping pong ball is much greater than the acceleration of the bowling ball.

If we use the relationship of momentum and momentum, assuming that the time for the two cases is the same and that both start from rest

Bowling ball

I = F t = Δp

I = M (v₁ - v₀)

Ping pong ball

I = F t = Δp

I = m (v₂ -v₀)

the impulse itself

M v₁ = m v₂

v₂ = $\frac{M }{ m}$ v₁

so we conclude that the speed of the ping pong ball is much greater than the speed of the bowling ball.

In conclusion the speed and aceleration of the ping pong ball is greater than that of the bowling ball.

4 0

2 years ago

This is one of those special bounce collisions that is perfectly elastic. It is worth knowing that when identical masses collide

Dmitrij [34]

Answer:

0.069 J

Explanation:

mass of first ball (M1) = 0.839 kg

mass of second ball (M2) = 0.839 kg

initial velocity of first ball (U1) = 0.406 m/s

initial velocity of second ball (U2) = 0 m/s

For head on elastic collisions of equal masses, the velocities of the masses always changes.and this has also been clearly stated in the question. Hence the velocity if mass A will interchange with that of mass B after collision.

therefore

final velocity of first ball (V1) = 0 m/s

final velocity of second ball (V2) = 0.406 m/s

kinetic energy of second ball after collision = 0.5m $v^{2}$

= 0.5 x 0.839 x 0.406 x 0.406 = 0.069 J

6 0

2 years ago