3 years ago

Discuss the relationship between acceleration and velocity. Are these scalar or vector quantities?

1 answer:

5 0

Acceleration is just speeding up, and velocity is speeding up in a certain direction. If you are changing your velocity you accelerating, as you are changing direction..

You might be interested in

Subduction occurs because the sea floor is: more dense than the continental floor less dense than the continental floor more buo

svetoff [14.1K]

A)more dense than the continental floor

7 0

3 years ago

Read 2 more answers

If you were on a decision making board with the task of choosing which innovation to fund, what criteria would you use to make y

faust18 [17]

Explanation:

The criteria for decision making would be

1. I would fund for the school of young diabetics, for the sole purpose of them leaning and being motivated for a healthy lifestyle.

2. I would also fund for new and improved insulin pumps as old ones cause multiple problems.

3 0

3 years ago

A person pushes a 10 kg box from rest and accelerates it to a speed of 4 m/s with a constant force. If the box is pushed for a t

lutik1710 [3]

The box is accelerated from rest to 4 m/s in a matter of 2.5 s, so its acceleration a is such that

4 m/s = a (2.5 s) → a = (4 m/s) / (2.5 s) = 1.6 m/s²

Then the force applied to the box has a magnitude F such that

F = (10 kg) (1.6 m/s²) = 16 N

7 0

3 years ago

When a laser shines on a screen after passing through two closely spaced slits, you see ___

Aleonysh [2.5K]

Answer:

diffraction

Explanation:

diffraction occurs when light passes sharp edges or goes through narrow slits the rays are deflected and produce fringes of light and dark bands

6 0

3 years ago

A 120-kg object and a 420-kg object are separated by 3.00 m At what position (other than an infinitely remote one) can the 51.0-

djverab [1.8K]

Answer:

1.045 m from 120 kg

Explanation:

m1 = 120 kg

m2 = 420 kg

m = 51 kg

d = 3 m

Let m is placed at a distance y from 120 kg so that the net force on 51 kg is zero.

By use of the gravitational force

Force on m due to m1 is equal to the force on m due to m2.

$\frac{Gm_{1}m}{y^{2}}=\frac{Gm_{2}m}{\left ( d-y \right )^{2}}$

$\frac{m_{1}}{y^{2}}=\frac{m_{2}}{\left ( d-y \right )^{2}}$

$\frac{3-y}{y}=\sqrt{\frac{7}{2}}$

3 - y = 1.87 y

3 = 2.87 y

y = 1.045 m

Thus, the net force on 51 kg is zero if it is placed at a distance of 1.045 m from 120 kg.

6 0

3 years ago