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3 years ago

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Is it possible to accelerate while moving constant speed

1 answer:

marusya05 [52]3 years ago

3 0

Answer:

Explanation:

An object which experiences either a change in the magnitude or the direction of the velocity vector can be said to be accelerating. This explains why an object moving in a circle at constant speed can be said to accelerate - the direction of the velocity changes.

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Prisms seperate __ light, such as that from the Sun by wavelength

natka813 [3]

Answer: white

Explanation:

prisms separate white light, such as that from the sun by wavelength.

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3 years ago

You are pushing a heavy box across a rough floor. when you are initially pushing the box and it is accelerating, (a) you exert a

kodGreya [7K]

The answer is C. If the box is accelerating, that means that the amount of force you are exerting is greater than the force of the box.

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According to the Big Bang Theory, the universe is __________________. A) constant B) depleting C) expanding D) exploding

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Enter an expression, in terms of defined quantities and g, for the force that the scale under left pillar shows

ella [17]

The expression, in terms of defined quantities and g is therefore Fu =((mg/2) +2 mp) g

<h3>What is a Scale?</h3>

This can be defined as a balance or any of various other instruments or devices for weighing.

The expression in terms of defined quantities and g, for the force that the scale under left pillar shows that Fu =((mg/2) +2 mp) g .

Read more about Force here brainly.com/question/4515354

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2 years ago

Find the magnitude of the electric field due to a charged ring of radius "a" and total charge "Q", at a point on the ring axis a

34kurt

Answer:

E= $\frac{KQ}{2\sqrt 2a^2}$

Explanation:

We are given that

Charge on ring= Q

Radius of ring=a

We have to find the magnitude of electric filed on the axis at distance a from the ring's center.

We know that the electric field at distance x from the center of ring of radius R is given by

$E=\frac{kQx}{(R^2+x^2)^{\frac{3}{2}}}$

Substitute x=a and R=a

Then, we get

$E=\frac{KQa}{(a^2+a^2)^{\frac{3}{2}}}$

$E=\frac{KQa}{(2a^2)^{\frac{3}{2}}}$

$E=\frac{KQa}{2\sqrt 2a^3}$

$E=\frac{KQ}{2\sqrt 2a^2}$

Where K= $9\times 10^9 Nm^2/C^2$

Hence, the magnitude of the electric filed due to charged ring on the axis of ring at distance a from the ring's center= $\frac{KQ}{2\sqrt 2a^2}$

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3 years ago