Ask question

Ask question

All categories

3 years ago

7

Explain a situation in which you can accelerate even thought your speed doesn’t change

1 answer:

Juli2301 [7.4K]3 years ago

7 0

Explanation:

An object's acceleration is the rate its velocity (speed and direction) changes. Therefore, an object can accelerate even if its speed is constant - if its direction changes. ... Light's acceleration is zero. Since it travels in a straight line, its direction does not change.

You might be interested in

How do the tension of the cord and the force of gravity affect a pendulum?

LuckyWell [14K]

Answer:

<em>Force of gravity may not affect a pendulum during its equilibrium state</em>. But the gravity can affect the pendulum when a force occurs in any direction of the bob connected to the cord that makes a swing sideways. The gravity of pendulum never stops, it always accelerates. So the gravity affects the pendulum acceleration and speed.

<em>Similarly the tension in the cord will not affect the pendulum</em><em> </em>but if change in the length of the pendulum while keeping other factors constant changes the length of the period of pendulum. longer pendulum swings with lower frequency than shorter pendulums.

6 0

3 years ago

Read 2 more answers

A jet of water of cross section area and velocity v impinges normally on a stationary flat plate the mass per unit volume of wat

Sergio039 [100]

<h2>F = kAρv²</h2>

Explained in the attachment !

<h3>Hope it helps you!!</h3>

4 0

2 years ago

A 31.0 kg crate, initially at rest, slides down a ramp 2.6 m long and inclined at an angle of14.0° with the horizontal. Using th

mart [117]

The answer is 3.5m/s

8 0

3 years ago

This is a measure of force per unit area.

Vlad1618 [11]

<h2>Answer: Pressure</h2>

<u>Pressure</u> is the force $F$ exerted by a gas, a liquid or a solid on a surface (or area) $A$ .

Its unit according to the International System of Units is Pascal $Pa$ which is equal to $N/m^{2}$ and its formula is:

$P=\frac{F}{A}$

5 0

3 years ago

A block of mass 0.1 kg is attached to a spring of spring constant 21 N/m on a frictionless track. The block moves in simple harm

bogdanovich [222]

Answer:

A) 2.75 m/s B) 0.1911 m C) 0.109 s

Explanation:

mass of block = M =0.1 kg

spring constant = k = 21 N/m

amplitude = A = 0.19 m

mass of bullet = m = 1.45 g = 0.00145 kg

velocity of bullet = vᵇ = 68 m/s

as we know:

Angular frequency of S.H.M = ω₀ = $\sqrt\frac{k}{M}$

= $\sqrt\frac{21}{0.1}$

= 14.49 rad/sec

<h3>A) Speed of the block immediately before the collision:</h3>

displacement of Simple Harmonic Motion is given as:

$x = A sin (\omega t + \phi)\\$

Differentiating this to find speed of the block immediately before the collision:

$v=\frac{dx}{dt}= A\omega_{o} cos (\omega_{o}t =\phi}\\$

As bullet strikes at equilibrium position so,

φ = 0

t= 2nπ

⇒ cos (ω₀t + φ) = 1

⇒ $v= A\omega_{o}$

$v=(.19)(14.49)\\v= 2.75 ms^{-1}$

<h3>B) If the simple harmonic motion after the collision is described by x = B sin(ωt + φ), new amplitude B:</h3>

S.H.M after collision is given as :

$x= Bsin(\omega t + \phi)$

To find B, consider law of conservation of energy

$K.E = P.E\\K.E= \frac{1}{2}(m+M)v^{2} \\P.E = \frac{1}{2} kB^{2}$

$\frac{m+M}{k} v^{2} = B^{2} \\B =\sqrt\frac{m+M}{k} v\\B = \sqrt\frac{.00145+0.1}{21} (2.75)\\B = .1911m$

<h3>C) Time taken by the block to reach maximum amplitude after the collision:</h3>

Time period S.H.M is given as:

$T=2\pi \sqrt\frac{m}{k}\\ for given case\\m= m=M\\then\\T=2\pi \sqrt\frac{m+M}{k}$

Collision occurred at equilibrium position so time taken by block to reach maximum amplitude is equal to one fourth of total time period

$T=\frac{\pi }{2}\sqrt\frac{m+M}{k} \\T=0.109 sec$

5 0

4 years ago