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bagirrra123 [75]

2 years ago

13

What is an example of Deceleration?

2 answers:

Elina [12.6K]2 years ago

5 0

That's a word often used to mean "slowing down". An example is what happens
when the brake is applied in a moving car.

But 'deceleration' is a very unofficial, unscientific word. If the forward speed
is decreasing, then a much more useful concept is "negative acceleration".

Luden [163]2 years ago

4 0

When the body starts moving with a slower speed than the initial speed and it slows down continuously, the this is called deceleration. Examples of deceleration -

When a ball is thrown upwards, its speed goes on decreasing as its height increases.
When a car applies brakes.

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Explain why streets and highways have speed limits and not velocity limits.

alexdok [17]

We know that speed is a scalar unit while velocity is a vector. The signs are mostly concerned with the magnitude that the car should travel in, the direction is not as important.

Hope I helped :)

4 0

2 years ago

A wheel rotates with a constant angular acceleration of  rad/s2. During a certain time interval its angular displacement is  r

Hatshy [7]

Answer:

The angular velocity at the beginning of the interval is $\pi\sqrt{2}\ rad/s$ .

Explanation:

Given that,

Angular acceleration $\alpha=\pi\ rad/s^2$

Angular displacement $\theta=\pi\ rad$

Angular velocity $\omega =2\pi\ rad/s$

We need to calculate the angular velocity at the beginning

Using formula of angular velocity

$\alpha =\dfrac{\omega^2-\omega_{0}^2}{2\theta}$

$\omega_{0}^2=\omega^2-2\alpha\theta$

Where, $\alpha$ = angular acceleration

$\omega$ = angular velocity

Put the value into the formula

$\omega_{0}^2=(2\pi)^2-2\times\pi\times\pi$

$\omega=\sqrt{2\pi^2}$

$\omega_{0}=\pi\sqrt{2}\ rad/s$

Hence, The angular velocity at the beginning of the interval is $\pi\sqrt{2}\ rad/s$ .

3 0

2 years ago

If you could help me out that would be fantastic.?!!!

Volgvan

It's either C. Or B.

3 0

2 years ago

A particle (charge = +0.8 mC) moving in a region where only electric forces act on it has a kinetic energy of 6.7 J at point A.

Maksim231197 [3]

Answer:

The kinetic energy of the particle as it moves through point B is 7.9 J.

Explanation:

The kinetic energy of the particle is:

$\Delta K = \Delta E_{p} = q\Delta V$

<u>Where</u>:

K: is the kinetic energy

$E_{p}$ : is the potential energy

q: is the particle's charge = 0.8 mC

ΔV: is the electric potential = 1.5 kV

$\Delta K = q \Delta V= 0.8 \cdot 10^{-3} C*1.5 \cdot 10^{3} V = 1.2 J$

Now, the kinetic energy of the particle as it moves through point B is:

$\Delta K = K_{f} - K_{i}$

$K_{f} = \Delta K + K_{i} = 1.2 J + 6.7 J = 7.9 J$

Therefore, the kinetic energy of the particle as it moves through point B is 7.9 J.

I hope it helps you!

8 0

2 years ago

A machine has an efficiency of 70%. How much work dose the machine do when 20,000 J of work is done on it ?

Murrr4er [49]

Answer: The machine does 14.000 J

3 0

2 years ago