All categories

3 years ago

Which of the following types of energy is not associated with a car engine?

Physics

1 answer:

soldi70 [24.7K]3 years ago

4 0

Answer:

Explanation:

Does not assosicate with Light

You might be interested in

Can anybody write a short poem about friction

Luden [163]

Answer:

you will be the clouds
and I will be the sky.
you will be the ocean
and I will be the shore.
you will be the trees
and I will be the wind.

whatever we are, you and I will always collide.

There you go! Let me know if it helped.
:)

8 0

2 years ago

skater spins over a point at a speed of 3.0 rotations per second then the momentum of inertia is 0.60 kg.M2, what is its angular

laiz [17]

Answer:

$L=11.3\ kg-m^2/s$

Explanation:

Given that,

Angular speed of a skater, $\omega=3\ rot/s=18.84\ rad/s$

The moment of inertia of the skater, I = 0.6 kg-m²

We need to find the angular momentum of the skater. The formula for the angular momentum of the skater is given by :

$L=I\omega$

Substitute all the values,

$L=0.6\times 18.84\\\\L=11.3\ kg-m^2/s$

So, its angular momentum is equal to $11.3\ kg-m^2/s$ .

8 0

3 years ago

HELP!!! Why is plasma considered the most common state of matter in the universe?

olasank [31]

Clue : There is plasma in sun

4 0

3 years ago

Read 2 more answers

A record player turntable initially rotating at 3313 rev/min is braked to a stop at a constant rotational acceleration. The turn

Rus_ich [418]

Answer:

(A) It will take 22 sec to come in rest

(b) Work done for coming in rest will be 0.2131 J

Explanation:

We have given the player turntable initially rotating at speed of $33\frac{1}{3}rpm=33.333rpm=\frac{2\times 3.14\times 33.333}{60}=3.49rad/sec$

Now speed is reduced by 75 %

So final speed $\frac{3.49\times 75}{100}=2.6175rad/sec$

Time t = 5.5 sec

From first equation of motion we know that '

$\alpha =\frac{\omega -\omega _0}{t}=\frac{2.6175-3.49}{4}=-0.158rad/sec^2$

(a) Now final velocity $\omega =0rad/sec$

So time t to come in rest $t=\frac{0-3.49}{-0.158}=22sec$

(b) The work done in coming rest is given by

$\frac{1}{2}I\left ( \omega ^2-\omega _0^2 \right )=\frac{1}{2}\times 0.035\times (0^2-3.49^2)=0.2131J$

4 0

3 years ago

Two electrodes, separated by a distance d, in a vacuum are maintained at a constant potential difference. An electron, accelerat

Alja [10]

Answer:

Explanation:

Given that, the distance between the electrode is d.

The electron kinetic energy is Ek when the electrode are at distance "d" apart.

So, we want to find the K.E when that are at d/3 distance apart.

K.E = ½mv²

Note: the mass doesn't change, it is only the velocity that change.

Also,

K.E = Work done by the electron

K.E = F × d

K.E = W = ma × d

Let assume that if is constant acceleration

Then, m and a is constant,

Then,

K.E is directly proportional to d

So, as d increase K.E increase and as d decreases K.E decreases.

So,

K.E_1 / d_1 = K.E_2 / d_2

K.E_1 = E_k

d_1 = d

d_2 = d/3

K.E_2 = K.E_1 / d_1 × d_2

K.E_2 = E_k × ⅓d / d

Then,

K.E_2 = ⅓E_k

So, the new kinetic energy is one third of the E_k

7 0

3 years ago