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2 years ago

9

A model shows a machine that works using electric fields. What would this machine need for the electrical field to function prop

erly

1 answer:

Wittaler [7]2 years ago

7 0

It would need at least two charges interacting parts

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Explain about kinetic theory

Luden [163]

Answer:

The model, called the kinetic theory of gases, assumes that the molecules are very small relative to the distance between molecules. ... The molecules are in constant random motion, and there is an energy (mass x square of the velocity) associated with that motion. The higher the temperature, the greater the motion.

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3 years ago

An object that is falling has the following type(s) of energy. Ignore air resistance.

Anton [14]

Potential and kinetic

6 0

3 years ago

Read 2 more answers

I NEED THIS ASAP!!! A ball is thrown straight up with an initial velocity of 4.40 m/s. Assuming there is no air friction, what i

Lilit [14]

Answer:

I think it is 80m/s

Explanation:

d = ½ g t2

= ½ (10 m/s2) (4 s)2

= (5 m/s2) (16 s2)

= 80 ms

75% sure

Hope this helps!!!

5 0

3 years ago

A high-speed flywheel in a motor is spinning at 450 rpm when a power failure suddenly occurs. The flywheel has mass 40.0 kg and

alexira [117]

Answer:

A) $\omega_f=17.503\ rad.s^{-1}$

B) $t=55.6822\ s$

C) $\theta=1312\ rad$

Explanation:

Given:

mass of flywheel, $m=40\ kg$

diameter of flywheel, $d=0.72\ m$

rotational speed of flywheel, $N_i=450\ rpm \Rightarrow \omega_i=\frac{450\times 2\pi}{60} =15\pi\ rad.s^{-1}$

duration for which the power is off, $t_0=35\ s$

no. of revolutions made during the power is off, $\theta=180\times 2\pi=360\pi\ rad$

<u>Using equation of motion:</u>

$\theta=\omega_i.t+\frac{1}{2} \alpha.t^2$

$360\pi=15\pi\times 35+\frac{1}{2} \times \alpha\times35^2$

$\alpha=-0.8463\ rad.s^{-2}$

Negative sign denotes deceleration.

A)

Now using the equation:

$\omega_f=\omega_i+\alpha.t$

$\omega_f=15\pi-0.8463\times 35$

$\omega_f=17.503\ rad.s^{-1}$ is the angular velocity of the flywheel when the power comes back.

B)

Here:

$\omega_f=0\ rad.s^{-1}$

Now using the equation:

$\omega_f=\omega_i+\alpha.t$

$0=15\pi-0.8463\times t$

$t=55.6822\ s$ is the time after which the flywheel stops.

C)

Using the equation of motion:

$\theta=\omega_i.t+\frac{1}{2} \alpha.t^2$

$\theta=15\pi\times 55.68225-0.5\times 0.8463\times 55.68225^2$

$\theta=1312\ rad$ revolutions are made before stopping.

3 0

3 years ago

a ball is tossed straight up in the air to a maximum height of 25 meters before returning to the point that it was thrown, where

boyakko [2]

Answer:

height=25

velocity=8

force of gravity=10

Answer =2000

5 0

2 years ago