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Aleks04 [339]
4 years ago
14

Which of the following are energy solutions that release pollution into the air? coal petroleum fuel cells wind energy oil nucle

ar energy tidal energy
Physics
1 answer:
kotykmax [81]4 years ago
7 0
What i think is that some things that release pollution is nuclear energy
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The components of a 10.8-meters-per-second velocity at an angle of 34.° above the horizontal are
Kisachek [45]

Answer:

6.0 m/s vertical and 9.0 m/s horizontal

Explanation:

For the vertical component, we use the formula:

  • Sin(34°) = <em>y</em> / 10.8

Then we <u>solve for </u><u><em>y</em></u>:

  • 0.559 = <em>y</em> / 10.8
  • <em>y </em>= 6.0

And for the horizontal component, we use the formula:

  • Cos(34°) = <em>x</em> / 10.8

Then we <u>solve for </u><u><em>x</em></u><u>:</u>

  • 0.829 = <em>x</em> / 10.8
  • <em>y </em>= 9.0

So the answer is " 6.0 m/s vertical and 9.0 m/s horizontal".

3 0
3 years ago
A child has a toy car on a horizontal platform. The car starts from rest and reaches a maximum speed in 4 s. If the mass of the
dmitriy555 [2]

Answer:

a=4m/s²

Explanation:

F=ma

0.4=0.1a

7 0
3 years ago
Read 2 more answers
In any electromagnetic wave,
puteri [66]

A). Both the energy and the wave travel in the same direction.

If they didn't, they'd wind up in different cities almost instantly.

6 0
3 years ago
Calculate the speed for a car that went a distance of 125 meters in 2 seconds time.
stiv31 [10]

Answer:

<h2>62.5 m/s</h2>

Explanation:

The speed of the car can be found by using the formula

s =  \frac{d}{t} \\

d is the distance

t is the time

From the question we have

s =  \frac{125}{2}  = 62.5 \\

We have the final answer as

<h3>62.5 m/s</h3>

Hope this helps you

5 0
2 years ago
When the play button is pressed, a CD accelerates uniformly from rest to 450 rev/min in 3.0 revolutions. If the CD has a radius
Marina CMI [18]

To solve this problem it is necessary to apply the kinematic equations of angular motion.

Torque from the rotational movement is defined as

\tau = I\alpha

where

I = Moment of inertia \rightarrow \frac{1}{2}mr^2 For a disk

\alpha = Angular acceleration

The angular acceleration at the same time can be defined as function of angular velocity and angular displacement (Without considering time) through the expression:

2 \alpha \theta = \omega_f^2-\omega_i^2

Where

\omega_{f,i} = Final and Initial Angular velocity

\alpha = Angular acceleration

\theta = Angular displacement

Our values are given as

\omega_i = 0 rad/s

\omega_f = 450rev/min (\frac{1min}{60s})(\frac{2\pi rad}{1rev})

\omega_f = 47.12rad/s

\theta = 3 rev (\frac{2\pi rad}{1rev}) \rightarrow 6\pi rad

r = 7cm = 7*10^{-2}m

m = 17g = 17*10^{-3}kg

Using the expression of angular acceleration we can find the to then find the torque, that is,

2\alpha\theta=\omega_f^2-\omega_i^2

\alpha=\frac{\omega_f^2-\omega_i^2}{2\theta}

\alpha = \frac{47.12^2-0^2}{2*6\pi}

\alpha = 58.89rad/s^2

With the expression of the acceleration found it is now necessary to replace it on the torque equation and the respective moment of inertia for the disk, so

\tau = I\alpha

\tau = (\frac{1}{2}mr^2)\alpha

\tau = (\frac{1}{2}(17*10^{-3})(7*10^{-2})^2)(58.89)

\tau = 0.00245N\cdot m \approx 2.45*10^{-3}N\cdot m

Therefore the torque exerted on it is 2.45*10^{-3}N\cdot m

3 0
3 years ago
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