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4 years ago

Will make brainliest!

Physics

2 answers:

guajiro [1.7K]4 years ago

7 0

Answer:

Reflection

Explanation:

The process whereby light bounces back from an object at the same angle at which it encounters a surface and with the same intensity its called: Reflection.

Lena [83]4 years ago

3 0

Reflection is when light bounces off an object. If the surface is smooth and shiny, like glass, water or polished metal, the light will reflect at the same angle as it hit the surface. This is called specular reflection. Light reflects from a smooth surface at the same angle as it hits the surface

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A heavy flywheel is accelerated (rotationally) by a motor that provides constant torque and therefore a constant angular acceler

ki77a [65]

Answer:

a) $t_1=\frac{w_1-w_o}{\alpha}=\frac{w_1}{\alpha}sec$

b) $\theta_1=\frac{w_1^2}{2\alpha}rad$

c) $t_2=\frac{\alpha t_1}{5\alpha}=\frac{t_1}{5}sec$

Explanation:

1) Basic concepts

Angular displacement is defined as the angle changed by an object. The units are rad/s.

Angular velocity is defined as the rate of change of angular displacement respect to the change of time, given by this formula:

$w=\frac{\Delat \theta}{\Delta t}$

Angular acceleration is the rate of change of the angular velocity respect to the time

$\alpha=\frac{dw}{dt}$

2) Part a

We can define some notation

$w_o=0\frac{rad}{s}$ ,represent the initial angular velocity of the wheel

$w_1=?\frac{rad}{s}$ , represent the final angular velocity of the wheel

$\alpha$ , represent the angular acceleration of the flywheel

$t_1$ time taken in order to reach the final angular velocity

So we can apply this formula from kinematics:

$w_1=w_o +\alpha t_1$

And solving for t1 we got:

$t_1=\frac{w_1-w_o}{\alpha}=\frac{w_1}{\alpha}sec$

3) Part b

We can use other formula from kinematics in order to find the angular displacement, on this case the following:

$\Delta \theta=wt+\frac{1}{2}\alpha t^2$

Replacing the values for our case we got:

$\Delta \theta=w_o t+\frac{1}{2}\alpha t_1^2$

And we can replace $t_1$ from the result for part a, like this:

$\theta_1-\theta_o=w_o t+\frac{1}{2}\alpha (\frac{w_1}{\alpha})^2$

Since $\theta_o=0$ and $w_o=0$ then we have:

$\theta_1=\frac{1}{2}\alpha \frac{w_1^2}{\alpha^2}$

And simplifying:

$\theta_1=\frac{w_1^2}{2\alpha}rad$

4) Part c

For this case we can assume that the angular acceleration in order to stop applied on the wheel is $\alpha_1 =-5\alpha \frac{rad}{s}$

We have an initial angular velocity $w_1$ , and since at the end stops we have that $w_2 =0$

Assuming that $t_2$ represent the time in order to stop the wheel, we cna use the following formula

$w_2 =w_1 +\alpha_1 t_2$

Since $w_2=0$ if we solve for $t_2$ we got

$t_2=\frac{0-w_1}{\alpha_1}=\frac{-w_1}{-5\alpha}$

And from part a) we can see that $w_1=\alpha t_1$ , and replacing into the last equation we got:

$t_2=\frac{\alpha t_1}{5\alpha}=\frac{t_1}{5}sec$

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3 years ago

what is measurement? why is it important to have regular supervision of the weights and measurement in the market?

saveliy_v [14]

Answer:

The comparison of unknown quantity with known quantity is called measurement.

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4 years ago

_____ can be thrown, caught, and used to provide resistance for a variety of movements, in a variety of planes of motion, and at

FromTheMoon [43]

Answer:

b. Medicine balls

Explanation:

A medicine ball (also called an exercise ball, or a health ball) is a weighted ball about the shoulder diameter (approx. 13.7 inches), often used for recovery and strength training.... In 1705 similar large balls had been used in Persia.So you can use your medicine ball as a projectile to boost the strength of your throws.

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3 years ago

You and a friend are looking at the stars and your friend says, "Stars must be shrinking because gravity is constantly pulling t

Lemur [1.5K]

Technically? It's not wrong. A star is a ball of gas that is held together by it's own gravity of which it constantly struggles against. Gravity constantly works to try and make the star collapse. However, stars are extremely hot at their cores which help the star push against it's own gravity creating a sort of equilibrium. So long as the star continues to push against gravity, it will not collapse, however eventually a star will lose its energy and eventually collapse into a black hole.

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3 years ago

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State the relative position for the sun and earth during the lunar eclipse (ASAP)

Mademuasel [1]

Answer:

Solar eclipses result from the Moon blocking the Sun relative to the Earth; thus Earth, Moon and Sun all lie on a line. Lunar eclipses work the same way in a different order: Moon, Earth and Sun all on a line. In this case the Earth's shadow hides the Moon from view.

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3 years ago