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3 years ago

About light and energy

1 answer:

7 0

E = MC^2. Albert Einstein's proven formula. When mass travels at the square of speed of light, the mass gets converted into energy

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Planet z has a rocky surface. at height 3240 km above the surface, the gravitational acceleration is what is the radius of the p

jolli1 [7]

Https://www.google.com/search?tbs=sbi:AMhZZiuc6mk5S438WBb7iizj56MvoVz9sxfbOaVcy_1SdDTQnJdcyo0Qu7CKsc...

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4 years ago

A string is wrapped several times around the rim of a small hoop with radius 8.00 cm and mass 0.180 kg. The free end of the stri

Dmitrij [34]

Answer:

(a). The angular speed of the rotating is 33.8 rad/s.

(b). The speed of its center is 2.7 m/s.

Explanation:

Given that,

Radius = 8.00 cm

Mass = 0.180 kg

Height = 75.0 m

We need to calculate the angular speed of the rotating

Using conservation of energy

$\dfrac{1}{2}I\omega_{1}^2+\dfrac{1}{2}mv_{1}^{2}+mgh_{1}=\dfrac{1}{2}I\omega_{2}^2+\dfrac{1}{2}mv_{2}^{2}+mgh_{2}$

Here, initial velocity and angular velocity are equal to zero.

$mgh_{1}=\dfrac{1}{2}I\omega_{2}^2+\dfrac{1}{2}mv_{2}^{2}+mgh_{2}$

$mg(h_{1}-h_{2})=\dfrac{1}{2}I\omega_{2}^2+\dfrac{1}{2}mv_{2}^{2}$

$mgH=\dfrac{1}{2}mr^2\omega_{2}^2+\dfrac{1}{2}m(r\omega_{2})^2$

Here, $H = h_{1}-h_{2}$

$gH=r^2\omega_{2}^2$

$\omega_{2}^2=\dfrac{gH}{r^2}$

$\omega_{2}=\sqrt{\dfrac{9.8\times75.0\times10^{-2}}{(8.00\times10^{-2})^2}}$

$\omega_{2}=33.8\ rad/s$

The angular speed of the rotating is 33.8 rad/s.

(b). We need to calculate the speed of its center

Using formula of speed

$v=r\omega$

Put the value into the formula

$v=8.00\times10^{-2}\times33.8$

$v=2.7\ m/s$

Hence, (a). The angular speed of the rotating is 33.8 rad/s.

(b). The speed of its center is 2.7 m/s.

3 0

3 years ago

Read 2 more answers

How do you think the formation of new stars is related to supernovas and planetary nebulae?

kogti [31]

By God wanting them to be the same kinda

5 0

4 years ago

What are two disadvantages of GMO food?

Lubov Fominskaja [6]

A and C because GMO's can give people a large rash

6 0

4 years ago

Heater supplies 80 J of energy to a block of metal. The temperature of the block rises by 20 °C.

Grace [21]

Answer:

Its internal energy decreases by 40 J.

Explanation:

The energy supplied to the metal block causes the temperature of the block to rise. Therefore,

$Heat\ Supplied = mC\Delta T\\80\ J = mC(20^oC)\\mC = 4\ J/^oC\\$

Now, we will use this value of mC to find out the energy change of block:

$Energy\ Change = mC\Delta T\\Energy\ Change = (4\ J/^oC)(-10^oC)\\Energy\ Change = -40J\\$

It shows that when the temperature of the metal block falls by 10°C,

<u>Its internal energy decreases by 40 J.</u>

5 0

3 years ago