An object stays at rest until what happens to it?

How is item A different from Item B?

iris [78.8K]

Explanation:

well there is nothing there and it could be different by diffrent objects, idk

8 0

3 years ago

Read 2 more answers

What are the seven different forms of energy? Give an example of each.

wel

Answer:

Form of energy: Example

1. Light energy: Electromagnetic radiation

2. Nuclear energy: Nuclear fission

3. Chemical energy: Energy stored in plant matter

4. Electrical energy: Lightning

5. Thermal energy: A hot surface

6. Sound energy: A tuning fork

7. Solar energy: Energy from the Sun

8. Mechanical energy: A moving vehicle

5 0

3 years ago

One of the waste products of a nuclear reactor is plutonium-239 . This nucleus is radioactive and decays by splitting into a hel

Gekata [30.6K]

Answer:

a) $v_{U-235} = 2.68 \cdot 10^{5} m/s$

$v_{He-4} = -1.57 \cdot 10^{7} m/s$

b) $E_{He-4} = 8.23 \cdot 10^{-13} J$

$E_{U-235} = 1.41 \cdot 10^{-14} J$

Explanation:

Searching the missed information we have:

E: is the energy emitted in the plutonium decay = 8.40x10⁻¹³ J

m(⁴He): is the mass of the helium nucleus = 6.68x10⁻²⁷ kg

m(²³⁵U): is the mass of the helium U-235 nucleus = 3.92x10⁻²⁵ kg

a) We can find the velocities of the two nuclei by conservation of linear momentum and kinetic energy:

Linear momentum:

$p_{i} = p_{f}$

$m_{Pu-239}v_{Pu-239} = m_{He-4}v_{He-4} + m_{U-235}v_{U-235}$

Since the plutonium nucleus is originally at rest, $v_{Pu-239} = 0$ :

$0 = m_{He-4}v_{He-4} + m_{U-235}v_{U-235}$

$v_{He-4} = -\frac{m_{U-235}v_{U-235}}{m_{He-4}}$ (1)

Kinetic Energy:

$E_{Pu-239} = \frac{1}{2}m_{He-4}v_{He-4}^{2} + \frac{1}{2}m_{U-235}v_{U-235}^{2}$

$2*8.40 \cdot 10^{-13} J = m_{He-4}v_{He-4}^{2} + m_{U-235}v_{U-235}^{2}$

$1.68\cdot 10^{-12} J = m_{He-4}v_{He-4}^{2} + m_{U-235}v_{U-235}^{2}$ (2)

By entering equation (1) into (2) we have:

$1.68\cdot 10^{-12} J = m_{He-4}(-\frac{m_{U-235}v_{U-235}}{m_{He-4}})^{2} + m_{U-235}v_{U-235}^{2}$

$1.68\cdot 10^{-12} J = 6.68 \cdot 10^{-27} kg*(-\frac{3.92 \cdot 10^{-25} kg*v_{U-235}}{6.68 \cdot 10^{-27} kg})^{2} +3.92 \cdot 10^{-25} kg*v_{U-235}^{2}$

Solving the above equation for $v_{U-235}$ we have:

$v_{U-235} = 2.68 \cdot 10^{5} m/s$

And by entering that value into equation (1):

$v_{He-4} = -\frac{3.92 \cdot 10^{-25} kg*2.68 \cdot 10^{5} m/s}{6.68 \cdot 10^{-27} kg} = -1.57 \cdot 10^{7} m/s$

The minus sign means that the helium-4 nucleus is moving in the opposite direction to the uranium-235 nucleus.

b) Now, the kinetic energy of each nucleus is:

For He-4:

$E_{He-4} = \frac{1}{2}m_{He-4}v_{He-4}^{2} = \frac{1}{2} 6.68 \cdot 10^{-27} kg*(-1.57 \cdot 10^{7} m/s)^{2} = 8.23 \cdot 10^{-13} J$

For U-235:

$E_{U-235} = \frac{1}{2}m_{U-235}v_{U-235}^{2} = \frac{1}{2} 3.92 \cdot 10^{-25} kg*(2.68 \cdot 10^{5} m/s)^{2} = 1.41 \cdot 10^{-14} J$

I hope it helps you!

3 0

3 years ago

A piece of steel expands 10 cm when ur is heated from 20 to 50 degrees Celsius. How much would it expand if it was heated from 2

faltersainse [42]

The steel would expand by 4. 8 * 10^-3 cm

<h3>How to determine the linear expansion</h3>

The change in length ΔL is proportional to length L. It is dependent on the temperature, substance, and length.

Using the formula:

ΔL= α LΔT

where ΔL is the change in length L = 10cm

ΔT is the change in temperature = 60° - 20° = 40° C

α is the coefficient of linear expansion = 1.2 x 10^-5 °C

Substitute into the formula

ΔL = $1.2 * 10^-5 * 10 * 40$

ΔL = $4.8 * 10 ^-3$ cm

Therefore, the steel would expand by 4. 8 * 10^-3 cm

Learn more about linear expansion here:

brainly.com/question/14325928

#SPJ1

8 0

1 year ago

Name one situation in which you might notice the reflection of a wave

Lady bird [3.3K]

<u>Answer:</u>

A perfect example of wave reflection is an <u>echo</u>.

<u>Explanation:</u>

A wave reflection takes place when waves cannot pass through a surface and in turn they bounce back. It is not necessary that wave reflections can only happen with sound waves, they can also take place in light waves. Also, the waves which are reflected have the same frequency as the original wave, but their direction is different. When a wave strikes an object in the same angle, they bounce back straight but when they hit an object with different angle, their direction changes.

4 0

3 years ago