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skad [1K]
3 years ago
7

Help please !!!! LIKE ASAP!

Physics
1 answer:
serg [7]3 years ago
3 0
The first diagram is Nuclear Fission which is the splitting of a heavy, unstable nucleus into 2 lighter nuclei.

The second diagram is Nuclear Fusion which is the process where 2 light nuclei combine together releasing a lot of energy.
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How much heat must be added to make a 5g substance with a specific heat of 2 J/gC that has its temperature go up 10 degrees? Q =
zlopas [31]

Answer:

100 Joule

Explanation:

Amount of heat in agiven body is given by Q = m•C•ΔT

where m is the mass of the body

c is the specific heat capacity of body. It is the amount of heat stored in 1 unit weight of body which raises raises the temperature of body by 1 unit of temperature.

ΔT is the change in the temperature of body

___________________________________________

coming back to problem

m = 5g

C = 2J/gC

since, it is given that temperature of body increases by 10 degrees, thus

ΔT = 10 degrees

Using the formula for heat as given

Q = m•C•ΔT

Q = 5* 2 * 10  Joule= 100 Joule

Thus, 100 joule heat must be added to  a  5g substance with a specific heat of 2 J/gC to raise its temperature go up by 10 degrees.

8 0
3 years ago
Alien A lifts a 500-newton child from the floor to a height of 0.40 meters in 2 seconds
vivado [14]
Strong alien you got there good luck bud you never asked a question
4 0
3 years ago
The maximum Compton shift in wavelength occurs when a photon isscattered through 180^\circ .
vlabodo [156]

Answer: 90\°

Explanation:

The Compton Shift \Delta \lambda in wavelength when the photons are scattered is given by the following equation:

\Delta \lambda=\lambda_{c}(1-cos\theta)     (1)

Where:

\lambda_{c}=2.43(10)^{-12} m is a constant whose value is given by \frac{h}{m_{e}c}, being h the Planck constant, m_{e} the mass of the electron and c the speed of light in vacuum.

\theta) the angle between incident phhoton and the scatered photon.

We are told the maximum Compton shift in wavelength occurs when a photon isscattered through 180\°:

\Delta \lambda_{max}=\lambda_{c}(1-cos(180\°))     (2)

\Delta \lambda_{max}=\lambda_{c}(1-(-1))    

\Delta \lambda_{max}=2\lambda_{c}     (3)

Now, let's find the angle that will produce a fourth of this maximum value found in (3):

\frac{1}{4}\Delta \lambda_{max}=\frac{1}{4}2\lambda_{c}(1-cos\theta)      (4)

\frac{1}{4}\Delta \lambda_{max}=\frac{1}{2}\lambda_{c}(1-cos\theta)      (5)

If we want \frac{1}{4}\Delta \lambda_{max}=\frac{1}{2}\lambda_{c}, 1-cos\theta   must be equal to 1:

1-cos\theta=1   (6)

Finding \theta:

1-1=cos\theta

0=cos\theta  

\theta=cos^{-1} (0)  

Finally:

\theta=90\°    This is the scattering angle that will produce \frac{1}{4}\Delta \lambda_{max}      

7 0
3 years ago
30 points plz help ill do anything... literally anything.
ruslelena [56]

Answer:

1. 2.5s

Explanation:

1. For time, divide Distance / speed

25m / 10

=2.5s

3 0
3 years ago
You measure the velocity of a drag racer that accelerates with constant acceleration. You want to plot the data and determine th
Ivenika [448]

Answer:

a) Linear equation

Explanation:

Definition of acceleration

a=\frac{dv}{dt}\\

if a=constant and we integrate the last equation

v(t)=v_{o}+a*t

So the relation between the time and the velocity is linear. If we plot the velocity in function of time, the plot is a line, and the acceleration is the slope of this line.

8 0
2 years ago
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