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Crazy boy [7]
4 years ago
15

A student writes the following statements about the relationship between mass defect and nuclear binding energy.

Physics
2 answers:
zaharov [31]4 years ago
8 0

Answer:

C

Explanation:

Option C is correct, because statement 1 and 2 are both correct, while statement 3 is incorrect. Statement 3 should say: <em>Statement 1 describes the isotope's </em><em>mass defect</em><em>, and statement 2 describes the isotope's </em><em>binding energy</em><em>.</em>

Both statement 1 and 2 are correct, because it is only logical that (in statement 1) energy is needed to form a nuclide, and the only way for that energy to be present is if some mass is converted into energy, and (in statement 2) it is only logical that the same amount of energy needed to form the bonds, would be needed to break the bonds.

monitta4 years ago
4 0
C.

Statement 2 describes the binding energy and statement 1 describes the mass defect. 
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In the diagram, a force of 20 newtons is applied to a block. The block is in dynamic equilibrium. What is the magnitude and dire
sladkih [1.3K]

Answer:B 20 newtons opposite to the direction of the applied force

Explanation:

5 0
3 years ago
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4. A car, initially traveling east with a speed of
PilotLPTM [1.2K]

Answer:

150 m

Explanation:

Given,

u=5m/s

a=2m/s2

t=10s

v=?

s=?

Now,

v=u+at

=5+2×10

=5+20

=25m/s

So,

s=u+v/2×t

=5+25/2×10

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5 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
W=90 kg×4n/kg<br>Please halp me i need it ​
kiruha [24]

I'm guessing that this is a problem to find the weight of a 90kg mass on a planet where the acceleration of gravity is 4 m/s^2. (Much less gravity than Earth, a little more than Mars.)

Just do the multiplication, and you get

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5 0
4 years ago
What kind of energy do most electric inventions give off?
krek1111 [17]

Answer:

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Explanation:

4 0
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