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svetlana [45]
3 years ago
12

Why is the sign of the emf of the second peak opposite to the sign of the first peak?

Physics
1 answer:
eimsori [14]3 years ago
7 0
Because AC emf is a sine wave
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A flea jumps straight up to a maximum height of 0.400 m . what is its initial velocity v0 as it leaves the ground?
timama [110]

For an object`s motion, the Kinematic equation is,

v^2=v_{0}^2+2ah

Here, v is the final velocity and h is stands for the height of the object and a is the acceleration of the object.

As according to question,

v=0m/s,a=g-9.8 m/s^2 and h = 0.400 m

Thus, putting these values in above equation, we get

0= v_{0}^2 -2gh

or

v_{0} =\sqrt{2 \times 9.8 \times 0.400 }

v_{0} = 2.8 m/s

Therefore, initial velocity is 2.8 m/s



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Read 2 more answers
1. Calculate the total binding energy of 12
ahrayia [7]

Answer:

1. B = 79.12 MeV

2. B = -4.39 MeV/nucleon

3. B = 2.40 MeV/nucleon

4. B = 7.48 MeV/nucleon

5. B = -18.72 MeV

6. B = 225.23 MeV            

Explanation:

The binding energy can be calculated using the followng equation:

B = (Zm_{p} + Nm_{n} - M)*931 MeV/C^{2}

<u>Where</u>:

Z: is the number of protons

m_{p}: is the proton's mass = 1.00730 u

N: is the number of neutrons

m_{n}: is the neutron's mass = 1.00869 u

M: is the mass of the nucleus

1. The total binding energy of ^{12}_{6}C is:

B = (Zm_{p} + Nm_{n} - M)*931.49 MeV/u

B = (6*1.00730 + 6*1.00869 - 12.011)*931.49 MeV/u = 79.12 MeV

 

2. The average binding energy per nucleon of ^{24}_{12}Mg is:

B = \frac{(Zm_{p} + Nm_{n} - M)}{A}*931.49 MeV/u

<u>Where</u>: A = Z + N

B = \frac{(12*1.00730 + 12*1.00869 - 24.305)}{(12 + 12)}*931.49 MeV/u = -4.39 MeV/nucleon                                  

   

3. The average binding energy per nucleon of ^{85}_{37}Rb is:

B = \frac{(Zm_{p} + Nm_{n} - M)}{A}*931.49 MeV/u

B = \frac{(37*1.00730 + 48*1.00869 - 85.468)}{85}*931.49 MeV/u = 2.40 MeV/nucleon

     

4. The binding energy per nucleon of ^{238}_{92}U is:

B = \frac{(92*1.00730 + 146*1.00869 - 238.03)}{238}*931.49 MeV/u = 7.48 MeV/nucleon

 

5. The total binding energy of ^{20}_{10}Ne is:

B = (Zm_{p} + Nm_{n} - M)*931.49 MeV/u

B = (10*1.00730 + 10*1.00869 - 20.180)*931.49 MeV/u = -18.72 MeV

6. The total binding energy of ^{40}_{20}Ca is:

B = (Zm_{p} + Nm_{n} - M)*931.49 MeV/u

B = (20*1.00730 + 20*1.00869 - 40.078)*931.49 MeV/u = 225.23 MeV

I hope it helps you!

8 0
3 years ago
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