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kari74 [83]
3 years ago
10

Help me with question 13 please (use the graph to answer)

Physics
1 answer:
Sladkaya [172]3 years ago
5 0

Explanation:

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Can I have some help please with any of the questions
vovangra [49]

6b: impulse is change in momentum. Change in momentum p=m[v(final)-v(initial). Final velocity is zero and initial velocity is the one you calculated before impact: -15.7 since it’s going down. Now plug in numbers and you get 78.5 in the upward direction.

6c: change in momentum p=Ft. we already calculated change in momentum. So plug it into equation and solve for t. 78.5/655= 0.119 s

Apply the same idea for question 7. Hope this helped

3 0
3 years ago
A 5.0-cm by 7.0-cm rectangular coil has 100 turns. Its axis makes an angle of 65° with a uniform magnetic field of 0.55 T. What
kotegsom [21]

Answer:

0.0814 Wb

Explanation:

L = length of the rectangular coil = 5 cm = 0.05 m

w = width of rectangular coil = 7 cm = 0.07 m

A = Area of the coil = L w = 0.05 x 0.07 = 35 x 10⁻⁴ m²

B = magnitude of uniform magnetic field = 0.55 T

θ = angle of axis of coil with magnetic field = 65°

Magnetic flux through the coil is given as

Φ = N B A Cosθ

Φ = (100) (0.55) (35 x 10⁻⁴ ) Cos65

Φ = 0.0814 Wb

4 0
4 years ago
PLEASE HELP! I'LL GIVE BRAINLEST​
DochEvi [55]

Answer:

1.62 m/s²

Explanation:

4 0
3 years ago
A potential difference exists between the inner and outer surfaces of the membrane of a cell. The inner surface is negative rela
kifflom [539]

Answer:

Explanation:

Given

Work required W=1.35\times 10^{-20}\ J

Work done to eject the sodium ion from interior of the cell is given by the product of charge and Potential difference between inner and outer surface of the cell.

W=q\times V

Charge on sodium ion q=1.6\times 10^{19}\ C

V=\frac{W}{q}

V=\frac{1.35\times 10^{-20}}{1.6\times 10^{19}}

V=0.0718\ V            

8 0
3 years ago
Match the half life and time information to the percentage of radioactive isotope left.
Kazeer [188]

Answer:

Explanation:

can i have brainliest pls im new

8 0
3 years ago
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