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pogonyaev
4 years ago
9

True or false? Three common voltage sources are batteries, solar cells, and generators.

Physics
2 answers:
Len [333]4 years ago
8 0
The answer to this question would be true
marysya [2.9K]4 years ago
3 0
True .
Hope this helped!!
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What are the large, dark basaltic plains on the moon called?
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A.) LUNAR MARIA..............
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Hewo thoties can ya'll do me a favor and go friend my bff on here plz and thank u
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lol sure

Explanation:

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3 years ago
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The California sea lion is capable of making extremely fast, tight turns while swimming underwater. In one study, scientists obs
anygoal [31]

Answer:

Acceleration of Sea Lion is 4.41 g

This is 49% of maximum jet acceleration given as a = 9g

Explanation:

As we know that the radius of the circular loop is given as

R = 0.37 m

The speed of the fish is given as

v = 4 m/s

Now the centripetal acceleration of the sea lion is given as

a_c = \frac{v^2}{R}

a_c = \frac{4^2}{0.37}

a_c = 43.2 m/s^2

as we know that

g = 9.8 m/s^2

so we have

a = 4.41 g

Now Percentage of this acceleration wrt maximum jet acceleration is given as

P = \frac{4.41 g}{9g} \times 100

P = 49%

6 0
3 years ago
A bowling ball of mass 5.8 kg moves in a straight line at 4.34 m/s How fast must a Ping-Pong ball of mass 2.214 g move in a stra
lilavasa [31]

Answer: 11369.46 m/s

Explanation:

We have the following data:

m_{1}=5.8 kg is the mass of the bowling ball

V_{1}=4.34 m/s is the velocity of the bowling ball

m_{2}=2.214 g \frac{1 kg}{1000 g}=0.002214 kg is the mass of the ping-pong ball

V_{2} is the velocity of the ping-pong ball

Now, the momentum p_{1} of the bowling ball is:

p_{1}=m_{1}V_{1} (1)

p_{1}=(5.8 kg)(4.34 m/s)  

p_{1}=25.172 kg m/s (2)

And the momentum p_{2} of the ping-pong ball is:

p_{2}=m_{2}V_{2} (3)

If the momentum of the bowling ball is equal to the momentum of the ping-pong ball:

p_{1}=p_{2} (4)

m_{1}V_{1}=m_{2}V_{2} (5)

Isolating V_{2}:

V_{2}=\frac{m_{1}V_{1}}{m_{2}} (6)

V_{2}=\frac{25.172 kg m/s}{0.002214 kg} (7)

Finally:

V_{2}=11369.46 m/s

6 0
3 years ago
Number of conducting plates of a multiplate capacitor is 5. The no. Of capacitors is
Ivahew [28]

Answer:

4 capacitors

Explanation:

Given

n = 5 --- conducting plates

Required

The number of capacitor (c)

This is calculated as:

c = n - 1

So, we have:

c = 5 - 1

c = 4

8 0
3 years ago
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