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4 years ago

9

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

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?

faltersainse [42]

A.) LUNAR MARIA..............

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4 years ago

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Hewo thoties can ya'll do me a favor and go friend my bff on here plz and thank u

pickupchik [31]

Answer:

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

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

$P = 49$ %

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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:

$V_{2}=11369.46 m/s$

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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