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3 years ago

A car has a 12-v battery. if we hook it up to a 36 ohm resistor, what current will flow?

Physics

1 answer:

ikadub [295]3 years ago

8 0

Current = voltage/resistance
Current
= 12/36
= 1/3 A

I think...

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Georgie was pulling her brother (of mass 16 kg) in a 9.4 kg sled with a constant force of 39 N for one block (128 m). How much w

Genrish500 [490]

Answer:

4992 J

Explanation:

Force, F = 39 N

Distance, s = 128 m

Work done = force x distance

W = 39 x 128

W = 4992 J

Thus, the work done is 4992 J.

7 0

3 years ago

What is this Regeneration

Mila [183]

Answer:

Regeneration means that an organism regrows a lost part, so that the original function is restored.

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3 years ago

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An unknown element is found to have two naturally occurring isotopes, 200X and 210X with atomic masses of 200.028 and 210.039 re

Kruka [31]

Answer:

The average atomic mass of X is 206.0346

Explanation:

Atomic mass of 200X = 200.028

% abundance of 200X = 40% = 40/100 = 0.4

Atomic mass of 210X = 210.039

% abundance of 210X = 100% - 40% = 60% = 60/100 = 0.6

Average atomic mass of X = (0.4×200.028) + (0.6×210.039) = 80.0112 + 126.0234 = 206.0346

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4 years ago

On the earth, an astronaut throws a ball straight upward; it stays in the air for a total time of 2.5 s before reaching the grou

lesya [120]

<h2>Answer: 15 s</h2>

Explanation:

This is a situation related to vertical motion with constant acceleration, where the equation that will be usefull is:

$y=y_{o}+V_{o}t-\frac{1}{2}gt^{2}$ (1)

Where:

$y=0$ is the final height of the ball (when it reaches the ground)

$y_{o}=0$ is the initial height of the ball (is also zero because is thrown from ground)

$V_{o}$ is the initial velocity of the ball

$t=2.5 s$ is the time the ball is in air (on Earth)

$g_{E}=9.8 m/s^{2}$ is the acceleration due to gravity on Earth

$g_{M}=1.62 m/s^{2}$ is the acceleration due to gravity on the Moon

Having this clear, let's solve (1) for the Earth:

$0=0+V_{o}t-\frac{1}{2}gt^{2}$ (2)

$0=t(V_{o}-\frac{1}{2}gt^{2})$ (3)

$V_{o}=\frac{1}{2}gt^{2})$ (4)

$V_{o}=\frac{1}{2}(9.8 m/s^{2})(2.5 s)^{2})$ (5)

$V_{o}=12.25 m/s$ (6) This is the initial velocity

Using this same velocity and equation (4) for the Moon:

$12.25 m/s=\frac{1}{2}(1.62 m/s^{2})t^{2})$ (7)

Finally finding $t$ :

$t=15.12 s \approx 15 s$

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3 years ago

Question 7

Drupady [299]

Power = energy/time=20/4=5.0

6 0

3 years ago

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