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Temka [501]
3 years ago
15

1) A person walks 15 feet to the left, and then 15 feet to the right. What is the person's

Physics
1 answer:
Alisiya [41]3 years ago
4 0
Note: I believe this requires a illustration to be shown.

Answer: Wherever the persons original spot was before he/she walked 15 feet to the left.
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How much work is done to move 2.10 μC of charge from the negative terminal to the positive terminal of a 3.50 V battery?
rjkz [21]

Answer:

6.3E-6

Explanation:

Workdone = V Q

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3 years ago
What is an example of radiation? check all that apply
Katen [24]
There are no appropriate examples in the list you provided with your question.

Examples of radiation:

... sunshine to tan your skin
... radio energy to bring you the news
... X-ray to check your teeth
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4 0
3 years ago
Part B
Katena32 [7]

Answer:

While slavery was the major issue separating the North and South, it was not slavery itself that sparked the conflict. The South wanted to secede from the Union, and the North refused. While President Abraham Lincoln personally opposed slavery, he recognized that it was legal under the U.S. Constitution at the time. He also recognized that few in the North were ready to go to war to free the slaves. For Lincoln and the northern majority, preservation of the Union was the foremost goal.

4 0
2 years ago
A particle with charge − 2.74 × 10 − 6 C −2.74×10−6 C is released at rest in a region of constant, uniform electric field. Assum
s2008m [1.1K]

Answer:

241.7 s

Explanation:

We are given that

Charge of particle=q=-2.74\times 10^{-6} C

Kinetic energy of particle=K_E=6.65\times 10^{-10} J

Initial time=t_1=6.36 s

Final potential difference=V_2=0.351 V

We have to find the time t after that the particle is released and traveled through a potential difference 0.351 V.

We know that

qV=K.E

Using the formula

2.74\times 10^{-6}V_1=6.65\times 10^{-10} J

V_1=\frac{6.65\times 10^{-10}}{2.74\times 10^{-6}}=2.43\times 10^{-4} V

Initial voltage=V_1=2.43\times 10^{-4} V

\frac{\initial\;voltage}{final\;voltage}=(\frac{initial\;time}{final\;time})^2

Using the formula

\frac{V_1}{V_2}=(\frac{6.36}{t})^2

\frac{2.43\times 10^{-4}}{0.351}=\frac{(6.36)^2}{t^2}

t^2=\frac{(6.36)^2\times 0.351}{2.43\times 10^{-4}}

t=\sqrt{\frac{(6.36)^2\times 0.351}{2.43\times 10^{-4}}}

t=241.7 s

Hence, after 241.7 s the particle is released has it traveled through a potential difference of 0.351 V.

6 0
3 years ago
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