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2 years ago

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Two objects gravitationally attract with a force of 70N. If the masses of both of the objects are increased by 15x and the dista

nce between their centers change by a factor of 12x, then the new force of attraction is ___ N.
a. 1.56 N
b. 109.38 N
c. 7.29 N
d. 1312.5 N

1 answer:

erastova [34]2 years ago

7 0

(B is the answer!!
I took the test and I got a 100! Hope this helps

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A +0.05 C charge is placed in a uniform electric field pointing downward with a strength of 100 Newtons over Coulombs.. Determin

shtirl [24]

The magnitude of the electric force on the charge is 5 N.

<h3>Magnitude of force on the charge</h3>

The magnitude of force on the charge is calculated as follows;

F = Eq

where;

E is electric field
q is magnitude of the charge

F = 100 N/C x 0.05 C

F = 5 N

Thus, the magnitude of the electric force on the charge is 5 N.

Learn more about electric force here: brainly.com/question/20880591

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1 year ago

1. Give three examples, from the lab, where potential energy was converted to kinetic energy:

Lubov Fominskaja [6]

Answer:

A book on a table before it falls.

A yoyo before it is released.

A raised weight.

Explanation:

These are all examples of potential energy. So I hope you can find something that is comparable from the lab.

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2 years ago

EASY MULTIPLE CHOICE QUESTIONS! PLEASE HELP!

skad [1K]

1.A and 2.B there the answers

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2 years ago

A. 36 mins 52.5 secs<br> b. 24mins 22.5 secs<br> c. 13 mins 0 secs<br> d. 9 mins 45 secs

DedPeter [7]

I think it might be A. I’m sorry if I’m wrong

5 0

3 years ago

Read 2 more answers

Help me please I can't get the final step

inna [77]

Answer:

$\displaystyle m=\frac{2}{3},\ n=\frac{4}{3}$

Explanation:

<u>Dimensional Analysis</u>

It's given the relation between quantities A, B, and C as follows:

$\displaystyle A=\frac{3}{2}B^mC^n$

and the dimensions of each variable is:

$A=L^2T^2$

$B=LT^{-1}$

$C=LT^2$

Substituting the dimensions into the relation (the coefficient is not important in dimension analysis):

$\displaystyle L^2T^2=\left(LT^{-1}\right)^m\left(LT^2\right)^n$

Operating:

$L^2T^2=\left(L^mT^{-m}\right)\left(L^nT^{2n}\right)$

$L^2T^2=L^{m+m}T^{-m+2n}$

Equating the exponents:

$m+n=2$

$-m+2n=2$

Adding both equations:

$3n=4$

Solving:

$n=4/3$

$m=2-4/3=2/3$

Answer:

$\mathbf{\displaystyle m=\frac{2}{3},\ n=\frac{4}{3}}$

6 0

2 years ago