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aliya0001 [1]
3 years ago
12

Two fixed charges, q1 = +1.07µC and q2 = -3.28µC, are 61.8cm apart. Where may a third charge be located so that no net force act

s on it?
Physics
2 answers:
AnnZ [28]3 years ago
7 0

Answer:

Can u re-write the question please, I dont understand?

Explanation:

egoroff_w [7]3 years ago
3 0

Answer:

12...lollll

Explanation:

.

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C.It will stay at rest due to inertia
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What type of machine is wire cutter pliers?
Sonja [21]
It seems that you have missed the necessary options for us to answer this question, but anyway, here is the answer. The type of machine that a wire cutter pliers is classified is a simple machine. When we say simple machine, this is the type of machine that is considered basic wherein you need to apply force for it to function. Hope this helps.
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The efficiency of a modern bicycle is 95% if you exert 200 Joules of input work pedaling a bicycle on level ground what is the u
Delvig [45]

Answer:

95 J

Explanation:

You can calculate efficiency by dividing useful output by total input, then multiplying it to 100.

So the foumula goes like:

Efficiency= (Useful output/Total input)x100

In this question,

Efficiency= 95%

Useful output= x

Total input= 200

Therefore;

95=(x/200)x100

0.95=x/100

x=0.95x100

x=95 Joules

8 0
2 years ago
How much would it cost to cover the entire land area of the U.S. in dollar bills?
agasfer [191]

Answer:

  $900 trillion

Explanation:

If Alaska is 20% of the contiguous US, then the approximate area of interest is ...

  1200 miles × 3000 miles = 3.6×10^6 square miles.

The size of a dollar bill is about ...

  (6.5 cm)·(15.5 cm) = 100.75 cm^2

One mile is 160,934.4 cm, so 1 square mile is about ...

  1 mi^2 = (160,934.4 cm)^2 ≈ 2.59·10^10 cm^2

The number of dollars of interest is then ...

  (3.6 · 10^6 mi^2)(2.59 · 10^10 cm^2)/(100.75 cm^2) ≈ 9.3·10^14

  ≈ 930 × 10^12 . . . dollars

It would cost about 900 trillion dollars to cover the land area of the US in $1 bills.

7 0
3 years ago
The graph represents the force applied on an 3.00kg crate while it moved 5.0m. A. How much total work is done on the crate? B. I
jeka57 [31]

a. We can calculate the amount of work by calculating the area under the graph.

first area (rectangular): 2.5 x 6 = 15

second area(trapezoid): 1/2 x (6+10) x 2.5 =20

total work done: 35 J

b. the force was first applied = 6 N

F = m.a

a = 6 : 3 = 2 m/s²

vf²=vi²+2as

vf²=6²+2.2.5

vf²=56

vf=7.5 m/s

8 0
2 years ago
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