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AfilCa [17]
3 years ago
7

3)

Physics
2 answers:
o-na [289]3 years ago
8 0
A Correcting each car to the same size as the first one invented could get in the way during work days.
Rzqust [24]3 years ago
3 0

I honestly think the answer is actually C).

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10,500 J of GPE, weight 539 N, how tall is the hill you are sitting on?
maks197457 [2]

Answer:

19.48 m

Explanation:

Gravitational potential energy = mgh

Current weight = 539 N

Weight = mg = 539 N

Mass x Acceleration = 539 N

Mass x 9.81 = 539

Mass = 54.94 kg

Gravitational potential energy = mgh = 10500 J

                    54.94 x 9.81 x h = 10500

                               h = 19.48 m

Height of sitting = 19.48 m

6 0
3 years ago
A test charge of +3 µC is at a point P where the electric field due to the other charges is directed to the right and has a magn
umka2103 [35]

Answer:

Electric field will remain same.

Explanation:

Given that

At initial condition ,charge q = +3μ C

Electric field E = 4106 N/c

As we know that

Electric field due to charge q

E=\dfrac{kq}{r^2}  

Now when charge is replaced by new charge q'= -3μ C

From the expression of electric field we can say that electric field will remain same from same quantity of electric charge.

So we can say that electric field will remain same.

5 0
4 years ago
How much of the earth's animal life exists in the oceans?<br><br> 40%<br> 60%<br> 80%<br> 99%
Alexxx [7]

How much of the earth's animal life exists in the oceans?

80% is the answer.

7 0
3 years ago
A 20~\mu F20 μF capacitor has previously charged up to contain a total charge of Q = 100~\mu CQ=100 μC on it. The capacitor is t
sertanlavr [38]

Explanation:

The given data is as follows.

       C = 20 \times 10^{-6} F

        R = 100 \times 10^{3} ohm

        Q_{o} = 100 \times 10^{-6} C

          Q = 13.5 \times 10^{-6} C

Formula to calculate the time is as follows.

          Q_{t}  = Q_{o} [e^{\frac{-t}{\tau}]

       13.5 \times 10^{-6} = 100 \times 10^{-6} [e^{\frac{-t}{2}}]

               0.135 = e^{\frac{-t}{2}}

         e^{\frac{t}{2}} = \frac{1}{0.135}

                         = 7.407

           \frac{t}{2} = ln (7.407)

                      t = 4.00 s

Therefore, we can conclude that time after the resistor is connected will the capacitor is 4.0 sec.

4 0
3 years ago
Vestobular receptors enable one to balance. <br><br> True or False
frozen [14]
The answer is actually true. i just took the test and i put false and it was wrong
5 0
3 years ago
Read 2 more answers
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