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3 years ago

8

How did you figure that out? (think of FORMULA!)

1 answer:

Minchanka [31]3 years ago

5 0

Answer:

the most potential energy would be at the highest point and the least it's at the lowest point so let's say and rollercoaster when it's like moving up its gaining potential energy while it's going down it loses potential energy (sorry if u get it wrong)

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Which factors describe a gas

Alekssandra [29.7K]

Answer:

B. changing shape and changing volume

Explanation:

*no definite shape (takes the shape of its container)

*no definite volume

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3 years ago

If you are traveling at 75 km/h how long will it take to travel 32 km?

Oksi-84 [34.3K]

Answer:

This would be 24 minutes

Explanation:

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3 years ago

What the kinetic Energy of 80kg<br> object this is Moving with a Speed<br> of 23 m/s

Katyanochek1 [597]

Explanation:

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2 years ago

Each plate of a parallel‑plate capacitor is a square of side 3.63 cm, and the plates are separated by 0.473 mm. The capacitor is

NARA [144]

Answer:

E= 55.53 x 10³ V/m

Explanation:

Given that

a= 3.63 cm

Area ,A= a²

distance ,d= 0.473 mm

Stored energy ,U = 8.49 nJ

Value of capacitor given as

$C=\dfrac{\varepsilon _oA}{d}$

By putting the values

$C=\dfrac{8.85\times 10^{-12}\times 3.63^2\times 10^{-4}}{0.473\times 10^{-3}}$

C=2.46 x 10⁻¹¹ F

$U=\dfrac{1}{2}CV^2$

V=Voltage difference

$V=\sqrt{\dfrac{2U}{C}}$

$V=\sqrt{\dfrac{2\times 8.49\times 10^{-9}}{2.46\times 10^{-11}}}$

V=26.27 V

V= E d

E=Electric filed

26.27 = E x 0.473 x 10⁻³

E= 55.53 x 10³ V/m

7 0

3 years ago

Three charged particles are positioned in the xy plane: a 50-nC charge at y = 6 m on the y axis, a −80-nC charge at x = −4 m on

disa [49]

Answer:

V = 48 Volts

Explanation:

Since we know that electric potential is a scalar quantity

So here total potential of a point is sum of potential due to each charge

It is given as

$V = V_1 + V_2 + V_3$

here we have potential due to 50 nC placed at y = 6 m

$V_1 = \frac{kQ}{r}$

$V_1 = \frac{(9\times 10^9)(50 \times 10^{-9})}{\sqrt{6^2 + 8^2}}$

$V_1 = 45 Volts$

Now potential due to -80 nC charge placed at x = -4

$V_2 = \frac{kQ}{r}$

$V_2 = \frac{(9\times 10^9)(-80 \times 10^{-9})}{12}$

$V_2 = -60 Volts$

Now potential due to 70 nC placed at y = -6 m

$V_3 = \frac{kQ}{r}$

$V_3 = \frac{(9\times 10^9)(70 \times 10^{-9})}{\sqrt{6^2 + 8^2}}$

$V_3 = 63 Volts$

Now total potential at this point is given as

$V = 45 - 60 + 63 = 48 Volts$

3 0

3 years ago