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3 years ago

Do you think energy can be “used up?” Why or why not?

Physics

1 answer:

My name is Ann [436]3 years ago

8 0

I dint think energy can be used up because energy cant be created nor destroyed

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ANSWER QUICK 30 POINTS

Lelu [443]

What do you need help on

4 0

3 years ago

Two electrons exert a force of repulsion of 1.2 N on each other. How far apart are they? The elementary charge is 1.602 × 10−19

sergejj [24]

The distance between the charges is 13.86 X 10⁴m

<u>Explanation:</u>

Given:

Force, F = 1.2N

Charge, q₁ = 1.602 X 10⁻¹⁹ C

k = 8.987 X 10⁹ Nm²/C²

Distance, d = ?

According to Coulomb's law:

$F = k\frac{q_1q_2}{r^2} \\\\$

Substituting the value in the formula we get:

$1.2 = 8.987X 10^9 X\frac{1.602 X 10^-^1^9 X 1.602 X 10^1^9}{r^2} \\\\1.2 = \frac{23.06 X 10^9}{r^2} \\\\r^2 = 19.22 X 10^9\\\\r = 13.86 X 10^4m$

Therefore, the distance between the charges is 13.86 X 10⁴m

5 0

3 years ago

Two automobiles are equipped with the same single-frequency horn. When one is at rest and the other is moving toward the first a

prisoha [69]

Explanation:

This is a good example of Doppler effect

Given

Source velocity Vs=20m/s

Apparent frequency Fa= 5.2 Hz

Recall speed of sound V=340m/s

The frequency of the horn Fh=?

But mathematically Fh=(V-Vs/V)*Fa

(340-20)/340*5.2=4.89Hz

6 0

3 years ago

Read 2 more answers

53. A recently discovered planet has a mass four times as great as Earth's and a radius twice as large as Earth's. What will be

marusya05 [52]

Answer:

$g' = g = 9.81 m/s^2$

so gravity will be same as that of surface of earth

Explanation:

As we know that the acceleration due to gravity is given as

$g = \frac{GM}{R^2}$

here we have

$M = 4M_e$

$R = 2R_e$

we know that for earth we have

$g = 9.81 = \frac{GM_e}{R_e^2}$

now if the radius and mass is given as above

$g' = \frac{G(4M_e)}{(2R_e)^2}$

$g' = \frac{GM_e}{R_e^2}$

$g' = g = 9.81 m/s^2$

so gravity will be same as that of surface of earth

6 0

3 years ago

A 2.1 ✕ 103-kg car starts from rest at the top of a 5.9-m-long driveway that is inclined at 19° with the horizontal. If an avera

Lina20 [59]

Answer:

3.9 m/s

Explanation:

We are given that

Mass of car,m= $2.1\times 10^3 kg$

Initial velocity,u=0

Distance,s=5.9 m

$\theta=19^{\circ}$

Average friction force,f= $4.0\times 10^3 N$

We have to find the speed of the car at the bottom of the driveway.

Net force, $F_{net}=mgsin\theta-f=2.1\times 10^3\times 9.8sin19-4.0\times 10^3$

Where $g=9.8 m/s^2$

Acceleration, $a=\frac{F_{net}}{m}=\frac{2.1\times 10^3\times 9.8sin19-4.0\times 10^3}{2.1\times 10^3}$

$v=\sqrt{2as}$

$v=\sqrt{2\times \frac{2.1\times 10^3\times 9.8sin19-4.0\times 10^3}{2.1\times 10^3}\times 5.9}$

v=3.9 m/s

7 0

3 years ago