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2 years ago

Gravitational potential energy (Eg) and kinetic energy (Ek)

Physics

2 answers:

STatiana [176]2 years ago

6 0

Answer:

hope it will help you

Explanation:

madreJ [45]2 years ago

4 0

Answer:

$\boxed {\boxed {\sf E_K= \ 5,040 \ J}}$

Explanation:

Kinetic energy is the energy an object possesses due to motion. It is calculated using the following formula:

$E_K= \frac{1}{2} mv^2$

The mass of the athlete is 70.0 kilograms. The velocity is 12 meters per second.

m= 70.0 kg
v= 12 m/s

Substitute the values into the formula.

$E_K= \frac{1}{2} (70.0 \ kg)(12 \ m/s)^2$

Solve the exponent.

(12 m/s)² = (12 m/s)(12 m/s) = 144 m²/s²

$E_K= \frac{1}{2} (70.0 \ kg)(144 \ m^2/s^2)$

Multiply the numbers together.

$E_K= \frac{1}{2} (10, 080 \ kg*m^2/s^2)$

$E_K= 5,040 \ kg*m^2/s^2$

Convert the units. 1 kilogram meter squared per second squared is equal to 1 Joule, so our answer is equal to 5,040 Joules.

$E_K= 5.040 \ J$

The athlete has <u>5,040 Joules</u> of kinetic energy.

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2 years ago

Two stationary point charges of 3.00 nC and 2.00 nC are separated by a distance of 50.0 cm. An electron is released from rest at

andrew-mc [135]

Answer:

1. the electric potential energy of the electron when it is at the midpoint is - 2.9 x $10^{-17}$ J

2. the electric potential energy of the electron when it is 10.0 cm from the 3.00 nC charge is - 5.04 x $10^{-17}$ J

Explanation:

given information:

$q_{1}$ = 3 nC = 3 x $10^{-9}$ C

$q_{2}$ = 2 nC = 2 x $10^{-9}$ C

r = 50 cm = 0.5 m

the electric potential energy of the electron when it is at the midpoint

potential energy of the charge, F

F = k $\frac{q_{e}q}{r}$

where

k = constant (8.99 x $10^{9} Nm^{2} /C^{2}$ )

electron charge, $q_{e}$ = - 1.6 x $10^{-19}$ C

since it is measured at the midpoint,

r = $\frac{0.5}{2}$

= 0.25 m

thus,

F = $F_{1}+ F_{2}$

= k $\frac{q_{e} q_{1} }{r}$ + k $\frac{q_{e} q_{2} }{r}$

= $\frac{kq_{e} }{r}$ ( $q_{1} +q_{2}$ )

= (8.99 x $10^{9}$ )( - 1.6 x $10^{-19}$ )(3 x $10^{-9}$ +2 x $10^{-9}$ )/0.25

= - 2.9 x $10^{-17}$ J

the electric potential energy of the electron when it is 10.0 cm from the 3.00 nC charge

$r_{1}$ = 10 cm = 0.1 m

$r_{2}$ = 0.5 - 0.1 = 0.4 m

F = k $\frac{q_{e} q_{1} }{r}$ + k $\frac{q_{e} q_{2} }{r}$

= $kq_{e}$ ( $\frac{q_{1} }{r_{1} }$ + $\frac{q_{2} }{r_{2} }$ )

= (8.99 x $10^{9}$ )( - 1.6 x $10^{-19}$ )(3 x $10^{-9}$ /0.1+2 x $10^{-9}$ /0.4)

= - 5.04 x $10^{-17}$ J

3 0

2 years ago

A water gun fires 5 squirts per second. The speed of the squirts is 15 m/s.

stira [4]

Answer: 3.75 m

Explanation:

5 squirts in 1 second

So, 1 squirt in 1/5 second which is 0.2 second.

The difference in timing of two consecutive squirt is 0.2 second, so

time (t) = 0.2 s.

speed (s) = 15 m/s

Distance of separation (d) = ?

Now, formula for distance is

d = s × t

d = 15 × 0.2

d = 3.75 m

4 0

2 years ago

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taurus [48]

Answer:

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Explanation:

8 0

2 years ago

Radio waves travel at a speed of 300,000,000 m/s. WFNX broadcasts radio waves at a

yKpoI14uk [10]

Answer:

Wavelength, $\lambda=2.94\ m$

Explanation:

It is given that,

Speed of radio waves is $v=3\times 10^8\ m/s$

Frequency of radio waves is f = 101,700,000 Hz

We need to find the wavelength of WFNX’s radio waves. The relation between wavelength, frequency and speed of a wave is given by :

$v=f\lambda$

$\lambda$ is wavelength

$\lambda=\dfrac{v}{f}\\\\\lambda=\dfrac{3\times 10^8}{101,700,000}\\\\\lambda=2.94\ m$

So, the wavelength of WFNX’s radio waves is 2.94 m.

3 0

3 years ago