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yaroslaw [1]
3 years ago
12

What is the electric potential of a 2.2 µC charge at a distance of 6.3 m from the charge? Recall that Coulomb’s constant is k =

8.99 × 109 N • ___ V What is the electric potential at a distance of 99 m from the charge?___ V
Physics
2 answers:
Doss [256]3 years ago
6 0
The first part of the question is 3,100 V.

The second part of the question is 200 V.
const2013 [10]3 years ago
3 0

Electric potential due to a point charge is given by

V = \frac{kQ}{r}

here we know that

Q = charge

r = distance from the charge

V = \frac{9*10^9* 2.2* 10^{-6}}{6.3}

V = 3142.8 Volts

now again by the same equation

V = \frac{kQ}{r}

V = \frac{9*10^9* 2.2* 10^{-6}}{99}

V = 200 Volts


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Jasper made a list of the properties of electromagnetic waves. Identify the mistake in the list. Electromagnetic Wave Properties
Nata [24]

Answer:

Statement 2 is wrong

Explanation:

To check the statements in this exercise, let's describe the main properties of electromagnetic waves. Let's describe the characteristics

* they are transverse waves

* formed by the oscillations of the electric and magnetic fields

* the speed of the wave is the speed of light

with these concepts let's review the final statements

1) True. Formed by the oscillation of the two fields

2) False. They are transverse waves

3) True. Can travel by vacuum as they are supported by oscillations of the electric and magnetic fields

4) True. They all have the same speed of light

Statement 2 is wrong

6 0
3 years ago
A motorcycle traveling at a speed of 44.0 mi/h needs a minimum of 44.0 ft to stop. If the same motorcycle is traveling 79.0 mi/h
Tasya [4]

Answer:

141.78 ft

Explanation:

When speed, u = 44mi/h, minimum stopping distance, s = 44 ft = 0.00833 mi.

Calculating the acceleration using one of Newton's equations of motion:

v^2 = u^2 + 2as\\\\v = 0 mi/h\\\\u = 44 mi/h\\\\s = 0.00833 mi\\\\=> 0^2 = 44^2 + 2 * a * 0.00833\\\\=> 1936 = -0.01666a\\\\a = -116206.48 mi/h^2 or -14.43 m/s^2

Note: The negative sign denotes deceleration.

When speed, v = 79mi/h, the acceleration is equal to when it is 44mi/h i.e. -116206.48 mi/h^2

Hence, we can find the minimum stopping distance using:

v^2 = u^2 + 2as\\\\v = 0 mi/h\\\\u = 79 mi/h\\\\a = -116206.48 mi/h\\\\=> 0^2 = 79^2 + (2 * -116206.48 * s)\\\\6241 = 232412.96s\\\\s = \frac{6241}{232412.96} \\\\s = 0.0268531 mi = 141.78 ft

The minimum stopping distance is 141.78 ft.

4 0
3 years ago
Rachel hits a golf ball into the air. What type of motion is the ball's path?
sladkih [1.3K]

Answer:

omg

the ground is BREAKING

Explanation:

5 0
2 years ago
Typically, water runs through the baseboard copper tubing and, therefore, fresh hot water is constantly running through the pipi
Bas_tet [7]

Answer: The temperature of the water falls by 3.3°C

Explanation:

The heat change is related to the change in temperature by the equation

dH = m Cp dT

In this example, -2665 J = 193 g x 4.184 J/g°C x dT

so dT = -3.3 °C

3 0
3 years ago
An object with a charge of 0.9 x 10^-5 C is separated from a second object with a chare of 2.5 x 10^-4 C by a distance of 0.5 m.
meriva

Answer: Force = 81 N

Explanation:

from Columbs law,

F = k(q1*q2)/r²

k = 9 x 10^9 Nm²/C²

F = (9 x 10^9)x (0.9x10^-5 x 2.5x10^-4)/(0.5)²

F = 81 Newtons

4 0
3 years ago
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