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san4es73 [151]
3 years ago
8

Two +1 C charges are separated by 30000 m, what is the magnitude of the force?

Physics
1 answer:
Kipish [7]3 years ago
7 0

Answer:

<em>The magnitude of the force is 10 N</em>

Explanation:

<u>Coulomb's Law</u>

The electrostatic force between two charged objects is directly proportional to the product of their charges and inversely proportional to the square of the distance between the two objects.

Written as a formula:

\displaystyle F=k\frac{q_1q_2}{d^2}

Where:

k=9\cdot 10^9\ N.m^2/c^2

q1, q2 = the objects' charge

d= The distance between the objects

We have two identical charges of q1=q2=1 c separated by d=30000 m, thus the magnitude of the force is:

\displaystyle F=9\cdot 10^9\frac{1*1}{30000^2}

\displaystyle F=9\cdot 10^9\frac{1*1}{30000^2}

F = 10 N

The magnitude of the force is 10 N

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Answer:

20 Watts

Explanation:

Work = force × distance

W = (20 N) (10 m)

W = 200 Joules

Power = work / time

P = 200 J / 10 s

P = 20 Watts

8 0
3 years ago
How do you do this? Plz help answer
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Answer: Really

Explanation:

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3 years ago
In trampoline competitions, a good jump is one that lasts about 1.8 seconds. (A) How high can an athlete who stays in the air 1.
KonstantinChe [14]

Answer:

3.97305 m

Explanation:

a = Acceleration due to gravity = 9.81 m/s²

If a jump lasts for 1.8 seconds this means that from the moment when the person leaves the ground till the person touches the ground again it takes 1.8 seconds. So, maximum height reached will be at half the time of the jump i.e., 0.9 seconds.

u = Initial velocity = 0

Equation of motion

s=ut+\frac{1}{2}at^2\\\Rightarrow s=0t+\frac{1}{2}9.81\times 0.9^2\\\Rightarrow s=3.97305\ m

So, height of the jump is 3.97305 m.

4 0
4 years ago
An electric heater containing two heating wires X and Y is connected to a power supply of electromotive force(emf) 9.0V and negl
mina [271]

Answer:

0.4 ohms.

Explanation:

From the circuit,

The voltage reading in the voltmeter = voltage drop across each of the parallel resistance.

1/R' = 1/R1+1/R2

R' = (R1×R2)/(R1+R2)

R' = (2.4×1.2)/(2.4+1.2)

R' = 2.88/3.6

R' = 0.8 ohms.

Hence the current flowing through the circuit is

I = V'/R'................ Equation 1

Where V' = voltmeter reading

I = 6/0.8

I = 7.5 A

This is the same current that flows through the variable resistor.

Voltage drop across the variable resistor = 9-6 = 3 V

Therefore, the resistance of the variable resistor = 3/7.5

Resistance = 0.4 ohms.

7 0
3 years ago
a bullet with a mass of 4.0g and a speed of 650m/s is fired at a block of wood with a mass of 0.095kg. the block rests on a fric
n200080 [17]

Part a)

Here in this we can use momentum conservation as there is no external force on it

m_1v_{1i} + m_2v_{2i} = m_1v_{1f} +m_2v_{2f}

here we know that

m_1 = 0.004 kg

v_{1i} = 650 m/s

m_2 = 0.095

v_{2i} = 0

v_{2f} = 23 m/s

now by above equation

0.004*650 + 0.095* 0 = 0.004*v + 0.095*23

2.6 + 0 = 0.004*v + 2.185

v = 103.75 m/s

Part b)

Final kinetic energy of the system

KE_f = \frac{1}{2}m_1v_{1f}^2 + \frac{1}{2}m_2v_{2f}^2

KE_f = \frac{1}{2}*0.004*(103.75^2) + \frac{1}{2}*(0.095)*23^2

KE_f = 46.65 J

Initial Kinetic energy of the system will be

KE_f = \frac{1}{2}m_1v_{1i}^2 + \frac{1}{2}m_2v_{2i}^2

KE_f = \frac{1}{2}*0.004*(650^2) + \frac{1}{2}*(0.095)*0^2

KE_f = 845 J

So here kinetic energy is decreased for this system

final energy is less than initial energy

6 0
3 years ago
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