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Gennadij [26K]
3 years ago
14

What is the magnitude of the electric force between two point charges with Q1 = -1.5 C and Q2 = 0.8 C at a distance of 1 km?

Physics
1 answer:
Ilia_Sergeevich [38]3 years ago
8 0

Answer:

F = -10800 N

Explanation:

Given that,

Charge 1, q_1=-1.5\ C

Charge 2, q_2=0.8\ C

Distance between the charges, d=1\ km=10^3\ m

We need to find the electric force acting between two point charges. Mathematically, it is given by :

F=k\dfrac{q_1q_2}{d^2}

F=-9\times 10^9\times \dfrac{1.5\times 0.8}{(10^3)^2}

F = -10800 N

So, the magnitude of electric force between two point charges is 10800 N. Hence, this is the required solution.

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3 years ago
Tubby and his twin brother Libby have a combined mass of 200 kg and are zooming along in a 100 kg amusement park bumper car at 1
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Answer: 14.1 m/s

Explanation:

We can solve this with the Conservation of Linear Momentum principle, which states the initial momentum p_{i} (before the elastic collision) must be equal to the final momentum p_{f} (after the elastic collision):

p_{i}=p_{f} (1)

Being:

p_{i}=m_{1}V_{i} + m_{2}U_{i}

p_{f}=m_{1}V_{f} + m_{2}U_{f}

Where:

m_{1}=200 kg +100 kg=300 kg is the combined mass of Tubby and Libby with the car

V_{i}=10 m/s is the velocity of Tubby and Libby with the car before the collision

m_{2}=25 kg + 100 kg=125 kg is the combined mass of Flubby with its car

U_{i}=0 m/s is the velocity of Flubby with the car before the collision

V_{f}=4.12 m/s is the velocity of Tubby and Libby with the car after the collision

U_{f} is the velocity of Flubby with the car after the collision

So, we have the following:

m_{1}V_{i} + m_{2}U_{i}=m_{1}V_{f} + m_{2}U_{f} (2)

Finding U_{f}:

U_{f}=\frac{m_{1}(V_{i}-V_{f})}{m_{2}} (3)

U_{f}=\frac{300 kg(10 m/s-4.12 m/s)}{125 kg} (4)

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8 0
3 years ago
The maximum allowable resistance for an underwater cable is one hundredth of an ohm per
Studentka2010 [4]

Answer:

A = 1.54 x 10⁻⁵ m² = 15.4 mm²

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The resistance of a wire can be given by the following formula:

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where,

A = smallest cross-sectional area = ?

ρ = resistivity of copper = 1.54 x 10⁻⁸ Ωm

\frac{R}{L} = resistance per unit length of wire = 0.001 Ω/m

Therefore,

A = \frac{1.54\ x\ 10^{-8}\ \Omega m}{0.001\ \Omega/m}

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6 0
3 years ago
A simple pendulum is 1.15 meters long and has a period of 6.29 seconds. The pendulum is on an unknown planet. What is the accele
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Answer:

The value of the acceleration of gravity of the Unknown Planet = 1.14 \frac{m}{s^{2} }

Explanation:

length of the pendulum (L)= 1.15 m

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We know that time period  of a simple pendulum is given by

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put the values in the above formula we get

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By solving the above equation we get

⇒ g = 1.14 \frac{m}{s^{2} }

This is the value of the acceleration of gravity of the Unknown Planet.

7 0
4 years ago
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