A 0.140 kg baseball is thrown horizontally with a velocity of 28.9 m/s. It is struck with a constant horizontal force that lasts
1 answer:
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Answer:
0.099C
Explanation:
First, we need to get the common potential voltage using the formula
Where V is the common voltage, C and V represent capacitance and charge respectively. Subscripts 1 and 2 to represent the the first and second respectively. Substituting the above with the following given values then
Therefore
Charge, Q is given by CV hence for the first capacitor charge will be
Here,
Answer:
The ratio is KE : TM = 0.75
Explanation:
from the question we are told that
The displacement of a mass on a spring in simple harmonic motion is A/2 from the equilibrium position
Generally the total mechanical energy of the mass is mathematically represented as
Here k is the spring constant , A is the total displacement of the the mass from maximum compression to maximum extension of the spring
Generally this total mechanical energy is mathematically represented as
=>
Here the potential energy of the mass is mathematically represented as
Here x is the displacement of the mass from maximum compression or extension of the spring to equilibrium position and the value is
So
So
=>
=>
So the ratio of is mathematically represented as
=>
Refer to the figure below.
R = resistance.
Case 1:
The voltage source is V₁ and the current is 10 mA. Therefore
V₁ = (10 mA)R
Case 2:
The voltage source is V₂ and the current is 8 mA. Therefore
V₂ = (8 mA)R
Case 3:
The voltage across the resistance is V₁ - V₂. Therefore the current I is given by
V₁ - V₂ = IR
10R - 8R = (I mA)R
2 = I
The current is 2 mA.
Answer: 2 mA
R = resistance.
Case 1:
The voltage source is V₁ and the current is 10 mA. Therefore
V₁ = (10 mA)R
Case 2:
The voltage source is V₂ and the current is 8 mA. Therefore
V₂ = (8 mA)R
Case 3:
The voltage across the resistance is V₁ - V₂. Therefore the current I is given by
V₁ - V₂ = IR
10R - 8R = (I mA)R
2 = I
The current is 2 mA.
Answer: 2 mA