A block with mass m = 4.5 kg is attached to two springs with spring constants kleft = 36 N/m and kright = 50 N/m. The block is p
1 answer:
Answer:
21.5 N
Explanation:
As the block is pulled to a distance of 0.25 m to the left of its equilibrium position, the left spring is compressed by x = 0.25 m and exerting an elastic force to the right, the right spring is stretched by x =0.25 m and also exerting an elastic force to the right. Knowing their spring constant, we can calculate the total net force at that instant
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Answer:
490,400 J
Explanation:
Mass of first car, m = 1000 kg
Mass of second car, M = 1200 kg
velocity of first car, u = 20 m/s east
velocity of second car, U = 22 m/s west
The formula for the kinetic energy is
where, m is the body and v be the velocity of the body.
Total kinetic energy is given by
k = 490,400 J
Thus, the total kinetic energy of the system is 490,400 J.
The strength of the electric field on the point charge at this distance will be 4000 V/m.
<h3>What is the strength of the electric field?</h3>
The strength of the electric field is the ratio of electric force per unit charge.
The given data in the problem is;
Qis the unit charge = 4.0 × 10⁻⁶ C
E is the strength of the electric field
R is the distance from point charge = 3 m
The strength of the electric field is;
Hence, the strength of the electric field on the point charge at this distance will be 4000 V/m.
To learn more about the strength of the electric field refer to the link;
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