The solution for this problem is computed by through this formula, F = kQq / d²Plugging in the given values above, we can now compute for the answer.
F = 8.98755e9N·m²/C² * -(7e-6C)² / (0.03m)² = -489N, the negative sign denotes attraction.
Answer:
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Explanation:
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D multiply force
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Answer:
A-Caclcuate the potential energy of the ball at that height
Explanation:
(a). Mass of the Body = 10 kg.
Height = 10 m.
Acceleration due to gravity = 9.8 m/s².
Using the Formula,Potential Energy = mgh
= 10 × 9.8 × 10 = 980 J.
(b). Now, By the law of the conservation of the Energy, Total amount of the energy of the system remains constant.
∴ Kinetic Energy before the body reaches the ground is equal to the Potential Energy at the height of 10 m.
∴ Kinetic Energy = 980 J.
(c). Kinetic Energy = 980 J.
Mass of the ball = 10 kg.
∵ K.E. = 1/2 × mv²
∴ 980 = 1/2 × 10 × v²
∴ v² = 980/5
⇒ v² = 196
∴ v = 14 m/s.
Answer:
Initial velocity, U = 4.5m/s
Explanation:
Given the following data;
Final velocity, v = 12m/s
Time, t = 5 seconds
Acceleration, a = 1.5m/s²
To find the initial velocity, we would use the first equation of motion.
Where;
V is the final velocity.
U is the initial velocity.
a is the acceleration.
t is the time measured in seconds.
Substituting into the equation, we have;
12 = U + 1.5*5
12 = U + 7.5
U = 12 - 7.5
Initial velocity, U = 4.5m/s
