Answer:
Let M1 = 8 kg and M2 = 34 kg
F = M a = (M1 + M2) a
F = M2 g the net force accelerating the system
M2 g = (M1 + M2) a
a = M2 / (M1 + M2) g = 34 / (42) g = .81 g = 7.9 m/s^2
Answer:

Explanation:
<u>Frictional Force
</u>
When the car is moving along the curve, it receives a force that tries to take it from the road. It's called centripetal force and the formula to compute it is:

The centripetal acceleration a_c is computed as

Where v is the tangent speed of the car and r is the radius of curvature. Replacing the formula into the first one

For the car to keep on the track, the friction must have the exact same value of the centripetal force and balance the forces. The friction force is computed as

The normal force N is equal to the weight of the car, thus

Equating both forces

Simplifying

Substituting the values


The formula for the energy in a capacitor , u in terms of q and c is q²/2c
<h3>What is the energy of a capacitor?</h3>
The energy of a capacitor u = 1/2qv where
- q = charge on capacitor and
- v = voltage across capacitor.
<h3>What is the capacitance of a capacitor?</h3>
Also, the capacitance of a capacitor c = q/v where
- q = charge on capacitor and
- v = voltage across capacitor.
So, v = q/c
<h3>
The formula for energy of the capacitor in terms of q and c</h3>
Substituting v into u, we have
u = 1/2qv
= 1/2q(q/c)
= q²/2c
So, the formula for the energy in a capacitor , u in terms of q and c is q²/2c
Learn more about energy in a capacitor here:
brainly.com/question/10705986
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P = density × gravity acceleration × height
P = 1200 × 9.81 × 15/100
P = 1765.8