g While hauling a log in the back of a flatbed truck, a driver is pulled over by the state police. Although the log cannot roll
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The potential difference across a and b is 15 v. determine the electrical charge on the 3 μf capacitor will be 45 * C
Capacitance, property of an electric conductor, or set of conductors, that is measured by the amount of separated electric charge that can be stored on it per unit change in electrical potential. Capacitance also implies an associated storage of electrical energy.
Charge (Q) stored in a capacitor is the product of its capacitance (C) and the voltage (V) applied to it. The capacitance of a capacitor should always be a constant, known value. So we can adjust voltage to increase or decrease the cap's charge. More voltage means more charge, less voltage... less charge.
charge = capacitance * voltage
Q = CV
= 3 * * 15 v
= 45 * C
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The impulse of a force is due to the change in the motion of an object
A. The persons speed after impact is approximately 59.38 mi/h
B. The expected speed is <u>29.89 mi/h</u> which is less than the findings
Reason:
Known parameters are;
The speed of the bus, v = 30 mi/h
The force with which the person was hit, F = 58,000 lbs
Mass of the bus, M = 40,000 lbs
Mass of the person, m = 150 lbs
Duration of the impact, Δt = 0.007 seconds
A. The speed of the person at the end of the impact, <em>v</em>, is given as follows;
The impulse of the force = F × Δt = m × Δv
For the person, we get;
58,000 lbf ≈ 1866094.816 lb·ft./s²
58,000 lbf × 0.007 s = 150 lbs × Δv
1,866,094.816 lb·ft./s²
Δv = v₂ - v₁
The initial speed of the person at the instant, can be as v₁ = 0
The final speed, v₂ = Δv - v₁
∴ v₂ ≈ 87.084 ft./s - 0 = 87.084 ft./s
≈ <u>87.084 ft./s</u>
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The speed of the person at the end of the impact, v₂ ≈ <u>59.38 mi/h</u>
B. Where the momentum is conserved, we have;
m₁·v₁ + m₂v₂ = (m₁ + m₂)·v
The expected speed of the person at the end of the impact is 29.89 mi/h, and therefore, <u>the findings does not agree with the expectation</u>
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