關於物理的簡單問題 - Ans.fyi 參考答案

關於物理的簡單問題

Tieria

2010-04-14 5:04 am

1. A force of 1000 N is needed to lift the hook of a crane at a steady velocity. The crane is then used to lift a load of mass 1000 kg at a velocity of 0.5 ms^-1. How much of the power developed by the motor of the crane is used in lifting the hook and the load?

The answer is 5.5 kW.

2. The half-life of a certain radioactive element is such that 7 / 8 of a given quantity decays in 12 days. What fraction remains undecayed after 24 days?

The answer is 1 / 64.

3. The rate of change of momentum of a body falling freely under gravity is equal to its
A. impulse
B. kinetic energy
C. power
D. weight

The answer is D.

4. Two wires P and Q, each of the same length and the same material, are connected in parallel to a battery. The diameter of P is half that of Q. What fraction of the total current passes through P?

The answer is 0.20.

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回答 (1)

天同

2010-04-14 5:47 am

✔ 最佳答案

1. Ttoal force developed by the motor = (1000 + 1000g) N = 11000 N
where g is the acceleration due to gravity, taken as 10 m/s2
Hence, power = 11000 x 0.5 w = 5500 w

2. Let T be the half-life
hence, (1-7/8) = 1/2^(12/T)
i.e. T = 4 days

Fraction remains after 24 days = 1/2^(24/4) = 1/2^6 = 1/64

3. Rate of change of momentum = force
When an object falls under gravity, the force pulling it down = weight of object

4. Let R(p) and R(q) be the resistances of P and Q respectively
Since resistance is inversely proportional to area, which in turn proportional to radius (or diameter) square. Hence, resistance is inversely proportional to diameter square.

Thus, R(p) = k/D(p)^2 and R(q) = k/D(q)^2
where D(p) and D(q) are the diameters of P and Q

Since the applied voltage to P and Q are the same, the current flowing through P and Q , I(p) and I(q) respectively, is inversely proportional to resistances of P and Q.
I(p) = V/R(p) = V.D(p)^2/k and I(q) = V/R(q) = V.D(q)^2/k
where V is the applied voltage

Therefore, fraction of total current that passes through P
= I(p)/[I(p) + I(q)] = D(p)^2/[D(p)^2 + D(q)^2]
given that D(p) = D(q)/2, i.e. D(q) = 2 x D(p)
hence, I(p)/[I(p) + I(q)] = D(p)^2/[D(p)^2 + 4xD(p)^2]
= 1/5 = 0.2

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