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kakasveta [241]
3 years ago
14

When the pendulum bob reaches the mean position, the net force acting on it is zero. Why then does it swing past the mean positi

on?
Physics
1 answer:
ryzh [129]3 years ago
6 0

Answer:

<u>The pendulum bob swing past the mean position because:</u>

When a pendulum's bob is accelerating at its extreme position its velocity is zero. Due to  the restoring toque the bob starts to accelerates towards its mean postion. The  maximum acceleration of the pendulum's bob  is -w^{2} Aand the the acceleration decreases as -w^{2} x  towards the mean position.

The acceleration at the mean position becomes zero but the velocity remains maximum. Hence the bob continues to move and does not stops.Thus it can summarised as the force decreases ,acceleration decreases and velocity increases at slow rate.

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5 0
3 years ago
Read 2 more answers
3. Una cuerda de guitarra tiene 60 cm de longitud y una masa de 0.05 kg de masa. Si se tensiona mediante una fuerza de 20 N. La
jok3333 [9.3K]

Answer:

f1 = 12.90 Hz

Explanation:

To calculate the first harmonic frequency you use the following formula for n = 1:

f_n=\frac{n}{2L}\sqrt{\frac{T}{M/L}}

f_1=\frac{1}{2L}\sqrt{\frac{T}{M/L}}    ( 1 )

It is necessary that the unist are in meters, then you have:

L: length of the string = 60cm = 0.6m

M: mass of the string = 0.05kg

T: tension on the string = 20 N

you replace the values of L, M and T in the expression (1) for getting f1:

f_1=\frac{1}{2(0.6m)}\sqrt{\frac{20N}{0.05kg/0.6m}}=12.90\ Hz

Hence, the first harmonic has a frequency of 12.90 Hz

4 0
3 years ago
You have a mass of 95 kg.<br> a. What is your weight on Earth?
svp [43]

Answer:

931.63N

Hope this helps :D

8 0
3 years ago
An archer pulls the bowstring back for a distance of 0.470 m before releasing the arrow. The bow and string act like a spring wh
never [62]

Answer:

(a) 46.94 J.

(b)  55.95 m/s

Explanation:

(a)

Potential Energy: This is the energy of a body, due to its position. The S.I unit of potential energy is Joules (J).

The formula of potential energy in a stretched spring is

Ep = 1/2ke² .......................... Equation 1

Where Ep = potential energy of the spring, k = Force constant of the spring, e = extension or compression.

Given: k = 425 N/m, e = 0.47 m.

Substitute into equation 1

Ep = 1/2(425×0.47²)

Ep = 46.94 J.

(b)

at the instant When the arrow leaves the bow, the potential energy of the arrow is converted kinetic energy of the bow.

I.e,

Ep = 1/2mv² ............. Equation 2

Where m = mass of the arrow, v = velocity of the arrow.

make v the subject of the equation

v = √(2Ep/m)............. Equation 3

Given: Ep = 46.94 J, m = 0.03 Kg.

Substitute into equation 3

v = √(2×46.96/0.03)

v = √(93.92/0.03)

v = √(3130.67)

v = 55.95 m/s

5 0
3 years ago
David runs 100 m in a time of 9.80 s. Calculate his average speed. State the unit.
fiasKO [112]

Answer:

His average speed is 980 units

3 0
3 years ago
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