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3 years ago

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What are the two factors that determine the period of a pendulum? acceleration of gravity and the mass of the bob mass of the bo

b and length of the string length of the string and acceleration of gravity acceleration of gravity and angle of the swing

2 answers:

V125BC [204]3 years ago

8 0

<span>The two factors which determine the period of a pendulum is:

A) Length of the string
B) Acceleration of gravity

In short, Your Answer would be Option C

Hope this helps!</span>

Artist 52 [7]3 years ago

7 0

<h3><u>Answer</u>;</h3><h3> length of the string and acceleration of gravity </h3><h3>Explanation;</h3>

Period, T, refers to the time taken for one complete oscillation.

The period T of a pendulum is affected by the following factors;

The length of the pendulum or the length of the string; such that the longer the length of the pendulum, the longer the period.
The acceleration of gravity or the gravitational field strength; such that the greater the gravitational acceleration the shorter the period.

You might be interested in

GenaCL600 [577]

Answer:

1. F = 45,458.17 N

2. P = 12,800,000 W

Explanation:

Part 1. The thrust force is the sum of the forces on the air and on the fuel.

For the air, 107 kg of air is accelerated from 281 m/s to 679 m/s in 1 second.

F = ma

F = (107 kg) (679 m/s − 281 m/s) / (1 s)

F = 42,586 N

For the fuel, 4.23 kg of fuel is accelerated from 0 m/s to 679 m/s in 1 second.

F = ma

F = (4.23 kg) (679 m/s − 0 m/s) / (1 s)

F = 2,872.17 N

So the thrust on the jet is:

F = 42,586 N + 2,872.17 N

F = 45,458.17 N

Rounded to three significant figures, the force is 45,500 N.

Part 2. Power = work / time, and work = force × distance, so:

Power = force × distance / time

Power = force × velocity

P = (45,458.17 N) (281 m/s)

P = 12,773,745.77 W

Rounded to three significant figures, the power is 12,800,000 W.

7 0

3 years ago

A ski gondola is connected to the top of a hill by a steel cable of length 620 m and diameter 1.5 cm. As the gondola comes to th

xz_007 [3.2K]

Answer:

(a) 89 m/s

(b) 11000 N

Explanation:

Note that answers are given to 2 significant figures which is what we have in the values in the question.

(a) Speed is given by the ratio of distance to time. In the question, the time given was the time it took the pulse to travel the length of the cable twice. Thus, the distance travelled is twice the length of the cable.

$v=\dfrac{2\times 620 \text{ m}}{14\text{ s}} = \dfrac{1240\text{ m}}{14\text{ s}}=88.571428\ldots \text{ m/s}= 89\text{ m/s}$

(b) The tension, $T$ , is given by

$v =\sqrt{\dfrac{T}{\mu}}$

where $v$ is the speed, $T$ is the tension and $\mu$ is the mass per unit length.

Hence,

$T = \mu\cdot v^{2}$

To determine $\mu$ , we need to know the mass of the cable. We use the density formula:

$\rho = \dfrac{m}{V}$

where $m$ is the mass and $V$ is the volume.

$m=\rho\cdot V$

If the length is denoted by $l$ , then

$\mu = \dfrac{m}{l} = \dfrac{\rho\cdot V}{l}$

$T = \dfrac{\rho\cdot V}{l} v^{2}$

The density of steel = 8050 kg/m3

The cable is approximately a cylinder with diameter 1.5 cm and length or height of 620 m. Its volume is

$V = \pi \dfrac{d^{2}}{4} l$

$T = \dfrac{\rho\cdot\pi d^2 l}{4l}v^2 = \dfrac{\rho\cdot\pi d^2}{4}v^2$

$T = \dfrac{8050\times\pi\times0.015^2}{4} \times 88.57^2$

$T = 11159.4186\ldots \text{ N} = 11000 \text{ N}$

4 0

4 years ago

Middle School Physics 5+3 pts

Murljashka [212]

A) The ball on the small ball is far smaller than the force on the basketball.

B) The total momentum before and after the collision remains constant.

C) We know momentum is conserved so we do:
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
0.1 x 5 + 0.6 x 0 = 0.1 x -4 + 0.6 x v₂
v₂ = 1.5 m/s

5 0

3 years ago

Fifteen identical particles have various speeds: one has a speed of 2.00 m/s, two have speeds of 3.00 m/s, three have speeds of

Evgesh-ka [11]

Answer:

a) $V=7.5m/s$

b) $rms=8.4m/s$

c) Generally the most probable speed is 8m/s as it the most posses by particles being the average

Explanation:

From the question we are told that:

Sample size N=15

$Speed 1 v_1=2m/s\\\\Speed 2 v_2=3m/s\\\\Speed 3 v_2=5m/s\\\\Speed 4 v_4=8m/s\\\\Speed 3 v_5=9m/s\\\\Speed 2 v_6=15m/s\\\\$

Generally the equation for Average speed is mathematically given by

$V_{avg}=\frac{\sum(nv)}{N}$

Therefore

$V_{avg}=\frac{(2+2(3)+3(5)+4(8)+3(9)+2(15))}{15}$

$V=7.5m/s$

b)

Generally the equation for RMS speed of the particle is mathematically given by

$rms=\sqrt{\frac{\sum(nv^2)}{N}}$

$rms=\sqrt{\frac{2^2+2(3)^2+3(5)^2+4(8)^2+3(9)^2+2(15)^2}{15}}$

$rms=\sqrt{69.73}$

$rms=8.4m/s$

c

Generally the most probable speed is 8m/s as it the most posses by particles being the average

4 0

3 years ago

The force required to start an object sliding across a uniform horizontal surface is larger than the force required to keep the

taurus [48]

The force required to start an object sliding across a uniform horizontal surface is larger than the force required to keep the object sliding at a constant velocity once it starts.

The magnitudes of the required forces are different in these situations because the force of kinetic friction is less than the force of static friction. <em>(d)</em>

3 0

3 years ago