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iogann1982 [59]
2 years ago
11

The period of a pendulum is the time it takes the pendulum to swing back and forth once. If the only dimensional quantities that

the period depends on are the acceleration of gravity, g, and the length of the pendulum, l, what combination of g and l must the period be proportional to
Physics
1 answer:
Vinvika [58]2 years ago
7 0

Explanation:

Let T is the period of a pendulum. The SI unit of time is seconds (s).

It depends on the acceleration of gravity, g, and the length of the pendulum, l.

The SI unit of acceleration of gravity, g and the length of the pendulum, l are m/s² and m respectively.

If we divide m and m/s², we left with s². If the square root of s² is taken, we get s only i.e. the SI unit of period of a pendulum.

So,

T\propto \sqrt{\dfrac{l}{g}}

Hence, this is the required solution.

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Answer:

x=2t^2+15t+5

Explanation:

x=\frac{1}{2}at^2+v_0t+x_0

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The largest single publication in the world is the 1112-volume set of British Parliamentary Papers for 1968 through 1972. The co
Marat540 [252]

Answer:

m_l=550\ kg is the mass of librarian.

Explanation:

Given:

  • mass of the system, m_s=3.3\times 10^{3}\ kg
  • velocity of librarian relative to the ground, v_l=2.5\ m.s^{-1}
  • velocity of the cart relative to the ground, v_c=0.5\ m.s^{-1}

N<u>ow using the principle of elastic collision:</u>

Net momentum of the system is zero.

m_l\times v_l=(3300-m_l)\times v_c

m_l\times 2.5=(3300-m_l)\times 0.5

m_l=550\ kg is the mass of librarian.

6 0
2 years ago
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A Turtle and a Snail are 360 meters apart, and they start to move towards each other at 3 p.m. If the Turtle is 11 times as fast
Neko [114]

Answer:

Snail's speed = \frac{30m}{2400s} = 0.0125m/s

Turtle's speed =  \frac{330m}{2400s} = 0.1375m/s

Explanation:

Let the snail's speed be x m/s

The turtle's speed then is 11x m/s

Speed = Distance ÷ Time

Since speed and distance are directly proportional;

The ratio of the distances snail and turtle cover before they meet is x:11x respectively.

Simplified, the ratio of snail distance : turtle distance = 1:11

So snail covers a distance of \frac{1}{12} × 360 = 30m

And turtle covers a distance of \frac{11}{12} × 360 = 330m

The time each took before they met is 40 × 60 = 2400 seconds

Snail's speed = \frac{30m}{2400s} = 0.0125m/s

Turtle's speed =  \frac{330m}{2400s} = 0.1375m/s

8 0
2 years ago
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A girl is shown at position A on a swing when the seat is directly below the support bar. The seat is then at height A as shown
MrRa [10]

Answer:

<u></u>

  • <u>1. The potential energy of the swing is the greatest at the position B.</u>

  • <u>2. As the swing moves from point B to point A, the kinetic energy is increasing.</u>

Explanation:

Even though the syntax of the text is not completely clear, likely because it accompanies a drawing that is not included, it results clear that the posittion A is where the seat is at the lowest position, and the position B is upper.

The gravitational <em>potential energy </em>is directly proportional to the height of the objects with respect to some reference altitude. Thus, when the seat is at the position A the swing has the smallest potential energy and when the seat is at the <em>position B the swing has the greatest potential energy.</em>

Regarding the forms of energy, as the swing moves from point B to point A, it is going downward, gaining kinetic energy (speed) at the expense of the potential energy (losing altitude). When the seat passes by the position A, the kinetic energy is maximum and the potential energy is miminum. Then the seat starts to gain altitude again, losing the kinetic energy and gaining potential energy, up to it gets to the other end,

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2 years ago
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If in a vernier callipers 10 VSD coincides with 8 MSD then the least count of varnier calliper is
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Explanation:

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The least count of the instrument is the difference between the length of one MSD and length of one VSD

The length of inebriated MSD=1mm

Therefore,

The least count=1-0.8=0.2mm

3 0
2 years ago
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