PHYSICS COLLEGE

As the captain of the scientific team sent to Planet Physics, one of your tasks is to measure g. You have a long, thin wire labeled 1.73 g/m and a 1.30 kg weight. You have your accurate space cadet chronometer but, unfortunately, you seem to have forgotten a meter stick. Undeterred, you first find the midpoint of the wire by folding it in half. You then attach one end of the wire to the wall of your laboratory, stretch it horizontally to pass over a pulley at the midpoint of the wire, then tie the 1.30 kg weight to the end hanging over the pulley. By vibrating the wire, and measuring time with your chronometer, you find that the wire's second harmonic frequency is 200 Hz . Next, with the 1.30 kg weight still tied to one end of the wire, you attach the other end to the ceiling to make a pendulum. You find that the pendulum requires 313 s to complete 200 oscillations. Pulling out your trusty calculator, you get to work.What value of g will you report back to headquarters?

Answers

Answer 1

Answer:

Answer:

The value of g is $g =76.2 m/s^2$

Explanation:

From the question we are told that

The mass of the weight is $m = 1.30 kg$

The spring constant $k = 1.73 g/m = 1.73 *10^(-3) \ kg/m$

The second harmonic frequency is $f = 100 \ Hz$

The number of oscillation is $N = 200$

The time taken is $t = 315 \ s$

Generally the frequency is mathematically represented as

$f = (v)/(\lambda)$

At second harmonic frequency the length of the string vibrating is equal to the wavelength of the wave generated

$l = \lambda$

Noe from the question the vibrating string is just half of the length of the main string so

Let assume the length of the main string is $L$

So $l = (L)/(2)$

The velocity of the vibrating string is mathematically represented as

$v = \sqrt{(T)/(\mu) }$

Where T is the tension on the string which can be mathematically represented as

$T = mg$

$v = \sqrt{(mg)/(k) }$

Then

$f = (v)/((L)/(2) )$

=> $v = (fL )/(2)$

=> $\sqrt{(mg)/(k) } = (fL)/(2)$

=> $g = (f^2 L^2 \mu)/(4m)$

substituting values

$g = ((100) * (1.73 *10^(-3) ))/((4 * 1.30)) L^2$

$g = 3.326 m^(-1) s^(-2) L^2$

Generally the period of oscillation is mathematically represented as

$T_p = 2 \pi \sqrt{(L)/(g) }$

=> $L = (T^2 g)/(4 \pi ^2)$

The period can be mathematically evaluated as

$T_p = (t)/(N)$

substituting values

$T_p = (315)/(200)$

$T_p = 1.575 \ s$

Therefore

$L = (1.575^2 * g )/(4 \pi ^2)$

$L = 0.0628 ^2 g$

$g = 3.326 m^(-1) s^(-2) L^2$

substituting for L

$g = 3.326 ((0.0628) g)^2$

=> $g = (1)/((3.326)* (0.0628)^2)$

$g =76.2 m/s^2$

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Sharece knows that wave peaks and valleys can add and subtract. What would be the net effect if she was able to cross Wave 1 (a large-amplitude wave in a valley phase) with Wave 2 (a wave with slightly smaller amplitude than Wave 2, in a peak phase)?Sharece knows that wave peaks and valleys can add and subtract. What would be the net effect if she was able to cross Wave 1 (a large-amplitude wave in a valley phase) with Wave 2 (a wave with slightly smaller amplitude than Wave 2, in a peak phase)?

Answers

Answer:

The two waves will add vectorially to produce a small amplitude wave in a valley phase.

Explanation:

The two waves will add vectorially to produce a small amplitude wave in a valley phase. This is because the amplitudes of the waves are slightly different and in opposite directions. When wave 1 cancels out all of wave 2, the resultant wave would be the slight difference between both waves, and it would be in the direction of wave 1 which is a valley phase.

The concept of photons applies to which regions of the electromagnetic spectrum?A. visible light only
B. infrared light, visible light, and UV light only
C. X-rays and gamma rays only
D. all regions of the spectrum

Answers

Answer:

D. all regions of the spectrum

Explanation:

I did some research ; )

In college softball, the distance from the pitcher's mound to the batter is 43 feet. If the ball leaves the bat at 110 mph , how much time elapses between the hit and the ball reaching the pitcher?

Answers

The time taken should be 0.000074 hours or 0.2664 seconds.

Calculation of the time taken:

Here we assume the time be t

And, The distance from the pitcher's mound to the batter is 43 feet, d = 43 feet = 0.00814 miles

So, the following formula should be used.

$= 0.00814 / 110$

= 0.000074 hours or 0.2664 seconds.

Learn more about the time here: brainly.com/question/15094745

Explanation:

It is given that,

The distance from the pitcher's mound to the batter is 43 feet, d = 43 feet = 0.00814 miles

Speed with which ball leaves the ball, v = 110 mph

Let t is the time elapses between the hit and the ball reaching the pitcher. It is given by :

$t=(d)/(v)$

$t=(0.00814)/(110)$

t = 0.000074 hours

t = 0.2664 seconds

So, the time between the hit and the ball reaching the pitcher is 0.2664 seconds. Hence, this is the required solution.

Crude oil is a mixture of many different components. The extraction of crude oil from the Earth is important, but its refinement into different substances is a key piece to obtaining as many uses as possible from the crude oil. Using the diagram, justify the source of data used to develop the technology for refining the crude oil.

Answers

Crude oil is a mixture of nitrogen, oxygen, sulphur, and hydrogen components

What is Crude oil?

A combination of hydrocarbons known as crude oil is one that is found in naturally occurring subsurface reservoirs in the liquid phase and continues to be liquid at atmospheric pressure after passing through surface separation equipment.

Refineries transform crude oil into useful products including gasoline, diesel, and aviation fuels for transportation. Gasoline: A fuel used in both personal and commercial vehicles that are made for internal combustion engines.

In addition to some nitrogen, sulphur, and oxygen, crude oil is a combination of very flammable liquid hydrocarbons (compounds mostly made of hydrogen and carbon).

More about the Crude oil link is given below.

brainly.com/question/351648

#SPJ6

Answer: Different fuel components boil at different temperatures, allowing them to be separated.

Explanation:

A Hooke's law spring is mounted horizontally over a frictionless surface. The spring is then compressed a distance d and is used to launch a mass m along the frictionless surface. What compression of the spring would result in the mass attaining double the kinetic energy received in the above situation?

Answers

Answer:

The compression is $√(2) \ d$ .

Explanation:

A Hooke's law spring compressed has a potential energy

$E_(potential) = (1)/(2) k (\Delta x)^2$

where k is the spring constant and $\Delta x$ the distance to the equilibrium position.

A mass m moving at speed v has a kinetic energy

$E_(kinetic) = (1)/(2) m v^2$ .

So, in the first part of the problem, the spring is compressed a distance d, and then launch the mass at velocity $v_1$ . Knowing that the energy is constant.