PHYSICS HIGH SCHOOL

two pendulums of lengths 100cm and 110.25cm start oscillating in phase. after how many oscillations will they again be in same phase?

Answers

Answer 1

Answer:

Angular frequency of pendulum is given by

$\omega = \sqrt{(g)/(l)}$

for both pendulum we have

$\omega_1 = \sqrt{(9.81)/(1.00)}$

$\omega_1 = 3.13 rad/s$

For other pendulum

$\omega_2 = \sqrt{(9.81)/(1.1025)}$

$\omega_2 = 2.98 rad/s$

now we have relate angular frequency given as

[tex\omega_1 - \omega_2 = 3.13 - 2.98 = 0.15 rad/s[/tex]

now time taken to become in phase again is given as

$t = (2\pi)/(\omega_1 - \omega_2)$

$t = (2\pi)/(0.15) = 41.88 s$

now number of oscillations complete in above time

$N = (t)/((2\pi)/(\omega_1))$

$N = (41.88)/((2\pi)/(3.13))$

$N = 21 oscillation$

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calculate the area that will experience a force of 6000N from liquid exerting pressure of 300Pa. 1kPa=1000Pa

Answers

Answer:

$A=20\ m^2$

Explanation:

Pressure

Is a measure of how much force is applied perpendicular over a given area.

It is calculated by dividing the amount of force being applied F by the area over which it is being applied A:

$\displaystyle P=(F)/(A)$

The SI unit for pressure is the Pascal (Pa), equivalent to $1\ Nw/m^2$

A pressure of 300 Pa is being exerted over an area that experiences a force of 6000 N. We are required to calculate the value of the area.

If we know the pressure and the force, we solve the above equation for A;

$\displaystyle A=(F)/(P)$

Substituting the given values:

$\displaystyle A=(6000)/(300)$

$\mathbf{A=20\ m^2}$

In this graph, calculate the speed of
segment A in m/s?

Answers

Answer:

The answer is Speed=2m/s

Explanation:

S=D/T

S=10m/5s

S=2m/s

A slingshot is used to launch a stone horizontally from the top of a 20.0 meter cliff. The stone lands 36.0 meters away.a. At what speed was the stone launched?
b. What is the speed and angle of impact?

Answers

As per the question a stone is projected horizontally from the top of a 20 meter cliff.

Hence the height [h] of the cliff = 20 m.

The stone lands 36.0 m away.

Hence horizontal distance R] = 36 m.

First we are asked to calculate the launching speed of the stone.

Let it be denoted as u.

The motion of particle is the force of gravity only.Hence it is a projectile motion.

The total time taken by a projectile when it is fired horizontally at a height h from the ground is -

$T=\sqrt{(2h)/(g) }$ [ where g is the acceleration due to gravity = 9.8 m/s^2 or 10 m/s^2]

$=\sqrt{(2*20)/(10) } s$

$= 2 s$

The range of the projectile will be-

$R= \ horizontal\ speed*\ total\ time\ of\ flight$

The horizontal speed for a projectile is uniform through out the motion of the projectile as long as gravity is constant.

In this situation the horizontal velocity is equal to the initial projected speed i.e u.

Hence horizontal distance R = u ×T

Putting the value or R and T in the above formula we get-

$R=u*\sqrt{(2h)/(g) }$

$36 = u*\sqrt{(2*20)/(10) }$

$36 = u*2$

$u =(36)/(2)$

$u =18 m/s$

Now we are asked to calculate the angle of impact.

Initially the vertical velocity of the particle is 0. The path of the trajectory of the particle will be parabolic due to the force of gravity.At any instant of time the horizontal component of the instantaneous velocity of the of the particle will be constant .

Let the horizontal component is denoted as $V_(x) \ where \ V_(x) = u$

The vertical component is calculated as follows-

We know that v = u +at

Here v is the final velocity, u is the initial speed, a is the acceleration and t is the instantaneous time.

For vertically downward motion under gravity u = 0 and a = -g

Let the vertical velocity at any instant is denoted as $V_(y)$

Hence

$V_(y) = 0 -gt$

$V_(y) = gt$ [Here we are taking only the magnitude]

Let the resultant velocity makes an angle α with the horizontal when it strikes the ground.

The horizontal and vertical component is denoted in the diagram below-

From the figure we see that -

$tan\alpha =(V_(y) )/(V_(x) )$

$tan\alpha =(gt)/(u)$

$tan\alpha =(10*2)/(18)$

$tan\alpha =1.11111$

$\alpha =48.01^0$ [ans]

Hence the angle of impact is close to 48 .1 degree .

A)

It is a launch oblique, therefore the initial velocity in the vertical direction is zero. Space Hourly Equation in vertical, we have:

$S=S_(o)+v_(o)t+ (at^2)/(2) \n 20= (10t^2)/(2) \n t=2s$

Through Definition of Velocity, comes:

$\Delta v= (\Delta S)/(\Delta t) \n v_x= (36)/(2) \n \boxed {v_(x)=18m/s}$

B)

Using the Velocity Hourly Equation in vertical direction, we have:

$v_(y)=v_{y_(o)}+gt \n v_(y)=10*2 \n \boxed {v_(y)=20m/s}$

The angle of impact is given by:

$cos(\theta) =(v_(x))/(v_(y)) \n cos(\theta) = (18)/(20) \n cos(\theta) =0.9 \n arccos(0.9)=\theta \n \boxed {\theta \approx 25.84}$

If you notice any mistake in my english, please let me know, because i am not native.

If the velocity of an object changes from 65 m/s to 98 m/s during a time interval of 12 s, what is the acceleration of the object?

Answers

a=acceleration
vf=final velocity
v₀=original velocity
t=time period

a=(vf-v₀)/t
a=(98 m/s-65 m/s)/12s=33 m/s / 12 s=2.75 m/s²

Answer: the acceleration of the object is 2.75 m/s²

   Size of acceleration = (change in speed) / (time for the change)

Change in speed = (end speed) - (start speed)

   =    (98 m/s) - (65 m/s) = 33 m/s .

Time for the change = 12 s

Size of acceleration = (33 m/s) / (12 s)

   = (33/12) (m/s²)

   = 2.75 m/s² .

Yellow light has a longer wavelength than green light. Which color of light has the higher frequency?

Answers

well wavelength=(speed of wave)/frequency
so if wavelength is longer the frequency is shorter
since yellow light has longer wavelength than green light
yellow light has a shorter frequency than green light
so green light has a higher frequency

An outfielder throws a baseball with an initial speed of 81.8mi/h. Just before an infielder catches the ball at the same level, the ball’s speed is 110 ft/s. In foot-pounds, by how much is the mechanical energy of the ball– Earth system reduced because of air drag? (The weight of a baseball is 9.0 oz.)

Answers

Answer:

-20.158ft-lb

Explanation:

Check the attached files for the explanation.

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