PHYSICS COLLEGE

An athlete swings a 6.50-kg ball horizontally on the end of a rope. The ball moves in a circle of radius 0.900 m at an angular speed of 0.700 rev/s. (a) What is the tangential speed of the ball

Answers

Answer 1

Answer:

Answer:

v = 3.951 m/s

Explanation:

Given that,

Mass of a ball, m = 6.5 kg

Radius of the circle, r = 0.9 m

Angular speed of the ball, $\omega=0.7\ rev/s=4.39\ rad/s$

Let v is the tangential speed of the ball. It is given in terms of angular speed is follows :

$v=r\omega\n\nv=0.9* 4.39\n\nv=3.951\ m/s$

So, the tangential speed of the ball is 3.951 m/s.

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Part 1) A cop car traveling at 25 m/s has a siren producing a frequency of 700 Hz. A felon jumps on his motorcycle and speed off in the opposite direction of 15 m/s. What frequency does the felon hear as he sped away (speed of sound is 343 m/s)?Part 2) The cop does a U-turn and speeds towards the felon at 30 m/s, while the felon speeds up to 20 m/s. What frequency does the felon hear as he sped away (speed of sound is 343 m/s)?
Part 3) What if the felon then sped up to 30 m/s and all other conditions remained the same?

Answers

1) 621.8 Hz

2) 719.3 Hz

3) 700 Hz

Explanation:

The Doppler effect occurs when there is a source of a wave in relative motion with respect to an observer.

When this happens, the frequency of the wave appears shifted to the observer, according to the equation:

$f'=(v\pm v_o)/(v \pm v_s)f$

where

f is the real frequency of the sound

f' is the apparent frequency of the sound

v is the speed of the sound wave

$v_o$ is the velocity of the observer, which is negative if the observer is moving away from the source, positive if the observer is moving towards the source

$v_s$ is the velocity of the source, which is negative if the source is moving towards the observer, positive if the source is moving away

In this problem we have:

f = 700 Hz is the frequency of the siren

v = 343 m/s is the speed of sound

$v_s=-25 m/s$ is the velocity of the car with the siren

$v_o = +15 m/s$ is the velocity of the felon (he's moving away from the siren)

So, the frequency heard by the felon is

$f=(343-25)/(343+15)(700)=621.8 Hz$

In this case, the cop does a U-turn and speeds towards the felon at 30 m/s.

This means that now the siren is moving towards the observer (so, $v_s$ becomes positive), while the sign of $v_o$ still remains positive.

So we have:

f = 700 Hz is the frequency of the siren

v = 343 m/s is the speed of sound

$v_s=+30 m/s$ is the velocity of the car with the siren

$v_o = +20 m/s$ is the velocity of the felon

So, the frequency heard by the felon is

$f=(343+30)/(343+20)(700)=719.3 Hz$

In this case, the felon speeds up to 30 m/s.

This means that now the felon and the siren are moving with the same relative velocity: so, it's like they are not moving relative to each other, so the frequency will not change.

In fact we have:

f = 700 Hz is the frequency of the siren

v = 343 m/s is the speed of sound

$v_s=+30 m/s$ is the velocity of the car with the siren

$v_o = +30 m/s$ is the velocity of the felon

So, the frequency heard by the felon is

$f=(343+30)/(343+30)(700)=700 Hz$

So, the frequency will not change.

A dentist causes the bit of a high-speed drill to accelerate from an angular speed of 1.72 x 10^4 rad/s to an angular speed of 5.42 x 10^4 rad/s. In the process, the bit turns through 1.72 x 10^4 rad. Assuming a constant angular acceleration, how long would it take the bit to reach its maximum speed of 8.42 x 10^4 rad/s, starting from rest?

Answers

Answer:

The bit take to reach its maximum speed of 8,42 x10^4 rad/s in an amount of 1.097 seconds.

Explanation:

ω1= 1.72x10^4 rad/sec

ω2= 5.42x10^4 rad/sec

ωmax= 8.42x10^4 rad/sec

θ= 1.72x10^4 rad

$\alpha = (w2^(2)-w1^(2) )/(2*(\theta2 - \theta1))$

α=7.67 x10^4 rad/sec²

t= ωmax / α

t= 8.42 x10^4 rad/sec / 7.67 x10^4 rad/sec²

t=1.097 sec

A grandfather clock keeps time using a pendulum consisting of a light rod connected to a small heavy mass. With a rod of length L, the period of oscillation is 2.00 s. What should the length of the rod be for the period of the oscillations to be 1.00 s?

Answers

Answer:

The length of the rod should be

$(L)/(4) \n$

Explanation:

Period of simple pendulum is given by

$T=2\pi\sqrt{(l)/(g)} \n$

We have

$(T_1^2)/(T_2^2)=(l_1)/(l_2)\n\n(2^2)/(1^2)=(L)/(l_2)\n\nl_2=(L)/(4) \n$

The length of the rod should be

$(L)/(4) \n$

wo balls have the same mass of 5.00 kg. Suppose that these two balls are attached to a rigid massless rod of length 2L, where L = 0.550 m. One is attached at one end of the rod and the other at the middle of the rod. If the rod is held by the open end and rotates in a circular motion with angular speed of 45.6 revolutions per second,

Answers

Answer:

$T_1 =677224.40\ N$

Explanation:

given,

mass of the both ball = 5 Kg

length of rod = 2 L

where L = 0.55 m

angular speed = 45.6 rev/s

ω = 45.6 x 2 π

ω = 286.51 rad/s

v₁ = r₁ ω₁

v₁ =0.55 x 286.51 = 157.58 m/s

v₂ = r₂ ω₂

v₂ = 1.10 x 286.51 = 315.161 m/s

finding tension on the first half of the rod

r₁ = 0.55 r₂ = 2 x r₁ = 1.10

$T_1 = m ((v_1^2)/(r_1)+(v_2^2)/(r_2))$

$T_1 = 5 ((157.58^2)/(0.55_1)+(315.161^2)/(1.1))$

$T_1 =677224.40\ N$

Instantaneous speed is...a) A speed of 1000 km/h
b) The speed attained at a particular instant in time.
c) The speed that can be reached in a particular amount of time.

PLEASE HURRY

Answers

Answer:

The speed attained at a particular instant in time.

Explanation:

Instantaneous speed is the speed attained at a particular instant in time.

It is given by :

$v=(dx)/(dt)$

It is equal to the rate of change of speed.

It can be also defined as when the speed of an object is constantly changing, the instantaneous speed is the speed of an object at a particular moment (instant) in time.

Hence, the correct option is (b).

Which statement describes how a machine can help make work easier? It can put out more force than the input force by decreasing the distance over which force is applied. It can increase the amount of work performed if the output force is in the same direction as the input force. It can apply a greater output force over a greater distance compared to the input force and distance. It can increase the distance over which output force acts by applying less output force than input force.

Answers

Answer:

It can apply a greater output force over a greater distance compared to the input force and distance.

Explanation:

A machine helps to multiply force input and makes it easier to do work. Simple and complex machines are need to make work easier.

Most machines are designed to increase the input force by a system of mechanics.
Machines allow force to be applied over great distances and this allows for the input work done to be duly compensated for.

The statement describes how a machine can help make work easier, (c) It can apply a greater output force over a greater distance compared to the input force and distance is correct option.

A machine is a device designed to perform a specific task or work by using mechanical, electrical, or other forms of energy. Machines are created to make tasks easier, more efficient, or possible to accomplish that would be difficult or impossible to achieve using human effort alone.

They can range from simple tools like levers and pulleys to complex systems like engines and computers. Machines typically involve the conversion of input energy (such as human effort or electricity) into some form of output energy (such as mechanical motion or computation) to achieve a desired result.

To know more about machine

brainly.com/question/19427898

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