PHYSICS COLLEGE

A small grinding wheel is attached to the shaft of an electric motor which has a rated speed of 3600 rpm. When the power is turned on, the unit reaches its rated speed in 5 s, and when the power is turned off, the unit coasts to rest in 70 s. Assuming uniformly accelerated motion, determine the number of revolutions that the motor executes (a) in reaching its rated speed, (b) in coasting to rest.

Answers

Answer 1

Answer:

(a) θ1 = 942.5rad, (b) θ2 = 13195 rad

Explanation:

(a) Given

ωo = 0 rad/s

ω = 3600rev/min = 3600×2(pi)/60 rad/s

ω = 377rad/s

t1 = 5s

θ1 = (ω + ωo)t/2

θ1 = (377 +0)×5/2

θ1 = 942.5 rads

(b) ωo = 377rad/s

ω = 0 rad/s

t2 = 70s

θ2 = (ω + ωo)t/2

θ2 = (377 +0)×70/2

θ2 = 13195 rad

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Answers

Answer: C

Frictional force

Explanation:

The description of the question above is an example of a circular motion.

For a car travelling in a curved path, the frictional force between the tyres and the road surface will provide the centripetal force.

Since the road is banked, and the cross section of the banked road is constructed like a ramp. The car drives transversely to the slope of the ramp, so that the wheels of one side of the car are lower than the wheels on the other side of the car, for cornering the banked road, the car will not rely only on the frictional force.

Therefore, the correct answer is option C - the frictional force.

A electromagnetic wave of light has a wavelength of 500 nm. What is the energy (in Joules) of the photon representing the particle interpretation of this light?

Answers

Answer:

Energy, $E=4.002* 10^(-19)\ J$

Explanation:

It is given that,

Wavelength of the photon, $\lambda=500\ nm=5* 10^(-7)\ m$

We need to find the photon representing the particle interpretation of this light. it is given by :

$E=(hc)/(\lambda)$

$E=(6.67* 10^(-34)* 3* 10^8)/(5* 10^(-7))$

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Guitar string has an overall length of 1.22 m and a total mass of 3.5 g before being strung on a guitar. Once it is used on the guitar, there is a distance of 70 cm between fixed end points. The guitar string is tightened to a tension of 255 N.What is the frequency of the fundamental wave on the guitar string?

Answers

Answer:

Fundamental frequency= 174.5 hz

Explanation:

We know

fundamental frequency= $(velocity)/(2 *length)$

velocity = $\sqrt{(tension)/(mass per unit length) }$

mass per unit length= $(3.5)/(1000*1.22)$ =0.00427 $(kg)/(m)$

Now calculating velocity v= $\sqrt{(255)/(0.00427) }$

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Question Part Points Submissions Used A car is stopped for a traffic signal. When the light turns green, the car accelerates, increasing its speed from 0 to 5.30 m/s in 0.812 s. (a) What is the magnitude of the linear impulse experienced by a 62.0-kg passenger in the car during the time the car accelerates? kg · m/s (b) What is the magnitude of the average total force experienced by a 62.0-kg passenger in the car during the time the car accelerates? N

Answers

(a) 328.6 kg m/s

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(b) 404.7 N

The linear impulse experienced by the passenger is also equal to the product between the average force and the time interval:

$I=F \Delta t$

where in this case

$I=328.6 kg m/s$ is the linear impulse

$\Delta t = 0.812 s$ is the time during which the force is applied

Solving the equation for F, we find the magnitude of the average force experienced by the passenger:

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A 81.0 kg diver falls from rest into a swimming pool from a height of 4.70 m. It takes 1.84 s for the diver to stop after entering the water. Find the magnitude of the average force exerted on the diver during that time.

Answers

Explanation:

The given data is as follows.

height (h) = 4.70 m, mass = 81.0 kg

t = 1.84 s

As formula to calculate the velocity is as follows.

$\nu$ = 2gh

= $2 * 9.8 m/s^(2) * 4.70 m$

= 92.12 $s^(2)$

As relation between force, time and velocity is as follows.

F = $(m * \nu)/(t)$

Hence, putting the given values into the above formula as follows.

F = $(m * \nu)/(t)$

= $(81.0 kg * 92.12 s^(2))/(1.84 s)$

= 4055.28 N

Thus, we can conclude that the magnitude of the average force exerted on the diver during that time is 4055.28 N.

A train station bell gives off a fundamental tone of 505 Hz as the train approaches the station at a speed of 27.6 m/s. If the speed of sound in air on that day is 339 m/s, what will be the apparent frequency of the bell to an observer riding the train

Answers

Answer:

Apparent frequency of the bell to the observer is 546.12 Hz

Explanation:

The frequency of train bell (frequency of source) = 505 Hz

The speed of train (observer) = 27.6 m/s

The speed of sound in the air is (velocity of sound) = 339 m/s

The apparent frequency of the bell to the observer is calculated as follows:

Apparent frequency of bell to the observer.

$= \text{frequency of source} * (Observer + velocity \ of \ sound )/( velocity \ of \ sound ) \n= 505 * (27.6 + 339)/(339) \n= 546.12 Hz$

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