PHYSICS COLLEGE

Consider different points along one spoke of a wheel rotating with constant angular velocity. Which of the following is true regarding the centripetal acceleration at a particular instant of time?a. The magnitude of the centripetal acceleration is greater for points on the spoke closer to the hub than for points closer to the rim
b. both the magnitude and the direction of the centripetal acceleration depend on the location of the point on the spoke.
c. The magnitude of the centripetal acceleration is smaller for points on the spoke closer to the hub than for points closer to the rim but the direction of the acceleration is the same at all points on this spoke.
d. The magnitude and direction of the centripetal acceleration is the same at all points on this spoke.

Answers

Answer 1

Answer:

Answer:

Option (a).

Explanation:

Let the angular velocity is w.

The centripetal acceleration is given by

$a = r w^2$

where, r is the distance between the axle and the spoke.

So, more is the distance more is the centripetal acceleration.

(a) For the points on the spoke closer to the hub than for points closer to the rim is larger distance, so the centripetal force is more.

The statement is true.

(b) The direction of centripetal acceleration is always towards the center, so the statement is false.

(d) It is false.

Option (a) is correct.

Related Questions

A. With what speed must a ball be thrown vertically from ground level to rise to a maximum height of 45m?b. How long will it be in the air?
How does the force of gravity change as the mass of one object doubles?
A particle located at the position vector m has a force N acting on it. The torque about the origin is
A small space probe of mass 170 kg is launched from a spacecraft near Mars. It travels toward the surface of Mars, where it will eventually land. At a time 22.9 seconds after it is launched, the probe is at location <5600, 7200, 0> m, and at this same instant its momentum is <51000, -7000, 0> kg·m/s. At this instant, the net force on the probe due to the gravitational pull of Mars plus the air resistance acting on the probe is <-4000, -780, 0> N.Assuming that the net force on the probe is approximately constant during this time interval, what is the change of the momentum of the probe in the time interval from 22.6 seconds after the probe is launched to 22.9 seconds after the launch?
Box 1 and box 2 are whirling around a shaft with a constant angular velocity of magnitude ω. Box 1 is at a distance d from the central axis, and box 2 is at a distance 2d from the axis. You may ignore the mass of the strings and neglect the effect of gravity. Express your answer in terms of d, ω, m1 and m2, the masses of box 1 and 2. (a) Calculate TB, the tension in string B (the string connecting box 1 and box 2). (b) Calculate TA, the tension in string A (the string connecting box 1 and the shaft).

An electron has a velocity of 3.2 x 10^6 m/s. What is its’ momentum? (b) What is its’ wavelength? (c) What other objects/materials have this space/size? (d) Assuming that we can measure the velocity to an accuracy of 10%. Use the Heisenberg uncertainty principle to calculate the uncertainty in the position.

Answers

Answer:

P = 2.91*10^{-24} kg m/s

$\lambda = 2.73 *10^(-10) m$

size of atom hat lie in range of 1 to 5 Angstrom

$\Delta x = 0.2272$ Angstrom

Explanation:

A) MOMENTUM

p = mv

where m is mass of electron

so momentum p can be calculated as

p = 9.11*10^{-31} *3.2*10^{6}

P = 2.91*10^{-24} kg m/s

b) wavelength

$\lambda = (h)/(mv)$

where h is plank constant

so $\lambda = (6.626*10^(-34))/(2.91*10^(-24))$

$\lambda = 2.73 *10^(-10) m$

c) size of atom hat lie in range of 1 to 5 Angstrom

d) from the information given in the question we have

$(\Delta v)/(v) = 0.1$

$\Delta v = 0.1 v$

we know that

$\Delta p *\Delta x = (h)/(4\pi)$

$m \Delta v \Delta x =(h)/(4\pi)$

$\Delta x = (h)/(m \Delta v)$

$\Delta x = (2.272)/(0.1)$ [ $\Delta v = 0.1 v$ ]

$\Delta x = 0.2272$ Angstrom

Using energy considerations, calculate the average force (in N) a 62.0 kg sprinter exerts backward on the track to accelerate from 2.00 to 6.00 m/s in a distance of 25.0 m, if he encounters a headwind that exerts an average force of 30.0 N against him.

Answers

Answer:

69.68 N

Explanation:

Work done is equal to change in kinetic energy

W = ΔK = Kf - Ki = $(1)/(2) mv^(2) _(f) - (1)/(2) mv^(2) _(i)$

W = $F_(total) .d$

where m = mass of the sprinter

vf = final velocity

vi = initial velocity

W = workdone

kf = final kinetic energy

ki = initial kinetic energy

d = distance traveled

Ftotal = total force

vf = 8m/s

vi= 2m/s

d = 25m

m = 60kg

inserting parameters to get:

W = ΔK = Kf - Ki = $(1)/(2) mv^(2) _(f) - (1)/(2) mv^(2) _(i)$

$F_(total) .d =(1)/(2) mv^(2) _(f) - (1)/(2) mv^(2) _(i)$

$F_(total) = ((1)/(2) mv^(2) _(f) - (1)/(2) mv^(2) _(i))/(d)$

$F_(total=) ((1)/(2) X 62 X6^(2) -(1)/(2) X 62 X2^(2) )/(25)$

= 39.7

we know that the force the sprinter exerted F sprinter, the force of the headwind Fwind = 30N

$F_(sprinter) = F_(total) + F_(wind) = 39.7 + 30 = 69.68 N$

Answer:

Force exerted by sprinter = 69.68 N

Explanation:

From work energy theorem, we know that, work done is equal to change in kinetic energy.

Thus,

W = ΔK = Kf - Ki = (1/2)m•(v_f)² - (1/2)m•(v_i)² - - - - eq(1)

Now,

Work done is also;

W = Force x Distance = F•d - - - (2)

From the question, we are given ;

v_f = 6 m/s

v_i = 2 m/s

d = 25m

m = 62 kg

Equating equation 1 and 2,we get;

(1/2)m•(v_f)² - (1/2)m•(v_i)² = F•d

Plugging in the relevant values to obtain ;

(1/2)(62)[(6)² - (2)²] = F x 25

31(36 - 4) = 25F

992 = 25F

F = 39.68 N

The force the sprinter exerts backward on the track will be the sum of this force and the headwind force.

Thus,

Force of sprinter = 39.68 + 30 = 69.68N

Which of the following types of waves is not part of the electromagnetic spectrum? A) microwaves
B) gamma rays
C) ultraviolet radiation
D) radio waves
E) sound waves

Answers

Answer: Sound Waves

Explanation:

Sound waves are the only waves on this list that are not part of the electromagnetic spectrum. This is because sound waves require a medium to travel (molecules to transmit the sound waves), while waves on the electromagnetic spectrum do not require a medium. They are able to travel through space for example, while sound would not be able to.

Sound waves (E) are not electromagnetic at all.

Microwaves, gamma rays, ultraviolet waves, and radio waves all are.

High-energy particles are observed in laboratories by photographing the tracks they leave in certain detectors; the length of the track depends on the speed of the particle and its lifetime. A particle moving at 0.993c leaves a track 1.15 mm long. What is the proper lifetime of the particle

Answers

Answer:

Lifetime = 4.928 x 10^-32 s

Explanation:

(1 / v2 – 1 / c2) x2 = T2

T2 = (1/ 297900000 – 1 / 90000000000000000) 0.0000013225

T2 = (3.357 x 10^-9 x 1.11 x 10^-17) 1.3225 x 10^-6

T2 = (3.726 x 10^-26) 1.3225 x 10^-6 = 4.928 x 10^-32 s

Final answer:

To find the proper lifetime of the particle, we can use the time dilation equation and the Lorentz factor. Plugging in the given values, we find that the proper lifetime of the particle is approximately 5.42 × 10^-9 seconds.

Explanation:

To find the proper lifetime of the particle, we can use the time dilation equation, which states that the proper time (time experienced in the frame of reference of the particle) is equal to the time observed in the laboratory frame of reference divided by the Lorentz factor. The Lorentz factor can be calculated using the equation γ = 1/√(1 - (v/c)^2), where v is the velocity of the particle and c is the speed of light. Given that the particle is moving at 0.993c, the Lorentz factor is approximately 22.82.

Next, we can use the equation Δx = βγcτ, where Δx is the length of the track, β is the velocity of the particle in units of the speed of light (v/c), γ is the Lorentz factor, c is the speed of light, and τ is the proper lifetime of the particle. Plugging in the given values, we have 1.15 mm = 0.993 * 22.82 * c * τ. Solving for τ, we find that the proper lifetime of the particle is approximately 5.42 × 10^-9 seconds.

Learn more about Proper lifetime of high-energy particles here:

brainly.com/question/31962225

#SPJ3

Find the average speed of the electrons in a 1.0 cm diameter, copper power line, when it carries a current of 20 A.

Answers

Answer:

Average speed $v_d=(i)/(neA)=(20)/(8.5* 10^(28)* 1.6* 10^(-19)* 7.8* 10^(-5))=1.87* 10^(-5)m/sec$

Explanation:

We have given current through power i = 20 A

Diameter d = 1 cm = 0.01 m

So radius r = 0.005 m

So area $A=\pi r^2=3.14* 0.005^2=7.8* 10^(-5)m^2$

Charge on electron e = $1.6* 10^(-19)C$

We know that current is given by $i=neAv_d$ , here n is nuber density of free electron, e is charge on electron, A is area and $v_d$ is average speed

We know that for copper n = $8.5* 10^(28)per\ m^3$

So average speed $v_d=(i)/(neA)=(20)/(8.5* 10^(28)* 1.6* 10^(-19)* 7.8* 10^(-5))=1.87* 10^(-5)m/sec$

Why We can’t Cure Aging? Support your answer?

Answers

We can’t cure aging because it is inevitable. Aging is a part of life and it leads to death which is also inevitable and inescapable. We can’t stop our aging because it is natural and normal to age

Other Questions

Calculate the speed (in m/sec) of a wave with a wavelength of 2.1 meters and a period of 9.4 second.

A boy and a girl are pulling a heavy crate at the same time with 7 units of firce each. What is the net force acts on the ibject? Is the object balanced or unbalanced?

Suppose that a barometer was made using oil with rho=900 kg/m3. What is the height of the barometer at atmospheric pressure?

A proton is moving horizontally halfway between two parallel plates that are separated by 0.60 cm. The electric field due to the plates has magnitude 720,000 N/C between the plates away from the edges. If the plates are 5.6 cm long, find the minimum speed of the proton if it just misses the lower plate as it emerges from the field.