A bicycle racer is going downhill at 11.0 m/s when, to his horror, one of his 2.25 kg wheels comes off when he is 75.0 m above the foot of the hill. We can model the wheel as a thin-walled cylinder 85.0 cm in diameter and neglect the small mass of the spokes. (a) How fast is the wheel moving when it reaches the bottom of hill if it rolled without slipping all the way down? (b) How much total kinetic energy does it have when it reaches bottom of hill?

Answers

Answer 1
Answer:

Answer:

a.) Speed V = 29.3 m/s

b.) K.E = 1931.6 J

Explanation: Please find the attached files for the solution

Answer 2
Answer:

Final answer:

The wheel's speed at the bottom of the hill can be found through the conservation of energy equation considering both translational and rotational kinetic energy, while the total kinetic energy at the bottom of the hill is a sum of translational and rotational kinetic energy.

Explanation:

These two questions address the physics concepts of conservation of energy, kinetic energy, and rotational motion. To answer the first question, (a) How fast is the wheel moving when it reaches the bottom of the hill if it rolled without slipping all the way down?, we need to consider the potential energy the wheel has at the top of the hill is completely converted into kinetic energy at the bottom. This includes both translational and rotational kinetic energy. Solving for the final velocity, vf, which would be the speed of the wheel, we get vf = sqrt((2*g*h)/(1+I/(m*r^2))), where g is the acceleration due to gravity, h is the height of the hill, I is the moment of inertia of the wheel, m is the mass of the wheel, and r is the radius of the wheel.

For the second question, (b) How much total kinetic energy does it have when it reaches bottom of the hill?, we use the formula for total kinetic energy at the bottom of the hill, K= 0.5*m*v^2+0.5*I*(v/r)^2. Substituting the value of v found in the first part we find the kinetic energy which we can use the formula provided in the reference information.

Learn more about Conservation of Energy and Rotational Motion here:

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In a cyclotron, the orbital radius of protons with energy 300 keV is 16.0 cm . You are redesigning the cyclotron to be used instead for alpha particles with energy 300 keV . An alpha particle has charge q=+2e and massm=6.64×10−27kg .If the magnetic field isnt changed, what will be the orbital radius of the alpha particles? Express your answer to three significant figures and include the appropriate units.

Answers

Answer:

16 cm

Explanation:

For protons:

Energy, E = 300 keV

radius of orbit, r1 = 16 cm

the relation for the energy and velocity is given by

E = (1)/(2)mv^(2)

So, v = \sqrt{(2E)/(m)}   .... (1)

Now,

r = (mv)/(Bq)

Substitute the value of v from equation (1), we get

r = (√(2mE))/(Bq)

Let the radius of the alpha particle is r2.

For proton

So, r_(1) = \frac{\sqrt{2m_(1)E}}{Bq_(1)}    ... (2)

Where, m1 is the mass of proton, q1 is the charge of proton

For alpha particle

So, r_(2) = \frac{\sqrt{2m_(2)E}}{Bq_(2)}    ... (3)

Where, m2 is the mass of alpha particle, q2 is the charge of alpha particle

Divide equation (2) by equation (3), we get

(r_(1))/(r_(2))=(q_(2))/(q_(1))* \sqrt{(m_(1))/(m_(2))}

q1 = q

q2 = 2q

m1 = m

m2 = 4m

By substituting the values

(r_(1))/(r_(2))=\frac{2q}}{q}}* \sqrt{\frac{m}}{4m}}=1

So, r2 = r1 = 16 cm

Thus, the radius of the alpha particle is 16 cm.

Answer:15.95 cm

Explanation:

Given

Energy=300 kev

radius of Proton=16 cm

mass of alpha particle=6.64* 10^(-27) kg

mass of proton=1.67* 10^(-27) kg

charge on alpha particle is twice of proton

radius of Proton is given by

r=(mv)/(|q|B)

and Kinetic energy K=(P^2)/(2m)

where P=momentum

P=√(2Km)

r=(√(2km))/(qB)---1

Radius for Alpha particle is

r_(alpha)=\frac{\sqrt{2k\cdot m_(alpha)}}{2qB}-----2

Divide 1 & 2 we get

(r)/(r_(alpha))=(√(m))/(q)* \frac{2q}{\sqrt{m_(alpha)}}

r_(alpha)=\sqrt{(6.64* 10^(-27))/(1.67* 10^(-27))}* 0.5

r_(alpha)=0.997* 16

r_(alpha)=15.95 cm

You are exiting a highway and need to slow down on the off-ramp in order to make the curve. It is rainy and the coefficient of static friction between your tires and the road is only 0.4. If the radius of the off-ramp curve is 36 m, then to what speed do you need to slow down the car in order to make the curve without sliding?

Answers

Answer:

11.87m/s

Explanation:

To solve this problem it is necessary to apply the concepts related to frictional force and centripetal force.

The frictional force of an object is given by the equation

F_r = \mu N

Where,

\mu =Friction Coefficient

N = Normal Force, given also as mass for acceleration gravity

In the other hand we have that centripetal force is given by,

F_c= (mv^2)/(R)

The force experienced to stay on the road through friction is equal to that of the centripetal force, therefore

F_r = F_c

\mu mg = (mv^2)/(R)

Re-arrange to find the velocity,

V = √(R\mu g)

V = √(36*0.4*9.8)

V = 11.87m/s

Therefore the speed that it is necessaty to slow down the car in order to make the curve without sliding is 11.87m/s

Find the intensity III of the sound waves produced by four 60-WW speakers as heard by the driver. Assume that the driver is located 1.0 mm from each of the two front speakers and 1.5 mm from each of the two rear speakers.

Answers

Given that,

Power = 60 W

Distance = 1.0 m

Distance between speakers = 1.5 m

We need to calculate the intensity

Using formula of intensity

I_(1)=(P)/(A)

I_(1)=(P)/(4\pi r^2)

Put the value into the formula

I_(1)=(60)/(4\pi*(1.0)^2)

I_(1)=4.77\ W/m^2

We need to calculate the intensity

Using formula of intensity

I_(2)=(P)/(A)

I_(2)=(P)/(4\pi r^2)

Put the value into the formula

I_(1)=(60)/(4\pi*(1.5)^2)

I_(1)=2.12\ W/m^2

We need to calculate the intensity of the sound waves produced by four speakers

Using formula for intensity

I'=(I_(1)+I_(2))*2

Put the value into the formula

I'=(4.77+2.12)*2

I'=13.78\ W/m^2

Hence, The intensity of the sound waves produced by four speakers is 13.78 W/m².

Determine whether the following statements are true and give an explanation or counterexample.(A) If the acceleration of an object remains constant, its velocity is constant.
(B) If the acceleration of object moving along a line is always 0, then its velocity is constant.
(C) It is impossible for the instantaneous velocity at all times a(D) A moving object can have negative acceleration and increasing speed.

Answers

Answer:

Explanation:(A)if a body is accelerating then it's velocity can't be constant since an object is said to be accelerating if it is changing velocity (B)if the acceleration of an object moving along a line is 0 then it's velocity will be constant since there is no change in direction or speed(C)No.it is not possible for a moving body to have an instantaneous velocity at all times since instantaneous velocity is the velocity of a body at a certain instant of time..(D)Yes a moving object can have a negative acceleration and increasing speed,it can also have a positive acceleration with decreasing speed.

A rocket is attached to a toy car that is confined to move in the x-direction ONLY. At time to = 0 s, the car is not moving but the rocket is lit, so the toy car accelerates in the +x-direction at 5.35 m/s2. At t; = 3.60 s, the rocket's fuel is used up, and the toy car begins to slow down at a rate of 1.95 m/s2 because of friction. A very particular physics professor wants the average velocity for the entire trip of the toy car to be +6.50 m/s. In order to make this happen, the physics professor plans to push the car (immediately after it comes to rest by friction) with a constant velocity for 4.50 sec. What displacement must the physics professor give the car (immediately after it comes to rest by friction) in order for its average velocity to be +6.50 m/s for its entire trip (measured from the time the rocket is lit to the time the physics professor stops pushing the car)?

Answers

Answer:

What displacement must the physics professor give the car

= 12.91 METERS

Explanation:

Check the attached file for explanation

An electric current in a conductor varies with time according to the expression i(t) = 110 sin (120πt), where i is in amperes and t is in seconds. what is the total charge passing a given point in the conductor from t = 0 to t = 1/180 s?

Answers

As we know that current is defined as rate of flow of charge

i = (dq)/(dt)

so by rearranging the equation we can say

q = \int i dt

here we know that

i(t) = 110 sin(120\pi t)

here we will substitute it in the above equation

q = \int 110 sin(120\pi t) dt

q = 110 [- (cos(120\pi t))/(120\pi)]

now here limits of time is from t = 0 to t = 1/180s

so here it will be given as

q = (110)/(120\pi)( -cos0 + cos((2\pi)/(3)))

q = 0.44 C

so total charge flow will be 0.44 C

Answer:

The total charge passing a given point in the conductor is 0.438 C.

Explanation:

Given that,

The expression of current is

i(t)=110\sin(120\pi t)

(dq(t))/(t)=110\sin(120\pi t)

dq(t)=110\sin(120\pi t)dt....(I)

We need to calculate the total charge

On integrating both side of equation (I)

\int_(0)^(q)dq(t)=\int_(0)^{(1)/(180)}110\sin(120\pi t)dt

q=110((-\cos(120\pi t))/(120\pi))_(0)^{(1)/(180)}

q=-(110)/(120\pi)(cos(120\pi((1)/(180)))-\cos120\pi(0))

q=-0.2918(-(1)/(2)-1)

q=0.438\ C

Hence,  The total charge passing a given point in the conductor is 0.438 C.