PHYSICS COLLEGE

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:

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.

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Two 10-cm-diameter charged rings face each other, 25.0cm apart. Both rings are charged to +20.0nC. What is the electric field strength at:a) the midpoint between the two rings?b) the center of the left ring?

Answers

The electric field strength at the midpoint of the rings is 0 N/C.

The electric field strength at the center of the left ring is 2710.84 N/C.

The given parameters:

Diameter of the rings, d = 10 cm
Distance between the rings, r = 25 cm
Charge of the rings, q = 20 nC

The electric field strength at the midpoint of the rings is calculated as follows;

$E_(net) = E_1 + E_2\n\nE_(net) = E_1(+ve) - E_2(-ve) = 0$

The electric field strength at the center of the left ring is calculated as follows;

$E = (kqL)/((R^2 + L^2)^(3/2)) \n\nE = (9* 10^9 * 20* 10^(-9) * 0.25 \ )/((0.05^2 + 0.25^2 )^(3/2)) \n\nE = 2710.84 \ N/C$

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Final answer:

The electric field strength at the midpoint between the two rings is zero, and at the center of the left ring, it is 2.88 * 10^4 N/C.

Explanation:

The electric field strength at the

midpoint between the two rings is zero. The electric fields from each ring cancel each other out at this point because they are equal in magnitude and opposite in direction.
At the center of the left ring, the electric field strength can be calculated using the formula for the electric field due to a charged ring. The formula is E = k * (Q / r²), where E is the electric field strength, k is the Coulomb's constant, Q is the charge, and r is the distance from the center of the ring. Plugging in the values, we get:

E = (8.99 * 10^9 Nm²/C²) * (20.0 * 10^-9 C) / (0.05 m)² = 2.88 * 10^4 N/C

You have a grindstone (a disk) that is 105.00 kg, has a 0.297-m radius, and is turning at 71.150 rpm, and you press a steel axe against it with a radial force of 46.650 N. Assuming the kinetic coefficient of friction between steel and stone is 0.451. How many turns will the stone make before coming to rest?

Answers

Answer:

3.27 turns

Explanation:

To find how many turns (θ) will the stone make before coming to rest we will use the following equation:

$\omega_(f)^(2) = \omega_(0)^(2) + 2\alpha*\theta$

Where:

$\omega_(f)$ : is the final angular velocity = 0

$\omega_(0)$ : is the initial angular velocity = 71.150 rpm

α: is the angular acceleration

First, we need to calculate the angular acceleration (α). To do that, we can use the following equation:

$\alpha = (\tau)/(I)$

Where:

I: is the moment of inertia for the disk

τ: is the torque

The moment of inertia is:

$I = (mr^(2))/(2)$

Where:

m: is the mass of the disk = 105.00 kg

r: is the radius of the disk = 0.297 m

$I = (105.00 kg*(0.297 m)^(2))/(2) = 4.63 kg*m^(2)$

Now, the torque is equal to:

$\tau = -F x r = -\mu*F*r$

Where:

F: is the applied force = 46.650 N

μ: is the kinetic coefficient of friction = 0.451

$\tau = -\mu*F*r = -0.451*46.650 N*0.297 m = -6.25 N*m$

The minus sign is because the friction force is acting opposite to motion of grindstone.

Having the moment of inertia and the torque, we can find the angular acceleration:

$\alpha = (-6.25 N*m)/(4.63 kg*m^(2)) = -1.35 rad/s^(2)$

Finally, we can find the number of turns that the stone will make before coming to rest:

$0 = \omega_(0)^(2) + 2\alpha*\theta$

$\theta = -((\omega_(0))^(2))/(2\alpha) = -((71.150 (rev)/(min))^(2))/(2*(-1.35 (rad)/(s^(2)))*(1 rev)/(2\pi rad)*((60 s)^(2))/((1 min)^(2))) = 3.27 rev = 3.27 turns$

I hope it helps you!

How does the geosphere interact with the hydrosphere

Answers

Plants (biosphere) draw water (hydrosphere) and nutrients from the soil (geosphere) and release water vapor into the atmosphere.

What form of braking is used to bring a vehicle to a smooth stop by applying smooth,steady pressure to the braketrail braking
controlled braking
threshold braking
coasting

Answers

Answer:

Controlled braking

Explanation:

CONTROLLED BRAKING occur in a situation where a person or an individual driving a vehicle releases the brake and slowly apply smooth as well as firmly pressure on the brake without the wheels been locked which is why CONTROLLED BRAKING are often used for emergency stops by drivers reason been that it helps to reduce speed when driving as fast as possible while the driver maintain the steering control of the vehicle.

Therefore the form of braking which is used to bring a vehicle to a smooth stop by applying smooth,steady pressure to the brake is called CONTROLLED BRAKING.

Final answer:

The method of braking that involves applying smooth, steady pressure to the brake to bring the vehicle to a smooth stop is called controlled braking. It helps prevent skidding and provides the driver with more control over the vehicle.

Explanation:

The form of braking used to bring a vehicle to a smooth stop by applying smooth, steady pressure to the brake is known as controlled braking. This method of braking involves applying consistent, even pressure to the brake pedal, which allows the car to slow down gently and gradually. It helps prevent uncontrolled skidding and provides the driver with more control over the vehicle's direction and speed during the stop. Unlike other methods like trail braking, threshold braking, or coasting, controlled braking is typically the safest and most effective method for daily driving conditions.

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Two microwave frequencies are authorized for use in microwave ovens: 895 and 2560 mhz. calculate the wavelength of each.

Answers

Wavelength = (speed) / (frequency)

The speeds of the two possible signals are equal, just like
all other forms of electromagnetic radiation.

Wavelength of 895 MHz = (3 x 10⁸ m/s) / (8.95 x 10⁸/s) = 0.335 m

Wavelength of 2560 MHz = (3 x 10⁸ m/s) / (2.56 x 10⁹/s) = 0.117 m

A 0.060 ???????? tennis ball, moving with a speed of 5.28 m/???? , has a head-on collision with a 0.080 ???????? ball initially moving in the same direction at a speed of 3.00 m/ ???? . Assume that the collision is perfectly elastic. Determine the velocity (speed and direction) of both the balls after the collision.