Which is a characteristic of thermal energy transfer through convection

Answers

Answer 1

Answer:

Answer: The thermal energy transfer is When a fluid, such as air or a liquid, is heated and then travels away from the source, it carries the thermal energy along.

Explanation: heat transfer is called convection. hopefully this was helpful.

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If the frequency of a system undergoing simple harmonic motion doubles, by what factor does the maximum value of acceleration change?a. 4
b. 2/pi
c. 2
d. (2)^1/2

Answers

Answer:

the answers the correct one is a 4

Explanation:

The centripetal acceleration is by

a = v² / R

angular and linear velocities are related

v = w R

let's substitute

a = w² R

for initial condition

a₀ = w₀² R

suppose the initial angular velocity is wo, suppose the angular velocity doubles

a = (2w₀)² R

a = 4 (w₀² R)

a = 4 a₀

when reviewing the answers the correct one is a

Each shot of the laser gun most favored by Rosa the Closer, the intrepid vigilante of the lawless 22nd century, is powered by the discharge of a 1.89 Fcapacitor charged to 60.9 kV. Rosa rightly reckons that she can enhance the effect of each laser pulse by increasing the electric potential energy of the charged capacitor. She could do this by replacing the capacitor's filling, whose dielectric constant is 431, with one possessing a dielectric constant of 947.Required:
a. Find the electric potential energy of the original capacitor when it is charged. (in Joules)
b. Calculate the electric potential energy of the upgraded capacitor when it is charged. ( In Joules)

Answers

Answer:

$U = 3.505 *10^9 \ J$

$U_1 = 7.696 *10^9 \ J$

Explanation:

From the question we are told that

The capacitance is $C = 1.89 \ F$

The voltage is $V = 60.9 \ k V = 60.9 *10^(3) \ V$

The first dielectric constant is $\epsilon_1 = 431$

The second dielectric constant is $\epsilon_2 = 947$

Generally the electric potential energy is mathematically represented as

$U = (1)/(2) * C * V^2$

=> $U = (1)/(2) * 1.89 * (60.9 *10^(3))^2$

=> $U = 3.505 *10^9 \ J$

Generally the capacitance when the capacitor's filling was changed is

$C_n = 1.89 * (947)/(431)$

=> $C_n = 4.15$

Generally the electric potential energy when the capacitor's filling was changed is

$U_1 = (1)/(2) * C_1 * V^2$

=> $U_1 = (1)/(2) * 4.15 * (60.9 *10^(3))^2$

=> $U_1 = 7.696 *10^9 \ J$

For the system shown below, what is the critical angle (angle at which the system just begins to move)? Assume that the coefficient of friction between all flat surfaces is 0.0500 and that the pulley is frictionless. The mass of m1 is 76.00 kg and the mass of m2 is 194.00 kg. Express your answer in radians.

Answers

THIS IS A PROBLEM OF PHYSICS MECHANIC, PLEASE READ CAREFULLY THE ATTACHED FILE.

Final answer:

To find the critical angle, we need to consider the forces acting on the system. The weight and frictional force must be taken into account. By equating the forces and solving for the critical angle, we can determine at what angle the system just begins to move.

Explanation:

To determine the critical angle for the system shown, we need to consider the forces acting on the objects. The force pulling m1 downwards is its weight, which is equal to its mass multiplied by the acceleration due to gravity. The force preventing m1 from moving is the frictional force, which is equal to the coefficient of friction multiplied by the normal force. The normal force is the force exerted by the surface perpendicular to it, which is equal to the weight of m2 minus the weight of the hanging part of the rope.

At the critical angle, the force of friction is at its maximum value, which is equal to the coefficient of friction multiplied by the normal force. The force pulling m1 downwards is equal to the force of friction. By equating these forces and solving for the critical angle, we can find the answer.

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Name the four forces in physics?

Answers

Answer:

Gravitational

Electrostatic

magnetic

Frictional

gravitational

electrostatic

magnetic

frictional

hope it helps

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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:

a.) Speed V = 29.3 m/s

b.) K.E = 1931.6 J

Explanation: Please find the attached files for the solution

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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A revolutionary war cannon, with a mass of 2260 kg, fires a 21 kg ball horizontally. The cannonball has a speed of 105 m/s after it has left the barrel. The cannon carriage is on a flat platform and is free to roll horizontally. What is the speed of the cannon immediately after it was fired?