In 10s a car accelerates 4m/s^2 to 60m/s. How fast was the car going before it accelerated?

An ideal air-filled parallel-plate capacitor has round plates and carries a fixed amount of equal but opposite charge on its pla

Vikentia [17]

Answer:

The new energy density is same as initial energy density u0 as it is independent of plate dimensions

Explanation:

As we know that energy density is total energy per unit volume

so we know that isolated plates of capacitor is placed here

$C = \frac{\epsilon_0 A}{d}$

here we have energy density given as

$u = \frac{Q^2}{2(\epsilon_0 A/d)} Ad$

so we have

$u = \frac{Q^2d^2}{2\epsilon_0}$

so it is independent of the dimensions of the plate while total charge on the plates is constant always

so energy density will not change and remains the same

8 0

3 years ago

the declaration of Independence discusses the protection of natural rights. These rights include the right life liberty and the

nikdorinn [45]

Answer:

i can't understand ur question

4 0

3 years ago

1. A 2000kg truck at rest accelerates to 45m/s in 7 seconds. What is its change in kinetic energy?

Gemiola [76]

Answer:

2025000 J

Explanation:

The formula for kinetic energy is KE=.5(m)(v²).

The initial kinetic energy is 0 because it is at rest. .5(m)(0) = 0.

To calculate the final kinetic energy, use the kinetic energy equation. KE = .5(2000)(45²) = 2025000 J.

To find the change in kinetic energy, you do KE(f) - KE(i). 2025000-0 = 2025000 J.

5 0

4 years ago

A proposed space station includes living quarters in a circular ring 50.0 m in diameter. At what angular speed should the ring r

mel-nik [20]

Given that the space station is free of gravitational force, it is required that it spins an certain speed to acquire centripetal acceleration.

In this case, you want that the centripetal acceleration, Ac, equals g (gravitational acceleration on the earth), becasue this will cause a centripetal force equal to the weight on earth.

The formula for centripetal acceleration is Ac = [angular velocity]^2*R

where R = [1/2]50.0m = 25.0 m

Ac = 9.81 m/s^2

=> [angular velocity]^2 = Ac/R = 9.81m/s^2v/ 25.0m = 0.3924 (rad/s)^2

[angular velocity] = √(0.3924) rad/s = 0.63 rad/s

Answer: 0.63 rad/s

5 0

3 years ago

Particle A of charge 3.06 10-4 C is at the origin, particle B of charge -5.70 10-4 C is at (4.00 m, 0), and particle C of charge

Alex

Answer:

F_net = 26.512 N

Explanation:

Given:

Q_a = 3.06 * 10^(-4 ) C

Q_b = -5.7 * 10^(-4 ) C

Q_c = 1.08 * 10^(-4 ) C

R_ac = 3 m

R_bc = sqrt (3^2 + 4^2) = 5m

k = 8.99 * 10^9

Coulomb's Law:

F_i = k * Q_i * Q_j / R_ij^2

Compute F_ac and F_bc :

F_ac = k * Q_a * Q_c / R^2_ac

F_ac = 8.99 * 10^9* ( 3.06 * 10^(-4 ))* (1.08 * 10^(-4 )) / 3^2

F_ac = 33.01128 N

F_bc = k * Q_b * Q_c / R^2_bc

F_bc = 8.99 * 10^9* ( 5.7 * 10^(-4 ))* (1.08 * 10^(-4 )) / 5^2

F_bc = - 22.137 N

Angle a is subtended between F_bc and y axis @ C

cos(a) = 3 / 5

sin (a) = 4 / 5

Compute F_net:

F_net = sqrt (F_x ^2 + F_y ^2)

F_x = sum of forces in x direction:

F_x = F_bc*sin(a) = 22.137*(4/5) = 17.71 N

F_y = sum of forces in y direction:

F_y = - F_bc*cos(a) + F_ac = - 22.137*(3/5) + 33.01128 = 19.72908 N

F_net = sqrt (17.71 ^2 + 19.72908 ^2) = 26.5119 N

Answer: F_net = 26.512 N

5 0

4 years ago