A charge of 28 nC is placed in a uniform electric field that is directed vertically upward and that has a magnitude of 4 x 104 N/C. What work is done by the electric force when the charge moves (a) 0.45 m to the right; (b) 0.67 m upward; (c) 2.6 m at an angle of 45o downward from the horizontal?

(a) 0 J

(b) 0.7504 mJ

(c) -2.0691 mJ

The work done (W) by an electric force (F) when a charge moves a given displacement (d) is given by;

W = F x d cos θ         (i)

Where;

d = displacement

θ = angle between the electric force and the displacement.

F = the electric force acting on the charge and is equal to the product of the charge (Q) and the electric field (E). i.e;

F = Q x E

Substitute F = Q x E into equation (i) to give;

W = Q x E x d cos θ   (ii)

(a) The work done when the charge moves 0.45m to the right

Since the electric field is directed vertically upwards and the displacement is directed rightwards, it means that the angle between them = 90°

Given;

E = 4 x 10⁴N/C

Q = 28 nC = 28 x 10⁻⁹ C

d = 0.45m

θ = 90°

Substitute these values into equation (ii)

=> W = 28 x 10⁻⁹ x 4 x 10⁴ x 0.45 x cos 90°

=> W = 28 x 10⁻⁹ x 4 x 10⁴ x 0.45 x 0

=> W = 0 J

Therefore, the work done by the electric force when the charge moves 0.45 to the right is 0 J

(b) The work done when the charge moves 0.67m upward

Since the electric field is directed vertically upwards and the displacement is also directed upwards, it means that the angle between them = 0°

Given;

E = 4 x 10⁴N/C

Q = 28 nC = 28 x 10⁻⁹ C

d = 0.67m

θ = 0°

Substitute these values into equation (ii)

=> W = 28 x 10⁻⁹ x 4 x 10⁴ x 0.67 x cos 0°

=> W = 28 x 10⁻⁹ x 4 x 10⁴ x 0.67 x 1

=> W = 75.04 x 10⁻⁵ J

=> W = 0.7504 x 10⁻³ J

=> W = 0.7504m J

Therefore, the work done by the electric force when the charge moves 0.67 upwards is 0.7504m J

(c) The work done when the charge moves 2.6m  at 45° downward from the horizontal

Since the electric field is directed vertically upwards and the displacement is directed 45° downward from the horizontal, it means that the angle between them = 135°

Given;

E = 4 x 10⁴N/C

Q = 28 nC = 28 x 10⁻⁹ C

d = 2.6m

θ = 135°

Substitute these values into equation (ii)

=> W = 28 x 10⁻⁹ x 4 x 10⁴ x 2.6 x cos 135°

=> W = 28 x 10⁻⁹ x 4 x 10⁴ x 2.6 x -0.7071

=> W =  -205.908 x 10⁻⁵ J

=> W = -2.0591 x 10⁻³ J

=> W = -2.0591 mJ

Therefore, the work done by the electric force when the charge moves 2.6m at 45° downwards from the horizontal is -2.0691 mJ

it's when something is going so fast it makes a loud noise, breaking the sound barrier, for example: a jet makes a loud noise, not only cause of the jet but because they reached a point where they went so fast that you can year the noise from various parts, kinda like a police car, lets say the car was going   so fast that you hear the siren of the cop car right next to you, but the car is probably a mile a way, it's cos the siren was next to you a few seconds if you understand me

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft