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ioda
2 years ago
11

Two parallel plates are 1 cm apart and are connected to a 500 V source. What force will be exerted on a single electron half way

between the plates?
Physics
1 answer:
VladimirAG [237]2 years ago
3 0

Answer: 8*10^-15 N

Explanation: In order to calculate the force applied on an electron in the middle of the two planes at 500 V we know that,  F=q*E

The electric field between  the plates is given by:

E = ΔV/d = 500 V/0.01 m=5*10^3 N/C

the force applied to the electron is: F=e*E=8*10^-15 N

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An ocean wave has an amplitude of 2.5 m. Weather conditions suddenly change such that the wave has an
Dovator [93]

Answer:

2.5

Explanation:2.5 +2.5 = 5.0

8 0
3 years ago
If a dog is mass iS 14.3 kg what is it’s weight on earth
BigorU [14]

Answer:

Weight of the dog on surface of earth is 140.14 Newton.

Given:

mass of the dog = 14.3 kg

To find:

Weight of the dog = ?

Formula used:

Weight of the dog is given by,

W = mg

Where, W = weight of the dog

m = mass of the dog

g = acceleration due to gravity

Solution:

Weight of the dog is given by,

W = mg

Where, W = weight of the dog

m = mass of the dog = 14.3 kg

g = acceleration due to gravity

W = 14.3 × 9.8

W = 140.14 Newton

Weight of the dog on surface of earth is 140.14 Newton.


8 0
3 years ago
After a great many contacts with the charged ball, how is the charge on the rod arranged (when the charged ball is far away)?
faust18 [17]

Answer: Option (b) is the correct answer.

Explanation:

Since, there is a negative charge present on the ball and a positive charge present on the rod. So, when the negatively charged metal ball will come in contact with the rod then positive charges from rod get conducted towards the metal ball.

Hence, the rod gets neutralized. But towards the metal ball there is a continuous supply of negative charges. Therefore, after the neutralization of positive charge from the rod there will be flow of negative charges from the metal ball towards the rod.

Thus, we can conclude that negative charge spread evenly on both ends.

8 0
3 years ago
A quarterback is set up to throw the football to a receiver who is running with a constant velocity v⃗ rv→rv_r_vec directly away
Artist 52 [7]

Answer:

a) V_o,y = 0.5*g*t_c

b) V_o,x = D/t_c - v_r

c) V_o = sqrt ( (D/t_c - v_r)^2 + (0.5*g*t_c)^2)

d)  Q = arctan ( g*t_c^2 / 2*(D - v_r*t_c) )

Explanation:

Given:

- The velocity of quarterback before the throw = v_r

- The initial distance of receiver = r

- The final distance of receiver = D

- The time taken to catch the throw = t_c

- x(0) = y(0) = 0

Find:

a) Find V_o,y, the vertical component of the velocity of the ball when the quarterback releases it.  Express V_o,y in terms of t_c and g.

b) Find V_o,x, the initial horizontal component of velocity of the ball.   Express your answer for V_o,x in terms of D, t_c, and v_r.

c) Find the speed V_o with which the quarterback must throw the ball.  

   Answer in terms of D, t_c, v_r, and g.

d) Assuming that the quarterback throws the ball with speed V_o, find the angle Q above the horizontal at which he should throw it.

Solution:

- The vertical component of velocity V_o,y can be calculated using second kinematics equation of motion:

                               y = y(0) + V_o,y*t_c - 0.5*g*t_c^2

                              0 = 0 + V_o,y*t_c - 0.5*g*t_c^2

                               V_o,y = 0.5*g*t_c

- The horizontal component of velocity V_o,x witch which velocity is thrown can be calculated using second kinematics equation of motion:

- We know that V_i, x = V_o,x + v_r. Hence,

                               x = x(0) + V_i,x*t_c

                               D = 0 + V_i,x*t_c

                               V_o,x + v_r = D/t_c

                                V_o,x = D/t_c - v_r

- The speed with which the ball was thrown can be evaluated by finding the resultant of V_o,x and V_o,y components of velocity as follows:

                           V_o = sqrt ( V_o,x^2 + V_o,y^2)

                          V_o = sqrt ( (D/t_c - v_r)^2 + (0.5*g*t_c)^2)

       

- The angle with which it should be thrown can be evaluated by trigonometric relation:

                            tan(Q) = ( V_o,y / V_o,x )

                            tan(Q) = ( (0.5*g*t_c)/ (D/t_c - v_r) )

                                   Q = arctan ( g*t_c^2 / 2*(D - v_r*t_c) )

                           

                               

6 0
2 years ago
What effect does the sun have on the earth. written response
maw [93]
It had a gravitational force that acts so that it pulls Earth in an orbital, which explains temperature variations throughout the years.
8 0
3 years ago
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