The pinion gear A rolls on the gear racks B and C.

The instruments carried by a spacecraft are called the

Sedaia [141]

I believe the answer is C- payload

6 0

3 years ago

two point charges of magnitude 4.0 μc and -4.0 μc are situated along the x-axis at x1 = 2.0 m and x2 = -2.0 m, respectively. wha

user100 [1]

The electric potential at the origin of the xy coordinate system is negative infinity

<h3>What is the electric field due to the 4.0 μC charge?</h3>

The electric field due to the 4.0 μC charge is E = kq/r² where

k = electric constant = 9.0 × 10 Nm²/C²,
q = 4.0 μC = 4.0 × 10 C and
r = distance of charge from origin = x₁ - 0 = 2.0 m - 0 m = 2.0 m

<h3>What is the electric field due to the -4.0 μC charge?</h3>

The electric field due to the -4.0 μC charge is E = kq'/r² where

k = electric constant = 9.0 × 10 Nm²/C²,
q' = -4.0 μC = -4.0 × 10 C and
r = distance of charge from origin = 0 - x₂ = 0 - (-2.0 m) = 0 m + 2.0 m = 2.0 m

Since both electric fields are equal in magnitude and directed along the negative x-axis, the net electric field at the origin is

E" = E + E'

= -2E

= -2kq/r²

<h3>What is the electric potential at the origin?</h3>

So, the electric potential at the origin is V = -∫₂⁰E".dr

= -∫₂⁰-2kq/r².dr

Since E and dr = dx are parallel and r = x, we have

= -∫₂⁰-2kqdxcos0/x²

= 2kq∫₂⁰dx/x²

= 2kq[-1/x]₂⁰

= -2kq[1/x]₂⁰

= -2kq[1/0 - 1/2]

= -2kq[∞ - 1/2]

= -2kq[∞]

= -∞

So, the electric potential at the origin of the xy coordinate system is negative infinity

Learn more about electric potential here:

brainly.com/question/26978411

#SPJ11

3 0

2 years ago

What is the equivalent resistance of a circuit that contains four 75.02

viktelen [127]

Answer:

option A is the correct answer

3 0

2 years ago

The electron gun in a television tube accelerates electrons (mass = 9.11x10^-31 kg, charge = 1.6 x 10^-19C) from rest to 3 x10^7

stich3 [128]

Answer:

Electric field acting on the electron is 127500 N/C.

Explanation:

It is given that,

Mass of an electron, $m=9.11\times 10^{-31}\ kg$

Charge on electron, $q=1.6\times 10^{-19}\ C$

Initial speed of electron, u = 0

Final speed of electron, $v=3\times 10^7\ m/s$

Distance covered, s = 2 cm = 0.02 m

We need to find the electric field required. Firstly, we will find the acceleration of the electron from third equation of motion as :

$a=\dfrac{v^2-u^2}{2s}$

$a=\dfrac{(3\times 10^7)^2-0}{2\times 0.02}$

$a=2.25\times 10^{16}\ m/s^2$

According to Newton's law, force acting on the electron is given by :

F = ma

$F=9.1\times 10^{-31}\times 2.25\times 10^{16}$

$F=2.04\times 10^{-14}\ N$

Electric force is given by :

F = q E, E = electric field

$E=\dfrac{F}{q}$

$E=\dfrac{2.04\times 10^{-14}}{1.6\times 10^{-19}}$

E = 127500 N/C

So, the electric field is 127500 N/C. Hence, this is the required solution.

8 0

3 years ago

Approximately how many degrees does high latitude ocean water change from the ocean surface to a depth of 1 km

saveliy_v [14]

There is no change in ocean water temperature when it is from the ocean surface to a depth of 1 km.

3 0

4 years ago