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An electron in a vacuum chamber is fired with a speed of 7400 km/s toward a large, uniformly charged plate 75 cm away. The electron reaches a closest distance of 15 cm before being repelled. What is the plate's surface charge density?

Answers

Answer 1

Answer:

2.29e-9C/m²

Explanation:

Using E = σ/ε₀ means the force on the electron is F = eE = eσ/ε₀.

The work done on the electron is W = Fd = deσ/ε₀. This equals the kinetic energy lost, ½mv².

½mv² = deσ/ε₀

d = 75cm – 15cm = 60cm = 0.6m

σ = mv²ε₀/(2de)

. .= 9.11e-31 * (7.4e6)² * 8.85e-12 / (2 * 0.6 * 1.6e-19)

. .= 2.29e-9 C/m² (i.e. 2.29x10^-9 C/m²)

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The speed of light is 3.00×108m/s. How long does it take for light to travel from Earth to the Moon and back again? Express your answer using two significant figures.

Answers

Answer:

v = 3×10^8 m/s

s= 384,400 km= 3.84×10^8 m/s

t = ?

v = s/t = 2s/t

t = 2s/v

t = (2×3.84×10^8) ÷ 3×10^8

t = 2.56 seconds

Explanation:

Earth's moon is the brightest object in our

night sky and the closest celestial body. Its

presence and proximity play a huge role in

making life possible here on Earth. The moon's gravitational pull stabilizes Earth's wobble on its axis, leading to a stable climate.

The moon's orbit around Earth is elliptical. At perigee — its closest approach — the moon comes as close as 225,623 miles (363,104 kilometers). At apogee — the farthest away it gets — the moon is 252,088 miles (405,696

km) from Earth. On average, the distance fromEarth to the moon is about 238,855 miles (384,400 km). According to NASA , "That means 30 Earth-sized planets could fit in between Earth and the moon."

You observe three carts moving to the right. Cart A moves to the right at nearly constant speed. Cart B moves to the right, gradually speeding up. Cart C moves to the right, gradually slowing down. Which cart or carts, if any, experience a net force to the right

Answers

Answer:

Explanation:

Cart A is moving to the right with constant speed i.e. net acceleration is zero

because acceleration is change in velocity in given time

Cart B is moving towards right with gradually speed up so there is net acceleration which helps to increase the velocity s

This indicates the net force acting on the cart towards right

For cart C there is gradual slow down of cart which indicates cart is decelerating and a net force is acting towards which opposes its motion.

An electric current in a conductor varies with time according to the expression i(t) = 110 sin (120πt), where i is in amperes and t is in seconds. what is the total charge passing a given point in the conductor from t = 0 to t = 1/180 s?

Answers

As we know that current is defined as rate of flow of charge

$i = (dq)/(dt)$

so by rearranging the equation we can say

$q = \int i dt$

here we know that

$i(t) = 110 sin(120\pi t)$

here we will substitute it in the above equation

$q = \int 110 sin(120\pi t) dt$

$q = 110 [- (cos(120\pi t))/(120\pi)]$

now here limits of time is from t = 0 to t = 1/180s

so here it will be given as

$q = (110)/(120\pi)( -cos0 + cos((2\pi)/(3)))$

$q = 0.44 C$

so total charge flow will be 0.44 C

Answer:

The total charge passing a given point in the conductor is 0.438 C.

Explanation:

Given that,

The expression of current is

$i(t)=110\sin(120\pi t)$

$(dq(t))/(t)=110\sin(120\pi t)$

$dq(t)=110\sin(120\pi t)dt$ ....(I)

We need to calculate the total charge

On integrating both side of equation (I)

$\int_(0)^(q)dq(t)=\int_(0)^{(1)/(180)}110\sin(120\pi t)dt$

$q=110((-\cos(120\pi t))/(120\pi))_(0)^{(1)/(180)}$

$q=-(110)/(120\pi)(cos(120\pi((1)/(180)))-\cos120\pi(0))$

$q=-0.2918(-(1)/(2)-1)$

$q=0.438\ C$

Hence, The total charge passing a given point in the conductor is 0.438 C.

Mudflows composed of soil, volcanic debris, and water can occur as the result of an explosive volcanic eruption. What are these mudflows called?

Answers

These mudflows are called Lahar.

A quickly moving mixture of rock debris and water that begins on a volcano's slopes is referred to as a lahar in Indonesian. Other names for lahars are volcanic mudflows and debris floods. The size, pace, and volume of material transported by a moving lahar can constantly alter as it rushes downstream. It resembles a swirling slurry of wet concrete.

The melting of snow and ice as well as the ingestion of river or lake water by the moving slurry may both add to its water consumption. A lahar's starting flow may be quite tiny, but as it entrains and integrates everything in its path, including rocks, dirt, vegetation, even structures like houses and bridges, it may increase in volume.

To learn more about lahar please visit-
brainly.com/question/14643369
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Two 10-cm-diameter charged rings face each other, 25.0cm apart. Both rings are charged to +20.0nC. What is the electric field strength at:a) the midpoint between the two rings?b) the center of the left ring?

Answers

The electric field strength at the midpoint of the rings is 0 N/C.

The electric field strength at the center of the left ring is 2710.84 N/C.

The given parameters:

Diameter of the rings, d = 10 cm
Distance between the rings, r = 25 cm
Charge of the rings, q = 20 nC

The electric field strength at the midpoint of the rings is calculated as follows;

$E_(net) = E_1 + E_2\n\nE_(net) = E_1(+ve) - E_2(-ve) = 0$

The electric field strength at the center of the left ring is calculated as follows;

$E = (kqL)/((R^2 + L^2)^(3/2)) \n\nE = (9* 10^9 * 20* 10^(-9) * 0.25 \ )/((0.05^2 + 0.25^2 )^(3/2)) \n\nE = 2710.84 \ N/C$

Learn more about electric field here: brainly.com/question/14372859

Final answer:

The electric field strength at the midpoint between the two rings is zero, and at the center of the left ring, it is 2.88 * 10^4 N/C.

Explanation:

The electric field strength at the

midpoint between the two rings is zero. The electric fields from each ring cancel each other out at this point because they are equal in magnitude and opposite in direction.
At the center of the left ring, the electric field strength can be calculated using the formula for the electric field due to a charged ring. The formula is E = k * (Q / r²), where E is the electric field strength, k is the Coulomb's constant, Q is the charge, and r is the distance from the center of the ring. Plugging in the values, we get:

E = (8.99 * 10^9 Nm²/C²) * (20.0 * 10^-9 C) / (0.05 m)² = 2.88 * 10^4 N/C

A cat leaps to catch a bird. If the cat's jump was at 60.0° off the ground and its initial velocity was point of its trajectory? 0.30 m 3.44 m/s, what is the highest O 13.76 m 0.45 m 0.90 m