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nataly862011 [7]
3 years ago
6

An electron is released from rest at a distance of 0.300 m from a large insulating sheet of charge that has uniform surface char

ge density 4.90×10^−12 C/m^2C/m^2.
a. How much work is done on the electron by the electric field of the sheet as the electron moves from its initial position to a point 0.050 m from the sheet?
b. What is the speed of the electron when it is 0.050 m from the sheet?
Physics
1 answer:
jarptica [38.1K]3 years ago
8 0
Yes that is the correct answer
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A searchlight is 210 ft from a straight wall. As the beam moves along the​ wall, the angle between the beam and the perpendicula
Ivenika [448]

Answer:

The length of the beam increasing is 9.64 ft/s.

Explanation:

Given that,

Height = 210 ft

Distance =290 ft

According to figure,

We need to calculate the angle

\cos\theta=\dfrac{210}{x}....(I)

Put the value of x in the equation

\cos\theta=\dfrac{210}{290}

\cos\theta=\dfrac{21}{29}=0.72

Now, \sin\theta=\dfrac{20}{29}

On differentiate of equation (I)

-\sin\theta\dfrac{d\theta}{dt}=-\dfrac{-210}{x^2}\dfrac{dx}{dt}

\sin\theta=\dfrac{210}{x^2}\dfrac{dx}{dt}

Put the value in the equation

\sin\dfrac{20}{29}\times2.0=\dfrac{210}{(290)^2}\dfrac{dx}{dt}

\dfrac{dx}{dt}=\sin\dfrac{20}{29}\times2.0\times\dfrac{290^2}{210}

\dfrac{dx}{dt}=9.64\ ft/s

Hence, The length of the beam increasing is 9.64 ft/s.

4 0
3 years ago
What is the probability of randomly selecting a z-score less than z = 2.25 from a normal distribution?
Vanyuwa [196]

The probability of randomly selecting a z-score less than z = 2.25 from a normal distribution is 0.8944

Probability is the branch of arithmetic concerning numerical descriptions of ways in all likelihood an occasion is to occur, or how probably it's far that a proposition is genuine. The possibility of an event is a number between zero and 1, wherein, more or less talking, zero suggests the impossibility of the event and 1 shows the truth.

Probability = the range of methods of achieving achievement. the total range of viable results. As an example, the chance of flipping a coin and it being heads is ½, due to the fact there's 1 way of having a head and the full number of possible effects is two (a head or tail). We write P(heads) = ½ .

A probability is a range that reflects the danger or chance that a particular event will occur. Chances can be expressed as proportions that vary from 0 to at least one, and they also can be expressed as probabilities ranging from zero% to 100%.

Learn more about probability here brainly.com/question/24756209

#SPJ4

4 0
1 year ago
Four +2 μC charges are placed at the positions (10 cm, 0 cm), (−10 cm, 0 cm), (0 cm, 10 cm), and (0 cm, −10 cm) such that they f
mylen [45]

Answer:

The force will be zero

Explanation:

Due to the symmetric location of the +2μC charges the forces the excert over the +5μC charge will cancel each other resulting in a net force with a magnitude of zero.However in this case it would be an unstable equilibrium, very vulnerable to a kind of bucking. If the central charge is not perfectly centered on the vertical axis the forces will have components in that axis that will add together instead of canceling each other.

7 0
3 years ago
Question 9 (28 points)
ladessa [460]
Im sure the answer is letter B
5 0
3 years ago
Read 2 more answers
65 POINTS! PLEASE ANSWER EVERY QUESTION! NEED HELP ASAP!
otez555 [7]
Maybe you can split up the questions. I will try to answer your first question.

1. In an elastic collision, momentum is conserved. The momentum before the collision is equal to the momentum after the collision. This is a consequence of Newton's 3rd law. (Action = Reaction)

2. Momentum: p = m₁v₁ + m₂v₂

m₁ mass of ball A
v₁ velocity of ball A
m₂ mass of ball B
v₂ velocity of ball B

Momentum before the collision:
p = 2*9 + 3*(-6) = 18 - 18 = 0

Momentum after the collision:
p = 2*(-9) + 3*6 = -18 + 18 = 0

3: mv + m(-v) = m(-v) + m(v)
the velocities would reverse.

4.This question is not factual since the energy of an elastic collision must also be conserved. The final velocities should be: v₁ = -1 m/s and v₂ = 5 m/s. That said assuming the given velocities were correct:
before collision
p = 10*3 + 5*(-3) = 30 - 15 = 15
after collision:
p = 10*(-2) + 5 * v₂ = 15
v₂ = 7

5.You figure out.



3 0
3 years ago
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