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3 years ago

14

What is the electric potential energy of a charge that experiences a force of 3.6 x 10^-4N when it is 9.8 x 10^-5 from the sourc

e of the electric field

1 answer:

Levart [38]3 years ago

7 0

The electric potential energy of the charge is equal to the potential at the location of the charge, V, times the charge, q:
$U=qV$
The potential is given by the magnitude of the electric field, E, times the distance, d:
$V=Ed$
So we have
$U=qEd$ (1)
However, the electric field is equal to the electrical force F divided by the charge q:
$E= \frac{F}{q}$
Therefore (1) becomes
$U=Fd$
And if we use the data of the problem, we can calculate the electrical potential energy of the charge:
$U=Fd=(3.6 \cdot 10^{-4}N)(9.8 \cdot 10^{-5} m)=3.53 \cdot 10^{-8} J$

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A 30.0 kg box hangs from a rope. What is the tension (in N) in the rope if the box has an initial velocity of 2.0 m/s and is slo

hjlf

Answer:

360 N

Explanation:

m = 30kg u = 2 m/s a = -2m/s/s

Since the object has an initial velocity of 2 m/s and acceleration of -2 m/s/s

the object will come to rest in 1 second but the force applied in that one second can be calculated by:

F = ma

F = 30 * -2

F = -60 N (the negative sign tells us that the force is acting downwards)

Now, calculating the force applied on the box due to gravity

letting g = -10m/s/s

F = ma

F = 30 * -10

F = -300 N (the negative sign tells us that the force is acting downwards)

Now, calculating the total downward force:

-300 + (-60) = -360 N

<em></em>

<em>Hence, a downward force of 360 N is being applied on the box and since the box did not disconnect from the rope, the rope applied the same amount of force in the opposite direction</em>

Therefore tension on the force = <u>360 N</u>

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3 years ago

A ball is dropped from a building taking 3sec to fall to the ground. Calculate:

GenaCL600 [577]

Answer:

Vf = 29.4 m/s

h = 44.1 m

Explanation:

Data:

Initial Velocity (Vo) = 0 m/s
Gravity (g) = 9.8 m/s²
Time (t) = 3 s
Final Velocity (Vf) = ?
Height (h) = ?

==================================================================

Final Velocity

Use formula:

Vf = g * t

Replace:

Vf = 9.8 m/s² * 3s

Multiply:

Vf = 29.4 m/s

==================================================================

Height

Use formula:

$\boxed{h=\frac{g*(t)^{2}}{2}}$

Replace:

$\boxed{h=\frac{9.8\frac{m}{s^{2}}*(3s)^{2}}{2}}$

Multiply time squared:

$\boxed{h=\frac{9.8\frac{m}{s^{2}}*9s^{2}}{2}}$

Simplify the s², and multiply in the numerator:

$\boxed{h=\frac{88.2m}{2}}$

It divides:

$\boxed{h=44.1\ m}$

What is the velocity when falling to the ground?

The final velocity is <u>29.4 meters per seconds.</u>

How high is the building?

The height of the building is <u>44.1 meters.</u>

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2 years ago

You are driving your car along a country road at a speed of 27.0 m/s. as you come over the crest of a hill, you notice a farm tr

Bezzdna [24]

speed of the car = 27 m/s

speed of truck ahead = 10 m/s

relative speed of car with respect to truck

$v_r = 27 - 10 = 17 m/s$

relative deceleration of car

$a_r = -7 m/s^2$

now the distance before they stop with respect to each other is given by

$v_f^2 - v_i^2 = 2 a d$

$0 - 17^2 = 2 *(-7)*d$

$d = 20.6 m$

so it will come at the same speed of truck after 20.6 m distance and hence it will not hit the truck as the distance of the truck is 25 m from car

Part b)

Distance traveled by car before it stops is given by

$v_f^2 - v_i^2 = 2 a s$

$0^2 - 27^2 = 2 * (-7)* s$

$s = 52.1 m$

so it will stop after it will cover total 52.1 m distance

Part c)

time taken by the car to stop

$v_f - v_i = at$

$0 - 27 = (-7) * t$

$t = 3.86 s$

now the distance covered by truck in same time

$d = 3.86 * 10 = 38.6 m$

now after the car will stop its distance from the truck is

$D = 25 + 38.6 - 52.1 = 11.5 m$

<em>so the distance between them is 11.5 m</em>

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2 years ago

or Ethernet, if an adapter determines that a frame it has just received is addressed to a different adapter it discards the fram

Artist 52 [7]

For Ethernet, if an adapter determines that a frame it has just received is addressed to a different adapter

a. it discards the frame without sending an error message to the network layer

b. it sends a NACK (not acknowledged frame) to the sending host

c. it delivers the frame to the network layer, and lets the network layer decide what to do

d. it discards the frame and sends an error message to the network layer

Answer:

Option A

Explanation:

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3 years ago

A force of 25 newtons moves a box a distance of 4 meters in 5 seconds.

Simora [160]

By definition, we have that the work done is given by:

$W = F * d$

Where,

F: force in the direction of displacement

d: displaced distance

Substituting values we have:

$W = (25) * (4)\\W = 100 Nm$

Then, the power is given by:

$P = \frac{W}{t}$

Where,

t: time

Substituting values we have:

$P = \frac{100}{5}$

$P = 20\frac{Nm}{s}$

Answer:

The work done on the box is 100 Nm, and the power is 20 Nm/s.

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3 years ago

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