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GREYUIT [131]
3 years ago
11

g Florence, who weighs 580 N, stands on a bathroom scale in an elevator. What will she see the scale read when the elevator is a

ccelerating while moving downward at 2.30 m/s²?(g = 9.8 m/s²)
Physics
1 answer:
Luda [366]3 years ago
8 0

Answer:

AW= m (g-a)   (1)

m= \frac{W}{g}   (2)

AW = \frac{W}{g} (g-a) = W - W \frac{a}{g}   (3)

And we know from the info given that:

W = 580 N, a = 2.3 m/s^2, g = 9.8 m/s^2

And replacing into equation (3) we got:

AW = 580 N - 580 N \frac{2.3 m/s^2}{9.8 m/s^2}= 443.878 N

So then we can conclude that the scale will read 443.878 N for Florence

Explanation:

For this case we have an accelerator moving downward with an acceleration a.

On this case the person fell like the weight is less than the normal since the required force on the elevator floor is lower than the normal force, for this reason the body accelerate downward.

We can use the formula of apparent weight given by:

AW= m (g-a)   (1)

Where AW represent the apparent weigth of the person in the elevator

We also know that the weigth is defined as:

W= mg

If we solve for m we got:

m= \frac{W}{g}   (2)

If we replace equation (2) into equation (1) we got:

AW = \frac{W}{g} (g-a) = W - W \frac{a}{g}   (3)

And we know from the info given that:

W = 580 N, a = 2.3 m/s^2, g = 9.8 m/s^2

And replacing into equation (3) we got:

AW = 580 N - 580 N \frac{2.3 m/s^2}{9.8 m/s^2}= 443.878 N

So then we can conclude that the scale will read 443.878 N for Florence

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an electron is released from rest in a region of space with a nonzero electric field.1. As the electron moves, does the electric
zvonat [6]

Answer:

1. a. increase

2. Because the electron has a negative charge its electric potential energy does not decrease as one might expect, but increases instead.

Explanation:

Lets first consider the relation between the electric field and electric potential.

E = -ΔV/Δs

As this equation indicates that the electric field is due to the change in potential and change in the the position of charge. Electric field is directed towards the decreasing potential and the electron moves in the opposite direction of the electric field  where potential increases. Thats why the best explanation is that the electron has a negative charge it moves towards the positive region where the electric potential energy increases.

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3 years ago
If the voltage across the first capacitor (the one with capacitance
RoseWind [281]
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Capacitors are like batteries in a way that they store power from the source. It has some rules depending on the type of circuit. For parallel circuits, the voltage across each capacitor is equal. Therefore, V₁=V₂=V₃.

On the other hand, if the capacitors are arranged in series, the voltage across each capacitor should add up to the total voltage of the source. Therefore, V₁+V₂+V₃ = Total Voltage.


8 0
3 years ago
Charlie pulls horizontally to the right on a wagon with a force of 37.2 N. Sara pulls horizontally to the left with a force of 2
Sidana [21]

Answer:

The work done on the wagon is 37 joules.

Explanation:

Given that,

The force applied by Charlie to the right, F = 37.2 N

The force applied by Sara to the left, F' = 22.4 N

We need to find the work done on the wagon after it has moved 2.50 meters to the right. The net force acting on the wagon is :

F_n=F-F'

F_n=37.2-22.4

F_n=14.8\ N

Work done on the wagon is given by the product of net force and displacement. It is given by :

W=F_n\times d

W=14.8\ N\times 2.5\ m

W = 37 Joules

So, the work done on the wagon is 37 joules. Hence, this is the required solution.

7 0
3 years ago
Read 2 more answers
When electrical energy is being used by an electric light, what really happens to the energy?
vitfil [10]

Energy is not created and not  destroyed it will only change form


So heres your answer ; It is given off as other forms of energy/light and heat !!


=answer 2nd one

7 0
3 years ago
Read 2 more answers
A car and a train move together along straight, parallel paths with the same constant cruising speed v0. At t=0 the car driver n
satela [25.4K]

Answer:

a) t1 = v0/a0

b) t2 = v0/a0

c) v0^2/a0

Explanation:

A)

How much time does it take for the car to come to a full stop? Express your answer in terms of v0 and a0

Vf = 0

Vf = v0 - a0*t

0 = v0 - a0*t

a0*t = v0

t1 = v0/a0

B)

How much time does it take for the car to accelerate from the full stop to its original cruising speed? Express your answer in terms of v0 and a0.

at this point

U = 0

v0 = u + a0*t

v0 = 0 + a0*t

v0 = a0*t

t2 = v0/a0

C)

The train does not stop at the stoplight. How far behind the train is the car when the car reaches its original speed v0 again? Express the separation distance in terms of v0 and a0 . Your answer should be positive.

t1 = t2 = t

Distance covered by the train = v0 (2t) = 2v0t

and we know t = v0/a0

so distanced covered = 2v0 (v0/a0) = (2v0^2)/a0

now distance covered by car before coming to full stop

Vf2 = v0^2- 2a0s1

2a0s1 = v0^2

s1 = v0^2 / 2a0

After the full stop;

V0^2 = 2a0s2

s2 = v0^2/2a0

Snet = 2v0^2 /2a0 = v0^2/a0

Now the separation between train and car

= (2v0^2)/a0 - v0^2/a0

= v0^2/a0

8 0
3 years ago
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