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patriot [66]
2 years ago
7

A 1400 kg car traveling in the positive direction takes 11.0 seconds to slow from 25.0 meters per second to 12.0 meters per seco

nd. What is the average force on the car during this time?
Physics
1 answer:
Alex777 [14]2 years ago
6 0

Answer:

–1652 N. The negative sign indicate that the force is in opposite direction to car.

Explanation:

The following data were obtained from the question:

Mass (m) of car = 1400 kg

Time (t) = 11 s

Initial velocity (u) = 25 m/s

Final velocity (v) = 12 m/s

Force (F) applied on the car =?

Next, we shall determine the acceleration of the car. This can be obtained as follow:

Time (t) = 11 s

Initial velocity (u) = 25 m/s

Final velocity (v) = 12 m/s

Acceleration (a) =?

a = (v – u) /t

a = (12 – 25) /11

a = –13/11

a = –1.18 m/s²

Finally, we shall determine the force on the car as follow:

Acceleration (a) = –1.18 m/s²

Mass (m) of car = 1400 kg

Force (F) applied on the car =?

F = ma

F = 1400 × –1.18

F = –1652 N

Thus, the force on the car is –1652 N. The negative sign indicate that the force is in opposite direction to car.

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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.

5 0
3 years ago
Find the distance from a point charge q=100nC where the field intensity is equal to E=6kN/C. please include description of the r
madam [21]

Answer: 0.3872m

Explanation:

q= 100nC -->  100x10^-9 C

k= 9x10^9 Nm^2/C^2

E= 6kN/C --> 600 N/C

r=?

E= K\frac{q}{r^{2} } --> r=\sqrt{\frac{kq}{E} } Despejas "r"

Resuelves

<h3>r=\sqrt{\frac{(9x10^9 Nm^2/C^2)(100x10x^{-9}  C)}{6000N/C} }  (la x es por, no es una variable)</h3><h3>r= 0.3872983346m</h3>
4 0
3 years ago
A man of weight Wman is standing on the second floor and is pulling on a rope to lift a box of weight Wbox from the floor below.
slavikrds [6]

Answer:

Explanation:

See the attached figure . See the forces acting on man pulling up the box .

Man is stationary so net force acting on man is zero .

T + R = Wman

R is the reaction force of the ground of second floor  .

R = Wman - T

3 0
3 years ago
The primary coil of a transformer contains 100 turns; the secondary has 200 turns. The primary coil is connected to a size aa ba
tester [92]

Answer:

3 volts

Explanation:

Number of turns in primary coil = N_{p} = 100

Number of turns in secondary coil = N_{s} = 200

Voltage across primary coil = V_{p} = 1.5 volts

Voltage across secondary coil = V_{s} = ?

In a transformer, the ratio of number of turns of primary to secondary coil is equal to the ratio of the respective voltages i.e.

\frac{N_{p} }{N_{s} } =\frac{V_{p} }{V_{s} }\\

Using the given values, we get:

\frac{100}{200}=\frac{1.5}{V_{s} }\\V_{s}=1.5 \times \frac{200}{100}\\V_{s}=3

Thus, the voltage measure across secondary coil would be 3 volts

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Answer:

29,999,999,970 words

Explanation:

9,999,999,990x3

3 0
3 years ago
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