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Helen [10]
2 years ago
13

The sum of all the forces acting on an object is zero.

Physics
2 answers:
sergey [27]2 years ago
8 0

Answer:

Equilibrium -The sum of all the forces acting on the object is zero, which is the first condition of it and it has two forms. Law of Inertia -The body will remain at rest or move at constant velocity unless acted. upon by an external net force or unbalanced force.

Explanation:

:) tell me if it's wrong please!

marysya [2.9K]2 years ago
3 0

Answer:

Newton's first law states that when the vector sum of all forces acting on an object (the net force) is zero, the object is in equilibrium. If the object is initially at rest, it remains at rest. If it is initially in motion, it continues to move with constant velocity.

Explanation:

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A 3900 kg truck is moving at 6.0 m/s what is the kinetic energy
Stolb23 [73]

Answer:

70200J

Explanation:

k.E = 1/2mv^2

K.E = 1/2(3900)(6)^2

3 0
3 years ago
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220V and 5A is supplied to the primary coil. The turns ratio is 500. What is the power on the secondary coil?
alisha [4.7K]

Answer: 1100 W

Explanation:

Input power  = 220(5) = 1100 W

The transformer will step up/down voltage, but will also step down/up current.

Neglecting hysteresis and other minor losses, the power will remain the same.

4 0
2 years ago
How would you find the horizontal net force for the free body diagram below
tiny-mole [99]

Answer:

Add Ff from Fa

Explanation:

Fnet = sum of all force

horizontal net force = Ff + Fa

7 0
2 years ago
Coulomb's law states that the force F of attraction between two oppositely charged particles varies jointly as the magnitude of
Mars2501 [29]

Answer: Force F will be one-sixteenth of the new force when the charges are doubled and distance halved

Explanation:

Let the charges be q1 and q2 and the distance between the charges be 'd'

Mathematical representation of coulombs law will be;

F1=kq1q2/d²...(1)

Where k is the electrostatic constant.

If q1 and q2 is doubled and the distance halved, we will have;

F2 = k(2q1)(2q2)/(d/2)²

F2 = 4kq1q2/(d²/4)

F2 = 16kq1q2/d²...(2)

Dividing equation 1 by 2

F1/F2 = kq1q2/d² ÷ 16kq1q2/d²

F1/F2 = kq1q2/d² × d²/16kq1q2

F1/F2 = 1/16

F1 = 1/16F2

This shows that the force F will be one-sixteenth of the new force when the charges are doubled and distance halved

4 0
3 years ago
Runner A is initially 5.7 km west of a flagpole and is running with a constant velocity of 8.9 km/h due east. Runner B is initia
Oksanka [162]

Answer:

B meet A 0.01 km east of flagpole

Explanation:

given data

distance A = 5.7 km  west

velocity V1 = 8.9 km/h

distance B = 4.5 km  east

velocity V2 = 7 km/h

to find out

How far runners from the flagpole, when paths cross

solution

we know A and B are 5.7 + 4.5 = 10.2 km apart

and we consider here B will run distance x km for meet

so time will be for B is

time B = distance / velocity

time B = x / 7     ...................1

and

for A distance for meet = ( 10.2 - x ) km

so time A = distance / velocity

time A = ( 10.2 - x )  / 8.9      .............2

now equating equation 1 and 2

time A = time B

x / 7  =   ( 10.2 - x )  / 8.9

x = 4.490

so distance of B run for meet is 4.490 km

so distance from the flagpole when their paths cross is 4.5 - 4.490 = 0.01 km

so B meet A 0.01 km east of flagpole

4 0
3 years ago
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