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mart [117]
3 years ago
10

Point charge 3.5 μC is located at x = 0, y = 0.30 m, point charge -3.5 μC is located at x = 0 y = -0.30 m. What are (a)the magni

tude and (b)direction of the total electric force that these charges exert on a third point charge Q = 4.5 μC at x = 0.40 m, y = 0?
Physics
1 answer:
IgorLugansk [536]3 years ago
3 0
PDF attached.....................
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a car with a mass of 2000 kilograms is moving around a circular curve at a uniform velocity of 25 meters per second. The curve h
melisa1 [442]
In the given problem, we say various information's that are going to help us reach the ultimate answer to the question. Let us first write the information's that have been presented in front of us.
Mass of the car = 2000 kg
Velocity of the car = 25 m/s^2
Radius of the circle = 80 m
Now we already know the equation for calculating the centripetal force and that is
Centripetal Force = [mass * (velocity)^2]/Radius
                            = [2000 * (25)^2]/80
                            = (2000 * 625)/80
                            = 1250000/80
                            = 15625
So the centripetal force on the car is 15625 Newtons
  
4 0
3 years ago
Read 2 more answers
If a star is 720 , 000 , 000 , 000 , 000 meters from Earth, how many seconds does it take light to travel from the Earth to the
Minchanka [31]

Answer:

anywhere between 100000 to about 400000 human years .

Explanation:

5 0
2 years ago
50 kg of water at 75o C is cooled to 25o C. How much heat was given off?
attashe74 [19]

Answer:

b the answer is b

Explanation:

b is the awnser because it cools after the heat on the water witch lets the steam out

8 0
2 years ago
The motion of an object undergoing constant acceleration can be modeled by the kinematic equations. One such equation is xf=xi+v
Arturiano [62]

Answer:

a = 1.72 m/s²

Explanation:

The given kinematic equation is the 2nd equation of motion. The equation is as follows:

xf = xi + (Vi)(t) + (1/2)(a)t²

where,

xf = the final position =  5000 m

xi = the initial position = 1000 m

Vi = the initial velocity = 15 m/s

t = the time taken = 60 s

a = acceleration = ?

Therefore,

5000 m = 1000 m + (15 m/s)(60 s) + (1/2)(a)(60 s)²

5000 m = 1000 m + 900 m + a(1800 s²)

5000 m = 1900 m + a(1800 s²)

5000 m - 1900 m = a(1800 s²)

a(1800 s²) = 3100 m

a = 3100 m/1800 s²

<u>a = 1.72 m/s²</u>

5 0
3 years ago
A person is pushing a box. The net external force on the 60-kg box is stated to be 90 N. If the force of friction opposing the m
Tems11 [23]

Answer:

b.1.5m/s^2

Explanation:

We are given that

Mass of box=60kg

Net external force applied on the box=90 N

Friction force =30N

We have to find the acceleration of the box.

We know that

Net external force=ma

Substitute the values then we get

90=60a

a=\frac{90}{60}=1.5m/s^2

Hence, option b is true.

5 0
2 years ago
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