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LuckyWell [14K]
3 years ago
7

How to calculate force without acceleration?

Physics
2 answers:
skelet666 [1.2K]3 years ago
8 0
First of all, your question makes no sense. There wouldn't be any force if acceleration is zero.

Secondly, if you want to solve question in which acceleration is not given, you can use following formulas:
1.) F = dp/dt(dp = change in momentum, dt = change in time). In other words, rate of change of momentum gives you acceleration.
2.) Pressure = Force/Area
3.) Gravitational force = Gm1m2/r^2 ( G = universal gravitational constant, m1,m2 = masses, r = distance between their center.
4.) Magnetic force = q(vxB) [ q = charge, v = velocity of the particle, B= magnetic field].
5.) Electric force = qE [ E = electric field, q = charge]
6.) Force = surface tension x length. 

These are the most important one's there are many methods by the way. 

Hope this helps you and have a nice day/night ahead!
Oksanka [162]3 years ago
6 0
There are many ways to calculate force. However, they are completely different aspects. The force on an area is pressure, so F = p*A. In magnetic fields you use another formule (forgot the formule atm). Unless you define what kind of force you are looking for, I can't help you any further.
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A 710 kg car drives at a constant speed of 23 m/s. It is subject to a drag force of 500 N. What power is required from the car's
inn [45]

Explanation:

Given that,

Mass of the car, m = 710 kg

Speed of the car, v = 23 m/s

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(a) Let P is the power is required from the car's engine to drive the car on the level ground. Power is given by :

P=F\times v\\\\P=500\ N\times 23\ m/s\\\\P=11500\ W

(b) Let P is the power is required from the car's engine to drive the car on up a hill with a slope of 2 degrees.

At this slope, force will be, F=mg\ \sin\theta

Total force will be :

F=710\times 9.8\times \sin(2)+500\\F=742.83\ N

Power is given by :

P=F\times v\\\\P=742.83\times 23\\\\P=17085.09\ W

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5 0
4 years ago
A constant current source supplies a electric current of 200 mA to a load of 2kΩ. When the load changed to 50Ω, the load current
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