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liberstina [14]
3 years ago
11

A 34n force is applied to a 213kg mass how much does the mass accelerate

Physics
1 answer:
motikmotik3 years ago
3 0
We Know, F = m*a
Here, F = 34 N
m = 213 Kg

Substitute their values in the equation,
34 = 213 * a
a = 34/213
a = 0.159 m/s²

So, your final answer & the acceleration of the object would be 0.159 m/s²

Hope this helps!
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An insulator can do which of the following? Conduct charge through it. Become positively or negatively charged. Transfer protons
ser-zykov [4K]
The correct answer is:
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w strong are you? Could you turn a moving car with just your own strength? No way! If a normal driver inputs about 50 N of force
malfutka [58]

The mechanical advantage of a simple machine is the measure of its amplified force gain.

The mechanical advantage  is defined as the force amplified by a machine to the force required to generate such output.

Mathematically\ mechanical\ advantage\ MA=\frac{F_{o}} {F_{i}}

F_{o} \ and\ F_{i} are the amplified force and applied force. We may also consider them as output and input force.          


In the given question, the force given to the steering wheel is 50 N.

The output force produced by the steering wheel is 3750 N.

Hence the mechanical advantage will be-

                               MA=\frac{F_{o}} {F_{i}}

                                       =\frac{3750\ N}{50\ N}

                                       =75      [ans]

3 0
2 years ago
During destructive interference, two waves moving through the same medium will
ddd [48]
They will amplify eachother.
6 0
2 years ago
Read 2 more answers
You place a point charge q = -4.00 nC a distance of 9.00 cm from an infinitely long, thin wire that has linear charge density 3.
valentinak56 [21]

Answer:

F=6\times 10^{-7}\ N

Explanation:

Given:

  • quantity of point charge, q=-4\times 10^{-9}\ C
  • radial distance from the linear charge, r=0.09\ m
  • linear charge density, \lambda=3\times 10^{-9}\ C.m^{-1}

<u>We know that the electric field by the linear charge  is given as:</u>

E=\frac{\lambda}{2\pi.\epsilon_0.r}

E=\frac{1}{2}\times 9\times 10^9\times \frac{3\times10^{-9}}{0.09}

E=150\ N.C^{-1}

<u>Now the force on the given charge can be given as:</u>

F=E.q

F=150\times 4\times 10^{-9}

F=6\times 10^{-7}\ N

3 0
3 years ago
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