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arsen [322]
3 years ago
15

How do you calculate mass using Newton’s 2nd Law?

Physics
2 answers:
deff fn [24]3 years ago
5 0
The 2nd Law says F=ma, where F is the force in Newtons, m is mass and a is acceleration.  Earth's gravity is an acceleration, 9.8m/s^2.  So you can solve the equation for mass, m=F/a, or m=F/9.8 where you've measured the force (weight) in Newtons.
egoroff_w [7]3 years ago
4 0
To calculate from the mass by the during Newton's 2nd law are:

Newton's second law of motions is a states that the total net force will be acting on an object is to equal by the mass times acceleration. 

With formula of m=f/a or a=f/m, and F=ma
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Which property of light would provide evidence for the idea that light is a wave?
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Color property of light would provide evidence for the idea that light is a wave

<h3><u>Explanation:</u></h3>

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5 0
3 years ago
2 equal charges are kept at x=a and x=-a. there can be only 1 point where the electric field will be 0? how, can anyone explain?
svet-max [94.6K]

Answer:

At the midpoint of the line joining the two equal charges

Explanation:

The midpoint of the line joining the two equal charges has equal distance from each charge. Since the charges are at x = a and x = -a, at the midpoint of the line joining the two equal charges, the magnitude of the electric field experienced as a result of each charge is the same. But these two fields are in opposite direction, hence the net resulting electric field would be equal to zero.

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3 years ago
A 10 g bullet is fired into, and embeds itself in, a 2 kg block attached to a spring with a force constant of 19.6 n/m and whose
dmitriy555 [2]

Answer:

x=0.478\ m is the compression in the spring

Explanation:

Given:

  • mass of the bullet, m=10\ g=0.01\ kg
  • mass of block, M=2\ kg
  • stiffness constant of the spring, k=19.6\ N.m^{-1}
  • initial velocity of the spring just before it hits the block, u=300\ m.s^{-1}

<u>Now since the bullet-mass gets embed into the block, we apply the conservation of momentum as:</u>

m.u=(M+m).v

0.01\times 300=(2+0.01)\times v

v=1.4925\ m.s^{-1}

Now this kinetic energy of the combined mass gets converted into potential energy of the spring.

\rm Kinetic\ energy=Spring\ potential\ energy

\frac{1}{2} (M+m).v^2=\frac{1}{2} k.x^2

\frac{1}{2} \times (2+0.01)\times 1.4925^2=\frac{1}{2} \times 19.6\times x^2

x=0.478\ m is the compression in the spring

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3 years ago
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