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3 years ago

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

2 answers:

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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3 years ago

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The conservation of momentum is most closely related to

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Answer:

<h2>B) Newton's 2nd law</h2>

Explanation:

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1 year ago

A 1060-kg car moving west at 16 m/s collides with and locks onto a 1830-kg stationary car. determine the velocity of the cars ju

iren [92.7K]

M1U1 + M2V2 = (M1+M2)V, where M1 is the mass of the moving car, M2 is the mass of the stationary car, U1 is the initial velocity, and V is the common velocity after collision.
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(1060× 16) + (1830 ×0) = (1060 +1830) V
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The velocity of the cars after collision will be 5.689 m/s

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3 years ago

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3 years ago

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2. A 20 cm object is placed 10cm in front of a convex lens of focal length 5cm. Calculate

adoni [48]

Answer:

<u> </u><u>»</u><u> </u><u>Image</u><u> </u><u>distance</u><u> </u><u>:</u>

${ \tt{ \frac{1}{v} + \frac{1}{u} = \frac{1}{f} }} \\$

v is image distance
u is object distance, u is 10 cm
f is focal length, f is 5 cm

${ \tt{ \frac{1}{v} + \frac{1}{10} = \frac{1}{5} }} \\ \\ { \tt{ \frac{1}{v} = \frac{1}{10} }} \\ \\ { \tt{v = 10}} \\ \\ { \underline{ \underline{ \pmb{ \red{ \: image \: distance \: is \: 10 \: cm \: \: }}}}}$

<u> </u><u>»</u><u> </u><u>Magnification</u><u> </u><u>:</u>

• Let's derive this formula from the lens formula:

${ \tt{ \frac{1}{v} + \frac{1}{u} = \frac{1}{f} }} \\$

» Multiply throughout by fv

${ \tt{fv( \frac{1}{v} + \frac{1}{u} ) = fv( \frac{1}{f} )}} \\ \\ { \tt{ \frac{fv}{v} + \frac{fv}{u} = \frac{fv}{f} }} \\ \\ { \tt{f + f( \frac{v}{u} ) = v}}$

• But we know that, v/u is M

${ \tt{f + fM = v}} \\ { \tt{f(1 +M) = v }} \\ { \tt{1 +M = \frac{v}{f} }} \\ \\ { \boxed{ \mathfrak{formular : } \: { \tt{ M = \frac{v}{f} - 1 }}}}$

v is image distance, v is 10 cm
f is focal length, f is 5 cm
M is magnification.

${ \tt{M = \frac{10}{5} - 1 }} \\ \\ { \tt{M = 5 - 1}} \\ \\ { \underline{ \underline{ \pmb{ \red{ \: magnification \: is \: 4}}}}}$

<u> </u><u>»</u><u> </u><u>Nature</u><u> </u><u>of</u><u> </u><u>Image</u><u> </u><u>:</u>

Image is magnified
Image is erect or upright
Image is inverted
Image distance is identical to object distance.

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2 years ago