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3 years ago

Someone please help with these 2

Physics

1 answer:

Ann [662]3 years ago

6 0

Answer:

Explanation:

The formula that you are working with is F = m*a

Since mass is one part of the formula if you increase the mass, you are going to increase the force.

The second one is much more difficult to answer because it is basically incomplete. This is one way to interpret it. If you start at a certain speed and increase during a known time period then effectively you are defining acceleration which is "a" in the formula.

Without those modifications, there is no answer.

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when all else remains the same, what effect would decreasing the focal have known a convex lens

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A weightlifter curls a 32 kg bar, raising it each time a distance of 0.50 m. How many times must he repeat this exercise to burn

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Read 2 more answers

A particle executes simple harmonic motion with an amplitude of 2.00 cm. At what positions does its speed equal one fourth of it

White raven [17]

Answer:

The positions are 0.0194 m and - 0.0194 m.

Explanation:

Given;

amplitude of the simple harmonic motion, A = 2.0 cm = 0.02 m

speed of simple harmonic motion is given as;

$v = \omega \sqrt{A^2-x^2}$

the maximum speed of the simple harmonic motion is given as;

$v_{max} = \omega A$

when the speed equal one fourth of its maximum speed

$v =\frac{v_{max}}{4}$

$\omega\sqrt{A^2-x^2} = \frac{\omega A}{4} \\\\\sqrt{A^2-x^2}= \frac{A}{4}\\\\A^2-x^2 = \frac{A^2}{16} \\\\x^2 = A^2 - \frac{A^2}{16} \\\\x^2 = \frac{16A^2 - A^2}{16} \\\\x^2 = \frac{15A^2}{16} \\\\x= \sqrt{\frac{15A^2}{16} } \\\\x = \sqrt{\frac{15(0.02)^2}{16} }\\\\x = 0.0194 \ m \ \ or\ - 0.0194 \ m$

Thus, the positions are 0.0194 m and - 0.0194 m.

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