3 years ago

What is the magnitude of the resultant vector? Round your answer to the nearest tenth

Physics

1 answer:

Harlamova29_29 [7]3 years ago

7 0

Answer:13.9 on edgy I just did it

Explanation:

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A 2.98-kg object oscillates on a spring with an amplitude of 8.05 cm. Its maximum acceleration is 3.55 m/s2. Calculate the total

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Answer:

a = ω^2 A formula for max acceleration (ignoring sign)

V = ω A formula for max velocity

V^2 = ω^2 A^2 = a A from first equation

E = 1/2 M V^2 = 1/2 * 2.98 * 3.55 * .0805 = .426 J

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2 years ago

Explain the conversion of energy in a roller coaster. Explain how energy is converted

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3 years ago

The total momentum of two marbles before a collision is 0.06 kg-m / s. no outside forces act on the marbles. what is the total m

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3 years ago

If it requires 2.0 J of work to stretch a particular spring by 2.0 cm from its equilibrium length, how much more work will be re

valina [46]

Answer:

16 J

Explanation:

It is given that,

Work done, W = 2 J

A spring is stretched by 2.0 cm from its equilibrium length

We need to find how much more work will be required to stretch it an additional 4.0 cm.

Let k is the spring constant of the spring. When W = 2J, and x = 2 cm, then energy required to stretch the spring is :

$U=\dfrac{1}{2}kx^2\\\\k=\dfrac{2U}{x^2}\\\\k=\dfrac{2(2)}{(0.02)^2}\\\\k=10000\ N/m$

The energy required to stretch the spring from 2 cm to additional 4 cm i.e. 2+4= 6 cm.

$W=\dfrac{1}{2}k(x_2^2-x_1^2)\\\\=\dfrac{1}{2}\times 10000\times ((0.06)^2-(0.02)^2)\\\\W=16\ J$

So, the required work done is 16 J.

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3 years ago

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Diano4ka-milaya [45]

The answer would be B.

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3 years ago