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3 years ago

Help plssssssssssss

Physics

1 answer:

grin007 [14]3 years ago

6 0

C. A little backward, 100 or 110 degrees

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In diving to a depth of 308 m, an elephant seal also moves 579 m due east of his starting point. What is the magnitude of the se

Kisachek [45]

Answer:

655.82m

Explanation:

By tracing the distances traveled from 308m down in the water and then 579m to the east, a right triangle can be formed and the displacement can be found by the Pythagorean theorem. This is shown in the attached image (the red line is the displacement).

One leg of the triangle will be the 308m

and the second leg of the triangle will be the 579m.

If we call the displacement $x$ , by the pythagorean theorem:

$x=\sqrt{308^2+579^2}$

$x=\sqrt{94,864+335,241}$

$x=\sqrt430,105}$

$x=655.82m$

The displacement is 655.82m

3 0

4 years ago

PLEASE CLICK ON THIS IMAGE I NEED HELP

Burka [1]

Answer:

Second option

Explanation:

"Uniform" pretty much means the same thing happens.

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3 years ago

Which word describes the amount of matter an object contains?

notka56 [123]

Answer: Mass is the correct answer. Explanation: As mass is defined as the amount of matter contained in the substances or an object.

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3 years ago

A heliocentric system is _____-centered. Milky Way Earth Moon Sun

madreJ [45]

Answer: Sun

Explanation:

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3 years ago

Read 2 more answers

I attach a 4.1 kg block to a spring that obeys Hooke's law and supply 3.8 J of energy to stretch the spring. I release the block

borishaifa [10]

Answer:

The amplitude of the oscillation is 2.82 cm

Explanation:

Given;

mass of attached block, m = 4.1 kg

energy of the stretched spring, E = 3.8 J

period of oscillation, T = 0.13 s

First, determine the spring constant, k;

$T = 2\pi \sqrt{\frac{m}{k} }$

where;

T is the period oscillation

m is mass of the spring

k is the spring constant

$T = 2\pi \sqrt{\frac{m}{k} } \\\\k = \frac{m*4\pi ^2}{T^2} \\\\k = \frac{4.1*4*(3.142^2)}{(0.13^2)} \\\\k = 9580.088 \ N/m\\\\$

Now, determine the amplitude of oscillation, A;

$E = \frac{1}{2} kA^2$

where;

E is the energy of the spring

k is the spring constant

A is the amplitude of the oscillation

$E = \frac{1}{2} kA^2\\\\2E = kA^2\\\\A^2 = \frac{2E}{k} \\\\A = \sqrt{\frac{2E}{k} } \\\\A = \sqrt{\frac{2*3.8}{9580.088} }\\\\A = 0.0282 \ m\\\\A = 2.82 \ cm$

Therefore, the amplitude of the oscillation is 2.82 cm

8 0

3 years ago