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3 years ago

Why does the area around the equator stay about the same temperature year round?

1 answer:

3 0

Axial Tilt and Sun Energy

This axial tilt means that during the Earth's journey around the sun the poles receive varying amounts of sunlight. The equator, however, receives relatively consistent sunlight all year. The consistency of energy means the equator's temperature stays relatively constant all year.

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An automobile traveling on a straight, level road has an initial speed v when the brakes are applied. In coming to rest with a c

Molodets [167]

Answer:

Explanation:

Use $v^{2} = u^{2} +2as$ to do the question.

For first instance,

$0 = v^{2} +2ax$ -------------------( 1 )

for second instance,

$0 = (2v)^{2} +2as$ -----------------( 2 )

So by (1) and (2),

s = 4x

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3 years ago

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Answer:

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3 years ago

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A small rock is thrown vertically upward with a speed of 21.0 m/sm/s from the edge of the roof of a 21.0-mm-tall building. The r

FromTheMoon [43]

Answer:

Explanation:

Given:

Vo = 21 m/s

Vf = 0 m/s

Using equation of Motion,

Vf^2 = Vo^2 - 2aS

S = (21^2)/2 × 9.8

= 22.5 m.

Given:

S = 22.5 + 21 mm

= 22.521 m

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S = Vo × t + 1/2 × a × t^2

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3 years ago

A 5.30kg block hangs from a spring with a spring constant 1700 N/m. The block is pulled down 4.50cm from the equilibrium positio

Andru [333]

To solve this problem it is necessary to apply the concepts related to the frequency in a spring, the conservation of energy and the total mechanical energy in the body (kinetic or potential as the case may be)

PART A) By definition the frequency in a spring is given by the equation

$f = \frac{1}{2\pi} \sqrt{\frac{k}{m}}$

Where,

m = mass

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Our values are,

k=1700N/m

m=5.3 kg

Replacing,

$f = \frac{1}{2\pi} \sqrt{\frac{1700}{5.3}}$

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PART B) To solve this section it is necessary to apply the concepts related to the conservation of energy both potential (simple harmonic) and kinetic in the spring.

$\frac{1}{2}kA^2 = \frac{1}{2}mv^2 + \frac{1}{2} kY^2$