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kotegsom [21]
4 years ago
12

What is an experience from everyday life that appears to support the geocentric model

Physics
1 answer:
fomenos4 years ago
3 0
<span>One everyday life experience that seems to support the geocentric model is the rising and setting of the Sun and Moon. The Moon rises and falls because it does revolve around the Earth and so it is easy to assume the same is true for the Sun.</span>
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Resting energy expenditure is a. slightly higher than basal energy expenditure. b. the same as basal energy expenditure. c. slig
kvv77 [185]

Answer:

B. Resting energy expenditure is the same with basal energy expenditure.

Explanation:

Basal Energy Expenditure can be explained as the energy required to execute essential metabolic functions e.g. coordination of enzymatic reactions in the body system.

On the other hand, Resting Energy Expenditure can be simply explained as the amount of energy expended or burnt when the body is resting.

Hence, in the context of definitions, and relating both definitions, it can be argued that Basal energy expenditure is simply the energy needed to execute essential metabolic functions e.g. coordination of enzymatic reactions in the body, with special emphassy on the body being at rest. Thus, in this context, Basal energy can be looked at through the prism of Resting energy expenditure. Consequently, this two definitions can be used interchangeably, with a special emphassy on perspective.

6 0
4 years ago
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gladu [14]

Answer:

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4 0
2 years ago
A football punter accelerates a football from rest to a speed of 15 m/s during the time in which his toe is in contact with the
KengaRu [80]
Use F=ma formula
F=m v/t. a=v/t
put values u will get answer.
answer should be around 30N .
7 0
4 years ago
A 1400.0 kg car crests a 3200.0 m pass in the mountains and briefly comes to rest. The car descends 1000 m before climbing and c
rjkz [21]

Answer:

a) v = 88.54 m/s

b) vf = 26.4 m/s

Explanation:

Given that;

m = 1400.0 kg

a)

by using the energy conservation

loss in potential energy is equal to gain in kinetic energy

mg × ( 3200-2800) = 1/2 ×m×v²

so

1400 × 9.8 × 400 = 0.5 × 1400 × v²

5488000 = 700v²

v² = 5488000 / 700

v² = 7840

v = √7840

v = 88.54 m/s

b)

Work done by all forces is equal to change in KE

W_gravity + W_non - conservative = 1/2×m×(vf² - vi²)

we substitute

1400 × 9.8 × ( 3200-2800) - (5 × 10⁶) = 1/2 × 1400 × (vf²  -0 )

488000 = 700 vf²

vf² = 488000 / 700

vf² = 697.1428

vf = √697.1428

vf = 26.4 m/s

4 0
4 years ago
A mass m at the end of a spring vibrates with a frequency
Wittaler [7]

Answer:

m = 0.59 kg.

Explanation:

First, we need to find the relation between the frequency and mass on a spring.

The Hooke's law states that

F = -kx

And Newton's Second Law also states that

F = ma = m\frac{d^2x}{dt^2}

Combining two equations yields

a = -\frac{k}{m}x

The term that determines the proportionality between acceleration and position is defined as angular frequency, ω.

\omega = \sqrt{\frac{k}{m}}

And given that ω = 2πf

the relation between frequency and mass becomes

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

Let's apply this to the variables in the question.

0.88 = \frac{1}{2\pi}\sqrt{\frac{k}{m}}\\0.60 = \frac{1}{2\pi}\sqrt{\frac{k}{m+0.68}}\\\frac{0.88}{0.60} = \frac{\frac{1}{2\pi}\sqrt{\frac{k}{m}}}{\frac{1}{2\pi}\sqrt{\frac{k}{m+0.68}}}\\1.4667 = \frac{\sqrt{m+0.68}}{\sqrt{m}}\\2.15m = m + 0.68\\1.15m = 0.68\\m = 0.59~kg

6 0
3 years ago
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