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1 year ago

This reaction represents cellular respiration. What happens to the energy released from glucose during this reaction?.

1 answer:

6 0

The energy released from glucose during this reaction is used and also wasted. Some of the energy is used as work while some amounts are used in other processes or stored for transfer to other organisms. Also, some of the energy is wasted in the form of heat.

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What is meant by an acceleration of negative 2metre per second square

soldi70 [24.7K]

means that a body is in motion, and its velocity is measured in meters per second. And, that velocity is increasing by two meters per second, every second.

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3 years ago

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How you use physics in your<br> everyday life.

stepan [7]

Answer:

Explanation:

Physics gets involved in your daily life right from you wake up in the morning. The buzzing sound of an alarm clock helps you wake up in the morning as per your schedule. The sound is something that you can't see, but hear or experience. Physics studies the origin, propagation, and properties of sound

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3 years ago

How long does water evaporation take? What factors influence it?

svetoff [14.1K]

There is no certain time on how long it takes. Because the factors will always be different and the factors heavily affect the evaporation time. Some factors include: humidity, heat, how the sun is visible (whether clouds are covering it or not)

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3 years ago

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A satellite circles the Earth in an orbit whose radius is twice the Earth’s radius. The Earth’s mass is 5.98 x 1024 kg, and its

gavmur [86]

Hello!

Recall the period of an orbit is how long it takes the satellite to make a complete orbit around the earth. Essentially, this is the same as 'time' in the distance = speed * time equation. For an orbit, we can define these quantities:

$d = 2\pi r$ ← The circumference of the orbit

speed = orbital speed, we will solve for this later

time = period

Therefore:

$T = \frac{2\pi r}{v}$

Where 'r' is the orbital radius of the satellite.

First, let's solve for 'v' assuming a uniform orbit using the equation:
$v = \sqrt{\frac{Gm}{r}}$

G = Gravitational Constant (6.67 × 10⁻¹¹ Nm²/kg²)

m = mass of the earth (5.98 × 10²⁴ kg)

r = radius of orbit (1.276 × 10⁷ m)

Plug in the givens:
$v = \sqrt{\frac{(6.67*10^{-11})(5.98*10^{24})}{(1.276*10^7)}} = 5590.983 m/s$

Now, we can solve for the period:

$T = \frac{2\pi (1.276*10^7)}{5590.983} =\boxed{ 14339.776 s}$

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2 years ago

kristine speeds past a parked police car at 32 m/s. The police car starts from rest with a uniform acceleration of 2.5 m/s^2. Ho

Digiron [165]

$32 = 0 \times t + \frac{1}{2} \times 2.5 \times t^{2} \\ 32 = 0 + 1.25 \times t {}^{2} \\ 32 = 1.25t {}^{2} \\ \frac{32}{1.25} = \frac{1.25t {}^{2} }{1.25} \\ t {}^{2} = 25.6 \\ \sqrt{t {}^{2} } = \sqrt{25.6} \\ t = 5.1seconds \\$

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2 years ago