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3 years ago

12

How are the sun, the moon, and Earth related during a solar eclipse?

1 answer:

german3 years ago

4 0

B. The moon is located between the Sun and Earth

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The relationship between power, force, and velocity is ______?

Kipish [7]

Answer: P = FV becuase all of them are equal in strength.

Explanation: In the straight forward cases where a constant force moves an object at constant velocity, the power is just P = Fv. In a more general case where the velocity is not in the same direction as the force, then the scalar product offorce and velocity must be used. P = Power, F = Force, V = Velocity.

Hoped This Helped!

5 0

1 year ago

Read 2 more answers

A hot-water bottle contains 787 g of water at 75∘C. If the liquid water cools to body temperature (37 ∘C), how many kilojoules o

IgorC [24]

Answer:

$Q = 787 gr * 1 \frac{cal}{gr C} *(37-75)C = -29906 cal$

So then the answer for this case would be 29906 cal but we need to convert this into KJ and we know that 1 cal = 4.184 J and if we convert we got:

$29906 cal *\frac{4.184 J}{1 cal}* \frac{1KJ}{1000 J}= 125.127 KJ$

Explanation:

For this case we know the mass of the water given :

$m = 787 gr$

And we know that the initial temperature for this water is $T_i =75 C$ .

We want to cool this water to the human body temperature $T_f = 37 C$

Since the temperatures given are not near to 0C (fusion point) or 100C (the boling point) we don't need to use latent heat, then the only heat involved for this case is the sensible heat given by:

$Q= m c_p \Delta T$

Where $c_p$ represent the specific heat for the water and this value from tables we know that $c_p =1 \frac{cal}{gr C}$ for the water.

So then we have everything in order to replace into the formula of sensible heat and we got:

$Q = 787 gr * 1 \frac{cal}{gr C} *(37-75)C = -29906 cal$

So then the answer for this case would be 29906 cal but we need to convert this into KJ and we know that 1 cal = 4.184 J and if we convert we got:

$29906 cal *\frac{4.184 J}{1 cal}* \frac{1KJ}{1000 J}= 125.127 KJ$

8 0

3 years ago

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A 56 kg astronaut stands on a bathroom scale inside a rotating circular space station. The radius of the space station is 250 m.

Zielflug [23.3K]

Answer:

The speed of space station floor is 49.49 m/s.

Explanation:

Given that,

Mass of astronaut = 56 kg

Radius = 250 m

We need to calculate the speed of space station floor

Using centripetal force and newton's second law

$F=mg$

$\dfrac{mv^2}{r}=mg$

$\dfrac{v^2}{r}=g$

$v=\sqrt{rg}$

Where, v = speed of space station floor

r = radius

g = acceleration due to gravity

Put the value into the formula

$v=\sqrt{250\times9.8}$

$v=49.49\ m/s$

Hence, The speed of space station floor is 49.49 m/s.

6 0

3 years ago

A wind turbine works by slowing the air that passes its blades and converting much of the extracted kinetic energy to electric e

ddd [48]

Answer:

2649600 Joules

Explanation:

Efficiency = 40%

m = Mass of air = 92000 kg

v = Velocity of wind = 12 m/s

Kinetic energy is given by

$K=\frac{1}{2}mv^2\\\Rightarrow K=\frac{1}{2}\times 92000\times 12^2\\\Rightarrow K=6624000\ J$

The kinetic energy of the wind is 6624000 Joules

The wind turbine extracts 40% of the kinetic energy of the wind

$E=0.4\times K\\\Rightarrow E=0.4\times 6624000\\\Rightarrow E=2649600\ J$

The energy extracted by the turbine every second is 2649600 Joules

8 0

3 years ago

The ray diagram shows a vase that is placed beyond the center of curvature of a concave mirror.

SIZIF [17.4K]

The image will form in the vicinity of F. Its nature will be small and inverted

8 0

3 years ago

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