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Fiesta28 [93]
3 years ago
10

How often do spring and neap tides occur?

Physics
2 answers:
balandron [24]3 years ago
8 0

Answer:

the answer is B. Twice a month

Explanation:

on Edge.

frez [133]3 years ago
3 0

Answer:

B twice a month

Explanation:

I did it on edge

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Honeybees acquire a positive charge while flying through the air. The rapid motion of the wings through the air provides the "fr
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A boy throws a ball straight up with a speed of 21.5 m/s. The ball has a mass of 0.19 kg. How much gravitational potential energ
astra-53 [7]

Answer:

Explanation:

The equation fo potential energy is PE = mgh, where m is the mass of the ball, g is the pull of gravity (constant at 9.8), and h is the max height of the ball. What we do not have here is that height. We need to first solve for it using one-dimensional equations. What we have to know above all else, is that the final velocity of an object at its max height is always 0. That allows us to use the equation

v_f=v_0+at where vf is the final velocity and v0 is the initial velocity. We will find out how long it takes for the object to reach that max height first and then use that time to find out what that max height is. Baby steps here...

0 = 21.5 + (-9.8)t and

-21.5 = -9.8t so

t = 2.19 seconds (Keep in mind that if I used the rules correctly for sig fig's, the answer you SHOULD get is not one shown, so I had to adjust the sig fig's and break the rules. But you know what they say about rules...)

Now we will use that time to find out the max height of the object in the equation

Δx = v_0t+\frac{1}{2}at^2 and filling in:

Δx = 21.5(2.19)+\frac{1}{2}(-9.8)(2.19)^2 which simplifies down a bit to

Δx = 47.1 - 23.5 so

Δx = 23.6 meters.

Now we can plug that in to the PE equation to find the PE of the object:

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PE = 43.9 J

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3 years ago
In the circuit diagram, what does the line segment with two circles at the end represent?
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If the volume of the balloon is 500 m3m3 and the surrounding air is at 15.0 ∘C∘C, what must the temperature of the air in the ba
Tresset [83]

Answer:

Temperature of the air in the balloon = 272°C

Explanation:

Given:

Volume of balloon = 500 m³

Air temperature = 15° C = 273 + 15 = 288 K

Total weight = 290 kg

Density of air = 1.23 kg/m³

Find:

Temperature of the air in the balloon

Computation:

Density of hot air = Density of air - [Total weight / Volume of balloon]

Density of hot air = 1.23 - [290 - 500]

Density of hot air = 0.65 kg/m³

[Density of hot air][Temperature of the air in the balloon] = [Density of air][Air temperature ]

Temperature of the air in the balloon = [(1.23)(288)]/(0.65)

Temperature of the air in the balloon = 544.98

Temperature of the air in the balloon = 545 K

Temperature of the air in the balloon = 545 - 273 = 272°C

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