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3 years ago

What does basketballs have inside of it

1 answer:

4 0

Hi there!

Great question!

Basketballs have air inside them. A special pump is used to insert the air. That's why you can lift the basketballs off the ground easily. If it was a solid, though, you'd hardly be able to lift the ball up! Basketballs can float, too, because anything with air inside can float. If it were solid, it would sink in the water easily.

Hope this helps! :D

You might be interested in

Optical astronomers need a clear, dark sky to collect good data. Explain why radio astronomers have no problem observing in the

gregori [183]

Answer:

In the clarification portion elsewhere here, the definition of the concern is mentioned.

Explanation:

So like optical telescopes capture light waves, introduce it to concentrate, enhance it, as well as make it usable through different instruments via study, so radio telescopes accumulate weak signal light waves, introduce that one to focus, enhance it, as well as make this information available during research. To research naturally produced radio illumination from stars, galaxies, dark matter, as well as other natural phenomena, we utilize telescopes.

Optical telescopes detect space-borne visible light. There are some drawbacks of optical telescopes mostly on the surface:

Mostly at night would they have been seen.
Unless the weather gets cloudy, bad, or gloomy, they shouldn't be seen.

Although radio telescopes monitor space-coming radio waves. Those other telescopes, when they are already typically very massive as well as costly, have such an improvement surrounded by optical telescopes. They should be included in poor weather and, when they travel through the surrounding air, the radio waves aren't obscured by clouds. Throughout the afternoon and also some at night, radio telescopes are sometimes used.

3 0

2 years ago

The frequency of a certain sound is 440 Mz. What is the wavelength of this sound when the temperature of the air is (a) 20°C; (b

Serggg [28]

Answer:

Explanation:

We know the frequency and the velocity, both of which have good units. All we have to do is rearrange the equation and solve for

Let's plug in our given values and see what we get!

340

440

−

0.773

3 0

2 years ago

A worker is thinking about two ways to get a box up 1.2 m onto a loading dock. He can use a force of 250 N to lift it straight u

harina [27]

Answer:

<em>The second option has a lower power output. P=30 W</em>

Explanation:

<u>Mechanical Power </u>

It is a physical magnitude that measures the rate a work W is done over time t.

$\displaystyle P=\frac{W}{t}$

Since W=F.d

$\displaystyle P=\frac{F.d}{t}$

The first option means the worker will lift the box by a distance of 1.2 meters in 3 seconds by applying 250 N of force. That produces a power of

$\displaystyle P=\frac{(250). (1.2)}{3}=100\ Watt$

The second option requires the worker applies 75 N of force and travel a distance of 4 meters for 10 seconds, thus the power is

$\displaystyle P=\frac{(75). (4)}{10}=30\ Watt$

The second option has a lower power output

7 0

3 years ago

A coin is dropped. What is its displacement after 0.30 s.

Morgarella [4.7K]

Assuming this coin is on earth and that it wasn’t dropped forcefully:
Use the formula d = 1/2at^2. Rewriting using a=g and solving for height h gets us h = 1/2(9.8)t^2.
In this case that would get that the change in height h is 0.5(9.8)(0.3^2) = 0.441 m.

3 0

3 years ago

Compare the catching of two different water balloons.

Stels [109]

Answer:

a. The volume V₁ and V₂

b. The case that involves the greatest momentum change = Case B

c. The case that involves the greatest impulse = Case B

d. b. The case that involves the greatest force = Case B

Explanation:

Here we have

Case A: V₁ = 150-mL, v₁ = 8 m/s

Case B: V₂ = 600-mL, v₁ = 8 m/s

a. The variable that is different for the two cases is the volume V₁ and V₂

b. The momentum change is by the following relation;

ΔM₁ = Mass, m × Δv₁

The mass of the balloon are;

Δv₁ = Change in velocity = Final velocity - Initial velocity

Mass = Density × Volume

Density of water = 0.997 g/mL

Case A, mass = 150 × 0.997 = 149.55 g

Case B, mass = 600 × 0.997 = 598.2 g

The momentum change is;

Case A: Mass, m × Δv₁ = 149.55 g/1000 × 8 m/s = 1.1964 g·m/s

Case B: Mass, m × Δv₁ = 598.2/1000 × 8 = 4.7856 g·m/s

Therefore Case B has the greatest momentum change

The case that has the gretest momentum change = Case B

c. The momentum change = impulse therefore Case B involves the greatest impulse

d. Here we have;

Impulse = Momentum change = $F_{average}$ × Δt = mΔV

∴ $F_{average}$ = m·ΔV/Δt

∴ For Case A $F_{average}$ = 149.55×8/Δt = 1196.4/Δt N

For Case B $F_{average}$ = 598.2×8/Δt = 4785.6/Δt

Where Δt is the same for Case A and Case B, $F_{average}$ for Case B >> $F_{average}$ for Case B

Therefore, Case B involves the greatest force.

4 0

3 years ago