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Anarel [89]
3 years ago
6

The center of gravity of a basketball is located _______.

Physics
1 answer:
zhenek [66]3 years ago
5 0

Answer:

at its centre

Explanation:

The centre of gravity is a point inside the body where entire weight os the body is said to be concentrated. For geometrical symmetric bodies it lies at the geometric centre of the body.

The base ball is a spherical body, so its centre of gravity lies at its centre.

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How does the scattering of tiny particles in the air affect the colors of the sky at sunrise and sunset
timofeeve [1]

It affects the colors of the sky at sunrise and sunset because only the longer-wavelengths colors are visible.

4 0
4 years ago
Read 2 more answers
Just how strong is the electric force? suppose you had two small boxes, each containing 1.0 g of protons. (a) if one were placed
vlabodo [156]
Let's use Newton's Law of Second Motion: F=ma. When no other direct force is acting on the system, the acceleration is due to the gravity. The modified equation becomes: F = mg. So, yes, you need to take into account the gravitational accelerations in the moon and on Earth. 

g,moon = 1.622 m/s²
g,Earth = 9.81 m/s²

The net force is the tension of the string:

F,Earth - F,moon = Tension
Tension = (1/1000 kg)(9.81 m/s²) - (1/1000 kg)(1.622 m/s²)
Tension = 8.188×10⁻³ N

To convert, 1 pound force is equal to 4.45 Newtons:

Tension = 8.188×10⁻³ N * 1 lbf/4.45 N
Tension = 1.84×10⁻³ lbf
3 0
3 years ago
I will give away all my points for whoever get's this right
romanna [79]

Answer:

Pencil lead.

Explanation:

no need for an explemation you can put two and two together XD

5 0
4 years ago
Read 2 more answers
A4.0 kg object is moving with speed 2.0 m/s. A 1.0 kg object is moving with speed 4.0 m/s. Both objects encounter the same const
Evgesh-ka [11]

Answer:

Both objects travel the same distance.

(c) is correct option

Explanation:

Given that,

Mass of first object = 4.0 kg

Speed of first object = 2.0 m/s

Mass of second object = 1.0 kg

Speed of second object = 4.0 m/s

We need to calculate the stopping distance

For first particle

Using equation of motion

v^2=u^2+2as

Where, v = final velocity

u = initial velocity

s = distance

Put the value in the equation

0= u^2-2as_{1}

s_{1}=\dfrac{u^2}{2a}....(I)

Using newton law

a=\dfrac{F}{m}

Now, put the value of a in equation (I)

s_{1}=\dfrac{8}{F}

Now, For second object

Using equation of motion

v^2=u^2+2as

Put the value in the equation

0= u^2-2as_{2}

s_{2}=\dfrac{u^2}{2a}....(I)

Using newton law

F = ma

a=\dfrac{F}{m}

Now, put the value of a in equation (I)

s_{2}=\dfrac{8}{F}

Hence, Both objects travel the same distance.

6 0
3 years ago
A sound wave, with speed of 340 m/s has a frequency of 17,000 Hz. What
Fiesta28 [93]
Wavelength = speed / frequency = 340 / 17000 = 0.02 m
6 0
3 years ago
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