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svetlana [45]
3 years ago
12

What is the volume in liters of the basketball?

Physics
2 answers:
babymother [125]3 years ago
5 0

he basket ball diameter used by NBA men players is around 9.55 inches

So diameter = 9.55 inch = 0.243 m

So radius of basket ball = 0.1215 m

Volume , V=\frac{4}{3} \pi r^3

V = \frac{4}{3}* \pi *0.1215^3=0.0075 m^3=7.5 L

So volume of basket ball used by men NBA players = 7.5 L

   

gtnhenbr [62]3 years ago
5 0

Answer:

7.31 L

Explanation:

A standard NBA basketball has a radius (r) of about 4.74 in. If we consider it to be an almost perfect sphere, its volume (V) is:

V = 4/3 × π × r³ = 4/3 × π × (4.74 in)³ = 446 in³

We know that 1 liter is equal to 61.02 in³. The volume, in liters, corresponding to 446 in³ is:

446 in³ × (1 L/ 61.02 in³) = 7.31 L

The volume of a basketball is about 7.31 L.

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Answer:

dont cry

Explanation:

4 0
3 years ago
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What is it asking MathPhys im sorry i dont know but i tried teaching myself physics but i cant
RSB [31]

Answer:

The puck B remains at the point of collision.

Explanation:

This is an elastic collision, so both momentum and energy are conserved.

The mass of both pucks is m.

The velocity of puck B before the collision is vb.

The velocity of puck A and B after the collision is va' and vb', respectively.

Momentum before = momentum after

m vb = m vb' + m va'

vb = vb' + va'

Energy before = energy after

½ m vb² = ½ m vb'² + ½ m va'²

vb² = vb'² + va'²

Substituting:

(vb' + va')² = vb'² + va'²

vb'² + 2 va' vb' + va'² = vb'² + va'²

2 va' vb' = 0

va' vb' = 0

We know that va' isn't 0, so:

vb' = 0

The puck B remains at the point of collision.

6 0
3 years ago
Carbon-14 has a half-life of 5,730 years. if the age of an object older than 50,000 years cannot be determined by radiocarbon da
Aleonysh [2.5K]
<h3><u>Answer;</u></h3>

Carbon-14 levels in a sample are undetectable after approximately 9 half lives

<h3><u>Explanation;</u></h3>
  • <em><u>The half life of Carbon-14 is 5,730 years . Half life is the time taken by a radioactive material to decay by half of its original mass.  Therefore, it  would take a time of 5730 years for a sample of 100 g of carbon-14 to decay to 50 grams</u></em>
  • <em><u>A period of 50,000 years, is equivalent to; </u></em>

<em><u>  50,000÷5,730 </u></em>

<em><u>= 8.73 half lives</u></em>

<em>Which is approximately equal to 9 half lives.</em>

  • Therefore, if the age of an object older than 50,000 years cannot be determined by radiocarbon dating, then <em><u>Carbon-14 levels in a sample are undetectable after approximately 9 half lives</u></em>.
6 0
3 years ago
Read 2 more answers
Car A with a mass of 725 kilograms is traveling east at an initial velocity of 15 meters/second. It collides head–on with car B,
ikadub [295]

Answer:

p_t_o_t_a_l=250kg\frac{m}{s}

Explanation:

<u>The total momentum of a system is defined by:</u>

(mv)_t_o_t=m_1v_1+m_2v_2+...

Where,

(mv)_t_o_t is the total momentum or it could be expressed also as p_t_o_t_a_l.

m_1 and m_2 represents the masses of the objects interacting in the system.

v_1 and v_2 are the velocities of the objects of the system.

<em>Remember: </em><em>The momentum is a fundamental physical magnitude of vector type.</em>

We have:

m_1=725 kg

v_1=15\frac{m}{s}\\m_2=625 kg

We are going to take the east side as positive, and the west side as negative. Then the velocity of the car B, has to be <u>negative</u>. It goes in a different direction from car A.

v_2=-17\frac{m}{s}

Then the total momentum of the system is:

p_t_o_t_a_l=m_1v_1+m_2v_2\\p_t_o_t_a_l=(725kg)(15\frac{m}{s})+(625kg)(-17\frac{m}{s})\\p_t_o_t_a_l=10875kg\frac{m}{s}-10625kg\frac{m}{s}\\p_t_o_t_a_l=250kg\frac{m}{s}

8 0
3 years ago
A force of 6600 N is exerted on a piston that has an area of 0.010 m2
sveticcg [70]

Answer:

Choice A: approximately 0.015\; \rm m^2, assuming that the two pistons are connected via some confined liquid to form a simple machine.

Explanation:

Assume that the two pistons are connected via some liquid that is confined. Pressure from the first piston:

\displaystyle P_1 = \frac{F_1}{A_1} = \frac{6.600\times 10^3\; \rm N}{1.0\times 10^{-2}\; \rm m^{2}} = 6.6\times 10^{5}\; \rm N \cdot m^{-2}.

By Pascal's Principle, because the first piston exerted a pressure of 6.6\times 10^{5}\; \rm N \cdot m^{-2} on the liquid, the liquid will now exert the same amount of pressure on the walls of the container.

Assume that the second piston is part of that wall. The pressure on the second piston will also be 6.6\times 10^{5}\; \rm N \cdot m^{-2}. In other words:

P_2 = P_1 = 6.6\times 10^{5}\; \rm N \cdot m^{-2}.

To achieve a force of 9.900 \times 10^3\; \rm N, the surface area of the second piston should be:

\displaystyle A_2 = \frac{F_2}{P_2} = \frac{9.900\times 10^{3}\; \rm N}{6.6\times 10^5\; \rm N \cdot m^{-2}} \approx 0.015\; \rm m^{2}.

4 0
3 years ago
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