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3 years ago

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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gizmo_the_mogwai [7]

Answer: is 13

Explanation:

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3 years ago

the amount of surface area of the block contact with the surface is 2.03*10^-2*m2 what is the average pressure exerted on the su

CaHeK987 [17]

Complete question:

A block of solid lead sits on a flat, level surface. Lead has a density of 1.13 x 104 kg/m3. The mass of the block is 20.0 kg. The amount of surface area of the block in contact with the surface is 2.03*10^-2*m2, What is the average pressure (in Pa) exerted on the surface by the block? Pa

Answer:

The average pressure exerted on the surface by the block is 9655.17 Pa

Explanation:

Given;

density of the lead, ρ = 1.13 x 10⁴ kg/m³

mass of the lead block, m = 20 kg

surface area of the area of the block, A = 2.03 x 10⁻² m²

Determine the force exerted on the surface by the block due to its weight;

F = mg

F = 20 x 9.8

F = 196 N

Determine the pressure exerted on the surface by the block

P = F / A

where;

P is the pressure

P = 196 / (2.03 x 10⁻²)

P = 9655.17 N/m²

P = 9655.17 Pa

Therefore, the average pressure exerted on the surface by the block is 9655.17 Pa

6 0

3 years ago

A tiger leaps horizontally out of a tree that is 6.00 m high. If he lands 2.00 m from the base of the tree, calculate his initia

Musya8 [376]

Answer:

The initial speed of the tiger is 1.80 m/s

Explanation:

Hi there!

The equation of the position vector of the tiger is the following:

r = (x0 + v0 · t, y0 + 1/2 · g · t²)

Where:

r = position vector at a time t.

x0 = initial horizontal position.

v0 = initial horizontal velocity.

t = time.

y0 = initial vertical position,

g = acceleration due to gravity.

Let´s place the origin of the frame of reference on the ground at the point where the tree is located so that the initial position vector will be:

r0 = (0.00, 6.00) m

We can use the equation of the vertical component of the position vector to obtain the time it takes the tiger to reach the ground.

y = y0 + 1/2 · g · t²

When the tiger reaches the ground, y = 0:

0 = 6.00 m - 1/2 · 9.81 m/s² · t²

2 · (-6.00 m) / -9.81 m/s² = t²

t = 1.11 s

We know that in 1.11 s the tiger travels 2.00 m in the horizontal direction. Then, using the equation of the horizontal component of the position vector we can find the initial speed:

x = x0 + v0 · t

At t = 1.11 s, x = 2.00 m

x0 = 0

2.00 m = v0 · 1.11 s

2.00 m / 1.11 s = v0

v0 = 1.80 m/s

The initial speed of the tiger is 1.80 m/s

4 0

3 years ago

Tres móviles, A, B y C, con las mismas características tienen los siguientes estados: A se encuentra inicialmente en reposo y ca

Morgarella [4.7K]

Answer:

La respuesta es sí, hay una fuerza que actúa sobre el móvil A y es la única fuerza ya que A cae libremente bajo la influencia de la fuerza.

Explanation:

Según la primera ley de movimiento de Newton, un cuerpo continuará en su estado de reposo o en un movimiento uniforme en línea recta a menos que actúen sobre él fuerzas impresas.

Dado que el móvil A cae libremente, desde su estado de reposo inicial, según la primera ley de movimiento de Newton, experimenta una fuerza que actúa sobre él para hacer que caiga y continúe en caída libre.

El móvil B se mueve con una velocidad constante, por lo tanto, de acuerdo con la primera ley de movimiento de Newton, no hay fuerzas impresas que actúen sobre él.

El móvil C está completamente en reposo en el suelo, por lo tanto, tampoco hay fuerzas que actúen sobre él.

La respuesta es sí, hay una fuerza actuando sobre el móvil A y es la única fuerza cuando A cae libremente bajo la influencia de la fuerza.

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3 years ago

A 5.92 g object moving to the right at 17.1 cm/s makes an elastic head-on collision with an 11.84 g object that is initially at

Lelechka [254]

Answer:

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Explanation:

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Rearranging terms, we have:

$m_{1} * (v_{10} - *v_{1f} ) = m_{2} * v_{2f} (2)$

We also know that the collision is elastic, so total kinetic energy must be conserved , as follows:

$\Delta K = 0 \\ \\ \frac{1}{2} * m_{1} *v_{10} ^{2} = \frac{1}{2}* m_{1} *v_{1f} ^{2} + \frac{1}{2}* m_{2} *v_{2f} ^{2} (3)$

Rearranging , and simplifying common terms, we have:

$m_{1}* (v_{10} ^{2} -v_{1f} ^{2} ) = m_{2} *v_{2f} ^{2} (4)$

Replacing by the givens, doing some algebra and dividing (4) by (2), we find the following relationship:

$v_{10} + v_{1f} = v_{2f}$

Replacing the expression above in (1), as m₂ = 2*m₁, we can find the value of v₁f, as follows:

$m_{1} * v_{10} = m_{1} * v_{1f} +2*m_{1} * (v_{10} + v_{1f})\\ \\ -(m_{1} * v_{10}) = 3* m_{1} *v_{1f} \\ \\ v_{1f} = - \frac{v_{10} }{3} = \frac{-17.1cm/s}{3} = -5.7 cm/s$

7 0

3 years ago