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kirza4 [7]
2 years ago
8

The force created when the court pushes LeBron James upwards is equal to which force?

Physics
2 answers:
nika2105 [10]2 years ago
8 0

Answer:

When LeBron James jumps, he is driving force into the court. How is this force created? His force is created by the energy stored inside his muscles.

Explanation:

i hope this helps you very much

max2010maxim [7]2 years ago
7 0

Answer:

When LeBron James jumps, he is driving force into the court. How is this force created? Suggested answer: This force is created by the energy stored inside his muscles.

Explanation:

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In a Hydrogen atom an electron rotates around a stationary proton in a circular orbit with an approximate radius of r =0.053nm.
leonid [27]

Answer:

(a): F_e = 8.202\times 10^{-8}\ \rm N.

(b): F_g = 3.6125\times 10^{-47}\ \rm N.

(c): \dfrac{F_e}{F_g}=2.27\times 10^{39}.

Explanation:

Given that an electron revolves around the hydrogen atom in a circular orbit of radius r = 0.053 nm = 0.053\times 10^{-9} m.

Part (a):

According to Coulomb's law, the magnitude of the electrostatic force of interaction between two charged particles of charges q_1 and q_2 respectively is given by

F_e = \dfrac{k|q_1||q_2|}{r^2}

where,

  • k = Coulomb's constant = 9\times 10^9\ \rm Nm^2/C^2.
  • r = distance of separation between the charges.

For the given system,

The Hydrogen atom consists of a single proton, therefore, the charge on the Hydrogen atom, q_1 = +1.6\times 10^{-19}\ C.

The charge on the electron, q_2 = -1.6\times 10^{-19}\ C.

These two are separated by the distance, r = 0.053\times 10^{-9}\ m.

Thus, the magnitude of the electrostatic force of attraction between the electron and the proton is given by

F_e = \dfrac{(9\times 10^9)\times |+1.6\times 10^{-19}|\times |-1.6\times 10^{-19}|}{(0.053\times 10^{-9})^2}=8.202\times 10^{-8}\ \rm N.

Part (b):

The gravitational force of attraction between two objects of masses m_1 and m_1 respectively is given by

F_g = \dfrac{Gm_1m_2}{r^2}.

where,

  • G = Universal Gravitational constant = 6.67\times 10^{-11}\ \rm Nm^2/kg^2.
  • r = distance of separation between the masses.

For the given system,

The mass of proton, m_1 = 1.67\times 10^{-27}\ kg.

The mass of the electron, m_2 = 9.11\times 10^{-31}\ kg.

Distance between the two, r = 0.053\times 10^{-9}\ m.

Thus, the magnitude of the gravitational force of attraction between the electron and the proton is given by

F_g = \dfrac{(6.67\times 10^{-11})\times (1.67\times 10^{-27})\times (9.11\times 10^{-31})}{(0.053\times 10^{-9})^2}=3.6125\times 10^{-47}\ \rm N.

The ratio \dfrac{F_e}{F_g}:

\dfrac{F_e}{F_g}=\dfrac{8.202\times 10^{-8}}{3.6125\times 10^{-47}}=2.27\times 10^{39}.

6 0
3 years ago
Paragraph about how the constellations were used by ancient civilizations.
nata0808 [166]

Answer:

What does this mean ?

Explanation:

8 0
3 years ago
Which item is not considered electromagnetic energy?
suter [353]
Sound waves are known to be the one that's not considered as a type of electromagnetic energy. As for microwaves and x-rays, they tend to share the same frequencies that can be considered as electromagnetic, and sound waves have a different frequency than them.
5 0
3 years ago
When the ratio of 2 variables is constant what can their relationship be described as
salantis [7]
It can be described as a constant variation
3 0
2 years ago
A uniform rod is 2.0 m long. The rod is pivoted about a horizontal, frictionless pin through one end. The moment of inertia of t
emmasim [6.3K]

Answer:

Angular acceleration = 6.37rad/sec²

Approximately, Angular acceleration =

6.4 rad/sec²

Explanation:

Length of the rod = 2.0m long

Inclination of the rod (horizontal) = 30°

Mass of the rod is not given so we would refer to it as = M

Rotational Inertia of the Rod(I) = 1/3ML²

Angular Acceleration = ?

There is an equation that shows us the relationship between Torque and Angular acceleration.

The equation is :

Torque(T) = Inertia × Angular Acceleration

Angular acceleration = Torque ÷ Inertia

Where:

Torque = L/2(MgCosθ)

Where M = Mass

L = Length = 2.0m

θ = Inclination of the rod (horizontal) = 30°

g = Acceleration due to gravity = 9.81m/s²

Inertia = 1/3ML²

Angular Acceleration =  (Mass × g × Cos (30°) × (L÷2)) ÷ 1/3ML²

Angular Acceleration =

(3 × g × cos 30°) ÷ 2× L

Angular Acceleration = (3 × 9.81m/s² × cos 30°) ÷ 2× L

Angular Acceleration = 3 × 9.81m/s² × cos 30°) ÷ 2× 2.0m

Angular Acceleration = 6.37rad/sec²

Approximately Angular Acceleration =

6.4rad/sec²

5 0
3 years ago
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