All categories

3 years ago

Which formula can be used to find the tangential speed of an orbiting object?

Physics

1 answer:

emmasim [6.3K]3 years ago

5 0

Answer: A.) v = 2πr/T

Explanation:

The tangential speed of an orbiting object can be obtained by the product of the radius of the orbit and the angular speed of the object in a circular motion.

This the tangential seed can be represented mathematically as :

Tangential speed (v) = angular speed(ω) × radius(r)

v = r × ω --------(1)

Recall:

ω = 2π/T

Substituting ω = 2π/T in equation (1)

v = r × 2π/T

v = 2πr/T

You might be interested in

suppose that a particle accelerator is used to move two beams of particles in opposite directions. in a particular region, elect

dusya [7]

The magnitude of average current is 2.02 * 10⁻¹⁴ A.

<h3 /><h3>What is current?</h3>

A stream of charged particles, such as electrons or ions, travelling through an electrical conductor or a vacuum is known as an electric current. The net rate of electric charge flowing through a surface or into a control volume is how it is calculated.

Number of electrons passing per second = 4910/ 0.0662 = 74169.18

Number of proton passing per second = 3485/0.0662 = 52643.50

current due to electrons,

I = ne

= 74169.18 * 1.6 * 10⁻¹⁹

= 1.18 * 10⁻¹⁴ A

current due to protons,

I = ne

= 52643.50 * 1.6 * 10⁻¹⁹

= 0.84 * 10⁻¹⁴ A

total current I = Ie + Ip

I =( 1.18 + 0.84 ) * 10⁻¹⁴ A

I = 2.02 * 10⁻¹⁴ A

to learn more about current go to -

brainly.com/question/1256531

#SPJ4

4 0

1 year ago

An electron is very far from the area. What is the ratio Ff/Fi of the electric force on the electron after the area is reduced t

timurjin [86]

The given question is incomplete. The complete question is as follows.

The irregularly shaped area of charge in the figure has surface charge density ηi. Each dimension (x and y) of the area is reduced by a factor of 3.68.

What is the ratio ηf/ηi where ηf is the final surface charge density? I found this value to be 13.5 and it is correct

An electron is very far from the area. What is the ratioFf/Fi of the electric force on the electronafter the area is reduced to the force before the area wasreduced? I found this to also be 13.5, but it is wrong.

Explanation:

Let us consider that expression for area will be as follows.

Area ( $\Delta A$ ) = $\Delta x \times \Delta y$

As, $\frac{\eta_{f}}{\eta_{i}} = (\frac{1}{3.68})^{2}$

= 13.5

Also, $\frac{F_{f}}{F_{i}}$ = 1

As the electron is very far from the area. Hence, it can be considered as a point charge. Whereas charge is also constant as the force is not changing.

3 0

4 years ago

A Van de Graaff generator is one of the original particle accelerators and can be used to accelerate charged particles like prot

Rzqust [24]

Answer:

595391.482946 m/s

$3.21875\times 10^{6}$

Explanation:

E = Energy = 1.85 keV

I = Current = 5.15 mA

e = Charge of electron = $1.6\times 10^{-19}\ C$

t = Time taken = 1 second

m = Mass of proton = $1.67\times 10^{-27}\ kg$

Velocity of proton is given by

$v=\sqrt{\dfrac{2E}{m}}\\\Rightarrow v=\sqrt{\dfrac{2\times 1.85\times 10^3\times 1.6\times 10^{-19}}{1.67\times 10^{-27}}}\\\Rightarrow v=595391.482946\ m/s$