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cupoosta [38]
3 years ago
8

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

Physics
2 answers:
Naily [24]3 years ago
8 0

Answer:

I think the answer is C. V= G  m central/r

Explanation:

Luden [163]3 years ago
6 0
The correct formula for calculating the tangential speed of an orbiting object is V(t)=wr.
V(t)= Tangential Speed
w= Angular Velocity
r= Radius of the Path

Hope this helps.
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Answer:44

Explanation:

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8 0
2 years ago
The strength of the force of friction depends on which two factors?
Stolb23 [73]

Options A and D are correct. The strength of the force of friction depends on the objects' sizes and weights and the heat generated by the friction and the types of surfaces involved.

<h3 /><h3>What is the friction force?</h3>

It is a type of opposition force acting on the surface of the body that tries to oppose the motion of the body. its unit is Newton (N).

Mathematically, it is defined as the product of the coefficient of friction and normal reaction.

On resolving the given force and acceleration in the different components and balancing the equation gets. Components in the x-direction.

The strength of the force of friction depends on the two factors, as;

A. The objects' sizes and weights.

D. The heat generated by the friction and the types of surfaces involved.

Hence, options A and D are correct.

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3 0
2 years ago
Name three elements that are good conductors of heat
Vera_Pavlovna [14]
Copper, gold and silver are three of them.
7 0
3 years ago
An electron and a 0.033 0-kg bullet each have a velocity of magnitude 495 m/s, accurate to within 0.010 0%. Within what lower li
lara31 [8.8K]

Answer:

1.170*10^-3 m

3.23*10^-32 m

Explanation:

To solve this, we apply Heisenberg's uncertainty principle.

the principle states that, "if we know everything about where a particle is located, then we know nothing about its momentum, and vice versa." it also can be interpreted as "if the uncertainty of the position is small, then the uncertainty of the momentum is large, and vice versa"

Δp * Δx = h/4π

m(e).Δv * Δx = h/4π

If we make Δx the subject of formula, by rearranging, we have

Δx = h / 4π * m(e).Δv

on substituting the values, we have

for the electron

Δx = (6.63*10^-34) / 4 * 3.142 * 9.11*10^-31 * 4.95*10^-2

Δx = 6.63*10^-34 / 5.67*10^-31

Δx = 1.170*10^-3 m

for the bullet

Δx = (6.63*10^-34) / 4 * 3.142 * 0.033*10^-31 * 4.95*10^-2

Δx = 6.63*10^-34 / 0.021

Δx = 3.23*10^-32 m

therefore, we can say that the lower limits are 1.170*10^-3 m for the electron and 3.23*10^-32 for the bullet

7 0
3 years ago
Any two waves that meet as they travel through the same medium will interact. If the two waves are in phase their crests will co
MrRissso [65]

Answer:

constructive

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6 0
3 years ago
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