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valentinak56 [21]
2 years ago
14

3. a) If the distance between two objects increases by a factor of 5 (5),

Physics
1 answer:
irina [24]2 years ago
3 0

Answer:

Answer for A

Explanation:

F1=GmM/r1^2

If r2 becomes r2=5r

F2=GmM/(25r^2)

Multiply with 25 gives to maintain the same force

I.e.,25F2=F1

F2=G(25m)M/25r^2=F1

By the factor 25 would change to increase to same.

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A car traveling on a flat (unbanked), circular track accelerates uniformly from rest with a tangential acceleration of 1.90 m/s2
Ahat [919]

Answer:

Approximately 0.608 (assuming that g = 9.81\; \rm N\cdot kg^{-1}.)

Explanation:

The question provided very little information about this motion. Therefore, replace these quantities with letters. These unknown quantities should not appear in the conclusion if this question is actually solvable.

  • Let m represent the mass of this car.
  • Let r represent the radius of the circular track.

This answer will approach this question in two steps:

  • Step one: determine the centripetal force when the car is about to skid.
  • Step two: calculate the coefficient of static friction.

For simplicity, let a_{T} represent the tangential acceleration (1.90\; \rm m \cdot s^{-2}) of this car.

<h3>Centripetal Force when the car is about to skid</h3>

The question gave no information about the distance that the car has travelled before it skidded. However, information about the angular displacement is indeed available: the car travelled (without skidding) one-quarter of a circle, which corresponds to 90^\circ or \displaystyle \frac{\pi}{2} radians.

The angular acceleration of this car can be found as \displaystyle \alpha = \frac{a_{T}}{r}. (a_T is the tangential acceleration of the car, and r is the radius of this circular track.)

Consider the SUVAT equation that relates initial and final (tangential) velocity (u and v) to (tangential) acceleration a_{T} and displacement x:

v^2 - u^2 = 2\, a_{T}\cdot x.

The idea is to solve for the final angular velocity using the angular analogy of that equation:

\left(\omega(\text{final})\right)^2 - \left(\omega(\text{initial})\right)^2 = 2\, \alpha\, \theta.

In this equation, \theta represents angular displacement. For this motion in particular:

  • \omega(\text{initial}) = 0 since the car was initially not moving.
  • \theta = \displaystyle \frac{\pi}{2} since the car travelled one-quarter of the circle.

Solve this equation for \omega(\text{final}) in terms of a_T and r:

\begin{aligned}\omega(\text{final}) &= \sqrt{2\cdot \frac{a_T}{r} \cdot \frac{\pi}{2}} = \sqrt{\frac{\pi\, a_T}{r}}\end{aligned}.

Let m represent the mass of this car. The centripetal force at this moment would be:

\begin{aligned}F_C &= m\, \omega^2\, r \\ &=m\cdot \left(\frac{\pi\, a_T}{r}\right)\cdot r = \pi\, m\, a_T\end{aligned}.

<h3>Coefficient of static friction between the car and the track</h3>

Since the track is flat (not banked,) the only force on the car in the horizontal direction would be the static friction between the tires and the track. Also, the size of the normal force on the car should be equal to its weight, m\, g.

Note that even if the size of the normal force does not change, the size of the static friction between the surfaces can vary. However, when the car is just about to skid, the centripetal force at that very moment should be equal to the maximum static friction between these surfaces. It is the largest-possible static friction that depends on the coefficient of static friction.

Let \mu_s denote the coefficient of static friction. The size of the largest-possible static friction between the car and the track would be:

F(\text{static, max}) = \mu_s\, N = \mu_s\, m\, g.

The size of this force should be equal to that of the centripetal force when the car is about to skid:

\mu_s\, m\, g = \pi\, m\, a_{T}.

Solve this equation for \mu_s:

\mu_s = \displaystyle \frac{\pi\, a_T}{g}.

Indeed, the expression for \mu_s does not include any unknown letter. Let g = 9.81\; \rm N\cdot kg^{-1}. Evaluate this expression for a_T = 1.90\;\rm m \cdot s^{-2}:

\mu_s = \displaystyle \frac{\pi\, a_T}{g} \approx 0.608.

(Three significant figures.)

7 0
3 years ago
Radio waves travel at the speed of light. What is the wavelength of a radio signal with a frequency of 9.45 x 10^7 Hz?
Klio2033 [76]

Answer:

So I never really knew you

God, I really tried to

Blindsided, addicted

Felt we could really do this

But really I was foolish

Hindsight, it's obvious

Explanation:

3 0
3 years ago
Which of the following is a type of natural disaster that has impacted Florida? A. Avalanche B. Flood C. Typhoon D. Volcano
snow_tiger [21]

Answer:

B-flood

Explanation:

The Sunshine State is famous all around the world for many great things but also is famous because of a big array of natural disasters.From hurricanes, tropical storms, tornadoes, to fires and floods many natural disasters in Florida affect the lives of residents.

7 0
3 years ago
Read 2 more answers
Jane placed two bar magnets on her desk next to each other with like poles facing. She then sprinkled some iron filings around t
viktelen [127]

Answer:

B.

Explanation:

3 0
3 years ago
Read 2 more answers
if a 3-kg object has a momentum of 33 kg��m/s, what's its velocity? a. 99 m/s b. 36 m/s c. 11 m/s d. 9.1 m/s
horsena [70]
M=3kg
p=33kg.m/s
p=m*v
v=p/m
=33/3
=11m/s
thus option (c)
6 0
3 years ago
Read 2 more answers
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