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cricket20 [7]
2 years ago
6

A 85.0kg man and a 65, 0kg woman are riding a Ferris wheel with a radius of 20.0m. What is the Ferris wheels tangential velocity

if the net centripetal force on the woman is 115N
Physics
1 answer:
zvonat [6]2 years ago
8 0

Answer:

The Ferris wheel's tangential (linear) velocity if the net centripetal force on the woman is 115 N is <u>3.92 m/s</u>.

Explanation:

Let's use <u>Newton's 2nd Law</u> to help solve this problem.

  • F = ma

The force acting on the Ferris wheel is the centripetal force, given in the problem: F_c=115 \ \text{N}.

The mass "m" is the <u>sum</u> of the man and woman's masses: 85+65= 150 \ \text{kg}.

The acceleration is the centripetal acceleration of the Ferris wheel: a_c=\displaystyle \frac{v^2}{r}.

Let's write an equation and solve for "v", the tangential (linear) acceleration.

  • \displaystyle 115=m(\frac{v^2}{r} )
  • \displaystyle 115 = (85+65)(\frac{v^2}{20})
  • \displaystyle 115=150(\frac{v^2}{20} )
  • .766667=\displaystyle(\frac{v^2}{20} )
  • 15.\overline{3}=v^2
  • v=3.9158

The Ferris wheel's tangential velocity is 3.92 m/s.

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Answer:

A) 1000 kg

Explanation:

vf = vi + at

0 = 20 + (a)(10)

a = -2.0 m/s^2

F = ma

2000 = (m)(2)

m = 1000 kg

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1 year ago
An astronomer is measuring the electromagnetic radiation emitted by two stars, both of which are assumed to be perfect blackbody
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What H &amp; N can friction produce?
Angelina_Jolie [31]
Friction produces heat hope this helps

5 0
3 years ago
Tim and Rick both can run at speed Vr and walk at speed Vw, with Vr &gt; Vw.
miss Akunina [59]

Answer:

Δt =  \frac{2D}{Vw+Vr} - \frac{D}{2Vr} - \frac{D}{2Vw}

Explanation:

Hi there!

Using the equation of speed for the whole trip, we can obtain the time each one needed to cover the distance D.

The speed (v) is calculated by dividing the traveled distance (d) over the time needed to cover that distance (t):

v = d/t

Rick traveled half of the distance at Vr and the other half at Vw. Then, when v = Vr, the distance traveled was D/2 and the time is unknown, Δt1:

Vr = D/ (2 · Δt1)

For the other half of the trip the expression of velocity will be:

Vw = D/(2 · Δt2)

The total time traveled is the sum of both Δt:

Δt(total) = Δt1 + Δt2

Then, solving the first equation for Δt1:

Vr = D/ (2 · Δt1)

Δt1 = D/(2 · Vr)

In the same way for the second equation:

Δt2 = D/(2 · Vw)

Δt + Δt2 = D/(2 · Vr) + D/(2 · Vw)

Δt(total) = D/2 · (1/Vr + 1/Vw)

The time needed by Rick to complete the trip was:

Δt(total) = D/2 · (1/Vr + 1/Vw)

Now let´s calculate the time it took Tim to do the trip:

Tim walks half of the time, then his speed could be expressed as follows:

Vw = 2d1/Δt  Where d1 is the traveled distance.

Solving for d1:

Vw · Δt/2 = d1

He then ran half of the time:

Vr = 2d2/Δt

Solving for d2:

Vr · Δt/2 = d2

Since d1 + d2 = D, then:

Vw · Δt/2 +  Vr · Δt/2 = D

Solving for Δt:

Δt (Vw/2 + Vr/2) = D

Δt = D / (Vw/2 + Vr/2)

Δt = D/ ((Vw + Vr)/2)

Δt = 2D / (Vw + Vr)

The time needed by Tim to complete the trip was:

Δt = 2D / (Vw + Vr)

Let´s find the diference between the time done by Tim and the one done by Rick:

Δt(tim) - Δt(rick)

2D / (Vw + Vr) - (D/2 · (1/Vr + 1/Vw))

\frac{2D}{Vw+Vr} - \frac{D}{2Vr} - \frac{D}{2Vw} = Δt

Let´s check the result. If Vr = Vw:

Δt = 2D/2Vr - D/2Vr - D/2Vr

Δt = D/Vr - D/Vr = 0

This makes sense because if both move with the same velocity all the time both will do the trip in the same time.

8 0
3 years ago
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Answer:

Caustic soda would be least affected by a gas curtailment

Explanation:

Let,

X1= ammonia

X2= ammonium phosphate

X3= ammonium nitrate

X4 = Urea

X5= hydro-fluoric acid

X6= Chlorine

X7= Caustic soda

X8= Vinyl chloride monomer.

Agri chem’s current natural gas usage

= (1200 × 8 + 540 × 10 + 490 × 12 + …)                  

= 85,680,000 cu. ft. per day  

When, the curtailment is 20%, availability is

= 0.8 X  85,680

= 68,554,000 cu. ft. per day

Therefore, the gas constant

= 8X1 + 10X2 + 12X3 + 12X4 + 7X5 + 18X6 + 20X7 + 14X8 ≤68,544

When, the curtailment is 40% availability is

= 0.6 X 85,680

= 51,408,000 cu. ft. per day

Constant = 8X1 + 10X2 + 12X3 + 12X4 + 7X5 + 18X6 + 20X7 + 14X8≤ 51,408

                     

7 0
3 years ago
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