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Mnenie [13.5K]
2 years ago
15

A big wheel at a theme park has a diameter of 14m and people on the ride complete one revolution in 24s. calculate the distance

a rider moves in 3.0 minutes​
Physics
1 answer:
just olya [345]2 years ago
7 0

Explanation:

We'll call the radius r and the diameter d:

We also assume that the riders are at a distance r = d/2 = 7m from the center of the wheel.

The period of the wheel is 24s. The tangent velocity of the wheel (and the riders) will be: (2pi/T)*r = 0.8 m/s (circa).

It means that in 3 minutes (180 seconds) they'll run 0.8 m/s * 180s = 144m.

Hopefully I understood the question. If yes, that's the answer.

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Water flows over a section of Niagara Falls at rate of 1.1×10^6 kg/s and falls 49.4 m. How much power is generated by the fallin
DerKrebs [107]

Answer:

<em>The power generated is =  5.33×10⁸ Watt. </em>

Explanation:

Power: Power can be defined as the time rate of doing work. The S.I unit of power is <em>Watt(W).</em>

<em>Mathematically,</em>

<em>Power (P) = Work done/time or Energy/time</em>

P = mgh/t............................... Equation 1

P = δgh............................. Equation 2

Where δ = fall rate, g = acceleration due to gravity, h = height.

<em>Given: </em>δ = 1.1×10⁶ kg/s, h = 49.4 m g = 9.81 m/s²

Substituting these values into equation 2

P = 1.1×10⁶×49.4×9.81

P = 533.08×10⁶

<em>P = 5.33×10⁸ Watt.</em>

<em>Thus the power generated is =  5.33×10⁸ Watt. </em>

5 0
3 years ago
Is the ratio between the sine of an angle of incidence to the sine of an angle of refraction is called the refractive index?
motikmotik

Answer:

Yes

Explanation:

Yes it is called the refractive index denoted by n

n=sin<i/sin<r

6 0
3 years ago
Read 2 more answers
A cat with a mass of 5.00 kg pushes on a 25.0 kg desk with a force of 50.0N to jump off. What is the force on the desk?
olya-2409 [2.1K]

Answer:

First of all the formula is F= uR,( force= static friction× reaction)

mass= 5+25=30

F= 50

R= mg(30×10)=300

u= ?

F=UR

u= F/R

u= 50/300=0.17N

3 0
3 years ago
Certain insects can achieve seemingly impossible accelerations while jumping. the click beetle accelerates at an astonishing 400
hichkok12 [17]

(a) The launching velocity of the beetle is 6.4 m/s

(b) The time taken to achieve the speed for launch is 1.63 ms

(c) The beetle reaches a height of 2.1 m.

(a) The beetle starts from rest and accelerates with an upward acceleration of 400 g and reaches its launching speed in a distance 0.53 cm. Here g is the acceleration due to gravity.

Use the equation of motion,

v^2=u^2+2as

Here, the initial velocity of the beetle is u, its final velocity is v, the acceleration of the beetle is a, and the beetle accelerates over a distance s.

Substitute 0 m/s for u, 400 g for a, 9.8 m/s² for g and 0.52×10⁻²m for s.

v^2=u^2+2as\\ = (0 m/s)^2+2 (400)(9.8 m/s^2)(0.52*10^-^2 m)\\ =40.768 (m/s)^2\\ v=6.385 m/s

The launching speed of the beetle is <u>6.4 m/s</u>.

(b) To determine the time t taken by the beetle for launching itself upwards is determined by using the equation of motion,

v=u+at

Substitute 0 m/s for u, 400 g for a, 9.8 m/s² for g and 6.385 m/s for v.

v=u+at\\ 6.385 m/s = (0 m/s) +400(9.8 m/s^2)t\\ t = \frac{6.385 m/s}{3920 m/s^2} = 1.63*10^-^3s=1.63 ms

The time taken by the beetle to launch itself upwards is <u>1.62 ms</u>.

(c) After the beetle launches itself upwards, it is acted upon by the earth's gravitational force, which pulls it downwards towards the earth with an acceleration equal to the acceleration due to gravity g. Its velocity reduces and when it reaches the maximum height in its path upwards, its final velocity becomes equal to zero.

Use the equation of motion,

v^2=u^2+2as

Substitute 6.385 m/s for u, -9.8 m/s² for g and 0 m/s for v.

v^2=u^2+2as\\ (0m/s)^2=(6.385 m/s)^2+2(-9.8m/s^2)s\\ s=\frac{(6.385 m/s)^2}{2(9.8m/s^2)} =2.08 m

The beetle can jump to a height of <u>2.1 m</u>



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HELP ASP PLEASED AND THANKS NO LINKS PLEASED AND NO FILES
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Answer:

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