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Marat540 [252]
3 years ago
9

Suppose the ring rotates once every 4.30 ss . If a rider's mass is 58.0 kgkg , with how much force does the ring push on her at

the top of the ride? Express your answer with the appropriate units.
Physics
1 answer:
Stells [14]3 years ago
6 0

Answer:

422.36 N

Explanation:

given,

time of rotation = 4.30 s

T = 4.30 s

Assuming the diameter of the ring equal to 16 m

radius, R = 8 m

v = \dfrac{2\pi R}{T}

v = \dfrac{2\pi\times 8}{4.30}

  v = 11.69 m/s

now, Force does the ring push on her at the top

- N - m g = \dfrac{-mv^2}{R}

N + m g = \dfrac{mv^2}{R}

N = \dfrac{mv^2}{R}- m g

N = m(\dfrac{v^2}{R}- g)

N = 58\times (\dfrac{11.69^2}{8}- 9.8)

N = 422.36 N

The force exerted by the ring to push her is equal to 422.36 N.

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forces F1 (east) and F2 are simultaneously applied to a 3.0kg mass. when F2 is east, a=5.0m/s² east and when F2 is west a=1.0m/s
zhenek [66]

Answer:

345453-5676

its the right answer

5 0
3 years ago
A bird is flying due east. Its distance from a tall building is given by x(t) = 28.0 m + (12.4 m/s)t – (0.0450 m/s3)t3. What is
Kruka [31]

Answer:

3.76 m/s

Explanation:

Instantaneous velocity: This can be defined as the velocity of an object in a non uniform motion. The S.I unit is m/s.

v' = dx(t)/dt..................... Equation 1

Where v' = instantaneous velocity, x = distance, t = time.

Given the expression,

x(t) = 28.0 m + (12.4 m/s)t - (0.0450 m/s³)t³

x(t) = 28 + 12.4t - 0.0450t³

Differentiating x(t) with respect to t.

dx(t)/dt = 12.4 - 0.135t²

dx(t)/dt = 12.4 - 0.135t²

When t = 8.00 s.

dx(t)/dt = 12.4 - 0.135(8)²

dx(t)/dt = 12.4 - 8.64

dx(t)/dt = 3.76 m/s.

Therefore,

v' = 3.76 m/s.

Hence, the instantaneous velocity = 3.76 m/s

8 0
3 years ago
A massless rod of length L has a small mass m fastened at its center and another mass m fastened at one end. On the opposite end
konstantin123 [22]

Answer:

onservation of energy

U top = K bottom

(m + m)*g*L = 1/2*I*?^2 where I = m*(L/2)^2 + m*L^2 = 1.25*m*L^2

So 2m*g*L = 1/2*1.25*m*L^2*?^2

So ? = sqrt(3*g*/(1.25*L) ) = sqrt(12g/5L)

3 0
3 years ago
Find the intensity of electromagnetic radiation at the surface of the sun (radius r=R=6.96×105kmr=R=6.96×105km). Ignore any scat
alisha [4.7K]

Answer:

I = 4.46*10^{16}W/m^2.

Explanation:

Intensity I of the electromagnetic radiation is given by

I = \dfrac{P}{4\pi r^2},

where r is the distance from the EM source (the center of the sun, in our case), and P is the power output of the sun and it has the value

P = 3.9 *10^{26}W.

Since the radius of the sun in meters is r = 6.96*10^8km, the intensity I of the electromagnetic radiation at the surface of the sun is

I = \dfrac{3.9*10^{26}W}{4\pi (6.96*10^8m)^2}\\\\\boxed{ I = 4.46*10^{16}W/m^2}

The intensity of the electromagnetic radiation at the surface of the sun is I = 4.46*10^{16}W/m^2.

7 0
3 years ago
A sinusoidal transverse wave travels along a long, stretched, string. the amplitude of this wave is 0.0885 m, it's frequency is
timofeeve [1]

Answer:

(a) 0.177 m

(b) 16.491 s

(c) 25 cycles

Explanation:

(a)

Distance between the maximum and the minimum of the  wave = 2A ............ Equation 1

Where A = amplitude of the wave.

Given: A = 0.0885 m,

Distance between the maximum and the minimum of the wave = (2×0.0885) m

Distance between the maximum and the minimum of the wave = 0.177 m.

(b)

T = 1/f ...................... Equation 2.

Where T = period, f = frequency.

Given: f = 4.31 Hz

T = 1/4.31

T = 0.23 s.

If 1 cycle pass through the stationary observer for 0.23 s.

Then, 71.7 cycles will pass through the stationary observer for (0.23×71.7) s.

= 16.491 s.

(c)

If  1.21 m contains  1 cycle,

Then, 30.7 m will contain (30.7×1)/1.21

= 25.37 cycles

Approximately 25 cycles.

6 0
3 years ago
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