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sp2606 [1]
3 years ago
14

A woman rides on a Ferris wheel of radius 16m that maintains the same speed throughout its motion. To better understand physics,

she takes along a digital bathroom scale (with memory) and sits on it. When she gets off the ride, she uploads the scale readings to a computer and creates a graph of scale reading versus time. (Figure 1) Note that the graph has a minimum value of 510N and a maximum value of 666N . The acceleration due to gravity, g=9.80m/s2.
Figure
Physics
1 answer:
Nadya [2.5K]3 years ago
7 0

Answer:

Mass of the woman is 60 kg and constant speed of the wheel is 4.56 m/s

Explanation:

The maximum reading of the scale is when wheel is at the bottom position

So it is given as

N_{max} = \frac{mv^2}{R} + mg

Similarly the minimum value is when she is at the top position of the wheel

So it is given as

N_{min} = mg - \frac{mv^2}{R}

so from above two equations we have

2mg = N_{max} + N_{min}

mg = \frac{510 + 666}{2}

mg = 588

m = 60 kg

also the speed of the wheel is given as

2\frac{mv^2}{R} = N_{max} - N_{min}

v = \sqrt{\frac{R(N_{max} - N_{min})}{2m}}

v = 4.56 m/s

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A block and tackle is used to lift an automobile engine that weighs 1800 N. The person exerts a force of 300 N to lift the engin
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Answer:

1800/300 = 6ropes

Explanation:

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4 years ago
In fission processes, which of the following statements is true? The total number of protons and the total number of neutrons bo
Amiraneli [1.4K]

Answer: A. The total number of protons and the total number of neutrons both remain the same.

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7 0
3 years ago
6) A pulsar is a rapidly rotating neutron star. The Crab nebula pulsar in the constellation Taurus has a period of 33.5 × 10−3 s
Romashka [77]

Answer:

a) L = 2.10x10⁴⁰ kg*m²/s

b) τ = 1.12x10²⁴ N.m

Explanation:

a) The angular momentum (L) of the pulsar can be calculated using the following equation:

L = I \omega

<u>Where:</u>

I: inertia momentum

ω: angular velocity

First we need to calculate ω and I. The angular velocity can be calculated as follows:

\omega = \frac{2 \pi}{T}

<u>Where:</u>

T: is the period = 33.5x10⁻³ s

\omega = \frac{2 \pi}{T} = \frac{2 \pi}{33.5 \cdot 10^{-3} s} = 187.56 rad/s

The inertia moment of the pulsar can be calculated using the following relation:

I = \frac{2}{5}mr^{2}

<u>Where</u>:

m: is the mass of the pulsar = 2.8x10³⁰ kg

r: is the radius = 10.0 km

I = \frac{2}{5}mr^{2} = \frac{2}{5}2.8\cdot 10^{30} kg*(10\cdot 10^{3} m)^{2} = 1.12 \cdot 10^{38} kg*m^{2}

Now, the  angular momentum of the pulsar is:

L = I \omega = 1.12 \cdot 10^{38} kg*m^{2}*187.56 rad/s = 2.10 \cdot 10^{40} kg*m^{2}*s^{-1}

b) If the angular velocity decreases at a rate of 10⁻¹⁴ rad/s², the torque of the pulsar is:

\tau = I*\alpha

<u>Where:</u>

α: is the angular acceleration = 10⁻¹⁴ rad/s²

\tau = I*\alpha = 1.12 \cdot 10^{38} kg*m^{2} * 10^{-14} rad*s^{-2} = 1.12 \cdot 10^{24} N.m

I hope it helps you!

5 0
3 years ago
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