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Paha777 [63]
3 years ago
12

A horizontal platform in the shape of a circular disk rotates on a frictionless bearing about a vertical axle through the center

of the disk. The platform has a radius of 3.70 m and a rotational inertia of 274 kg·m2 about the axis of rotation. A 67.8 kg student walks slowly from the rim of the platform toward the center. If the angular speed of the system is 2.62 rad/s when the student starts at the rim, what is the angular speed when she is 0.456 m from the center?
Physics
1 answer:
Bess [88]3 years ago
8 0

Answer:

10.93 rad/s

Explanation:

If we treat the student as a point mass, her moment of inertia at the rim is

I_r = mr^2 = 67.8*3.7^2 = 928.182 kgm^2

So the system moment of inertia when she's at the rim is:

I_1 = I_d + I_r = 274 + 928.182 = 1202.182 kgm^2

Similarly, we can calculate the system moment of inertia when she's at 0.456 m from the center

I_2 = I_d + 67.8*0.456^2 = 274 + 14.1 = 288.1 kgm^2

We can apply the law of angular momentum conservation to calculate the post angular speed when she's 0.456m from the center:

I_1\omega_1 = I_2\omega_2

\omega_2 = \omega_1\frac{I_1}{I_2} = 2.62*\frac{1202.182}{288.1} = 10.93 rad/s

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xeze [42]

Answer:

Hope it helps you :)

Explanation in the pic above.

8 0
2 years ago
A satellite is put into an orbit at a distance from the center of the Earth equal to twice the distance from the center of the E
Sveta_85 [38]

Answer:

995 N

Explanation:

Weight of surface, w= 4000N

Gravitational constant, g, is taken as 9.81 hence mass, m of surface is W/g where W is weight of surface

m= 4000/9.81= 407.7472

Using radius of orbit of 6371km

The force of gravity of satellite in its orbit, F=\frac {GMm}{(2r)^{2}}=\frac {GMm}{4(r)^{2}}

Where G=6.67*10^{-11} and M=5.94*10^{24}

F=\frac {(6.67*10^{-11}*5.94*10^{24}*407.7472)}{4*({6.371*10^{6}m)}^{2}}

F= 995.01142 then rounded off

F=995N

6 0
2 years ago
Normal atmospheric pressure is 1.013 105 Pa. The approach of a storm causes the height of a mercury barometer to drop by 27.1 mm
Burka [1]

Answer:

The atmospheric pressure is 0.97622\times10^{5}\ Pa.

Explanation:

Given that,

Atmospheric pressure P_{atm}= 1.013\times10^{5}\ Pa    

drop height h'= 27.1 mm

Density of mercury \rho= 13.59 g/cm^3

We need to calculate the height

Using formula of pressure

p = \rho g h

Put the value into the formula

1.013\times10^{5}=13.59\times10^{3}\times9.8\times h

h =\dfrac{1.013\times10^{5}}{13.59\times10^{3}\times9.8}

h=0.76\ m

We need to calculate the new height

h''=h - h'

h''=0.76-27.1\times10^{-3}

h''=0.76-0.027

h''=0.733\ m

We need to calculate the atmospheric pressure

Using formula of atmospheric pressure

P=\rho g h

Put the value into the formula

P= 13.59\times10^{3}\times9.8\times0.733

P=0.97622\times10^{5}\ Pa

Hence, The atmospheric pressure is 0.97622\times10^{5}\ Pa.

7 0
3 years ago
A microwave oven operating at 1.22 × 108 nm is used to heat 165 mL of water (roughly the volume of a teacup) from 23.0°C to 100.
ANTONII [103]

<u>Answer:</u> The number of photons are 3.7\times 10^8

<u>Explanation:</u>

We are given:

Wavelength of microwave = 1.22\times 10^8nm=0.122m    (Conversion factor:  1m=10^9nm  )

  • To calculate the energy of one photon, we use Planck's equation, which is:

E=\frac{hc}{\lambda}

where,

h = Planck's constant = 6.625\times 10^{-34}J.s

c = speed of light = 3\times 10^8m/s

\lambda = wavelength = 0.122 m

Putting values in above equation, we get:

E=\frac{6.625\times 10^{-34}J.s\times 3\times 10^8m/s}{0.122m}\\\\E=1.63\times 10^{-24}J

Now, calculating the energy of the photon with 88.3 % efficiency, we get:

E=1.63\times 10^{-24}\times \frac{88.3}{100}=1.44\times 10^{-24}J

  • To calculate the mass of water, we use the equation:

Density=\frac{Mass}{Volume}

Density of water = 1 g/mL

Volume of water = 165 mL

Putting values in above equation, we get:

1g/mL=\frac{\text{Mass of water}}{165mL}\\\\\text{Mass of water}=165g

  • To calculate the amount of energy of photons to raise the temperature from 23°C to 100°C, we use the equation:

q=mc\Delta T

where,

m = mass of water = 165 g

c = specific heat capacity of water = 4.184 J/g.°C

\Delta T = change in temperature = T_2-T_1=100^oC-23^oC=77^oC

Putting values in above equation, we get:

q=165g\times 4.184J/g.^oC\times 77^oC\\\\q=53157.72J

This energy is the amount of energy for 'n' number of photons.

  • To calculate the number of photons, we divide the total energy by energy of one photon, we get:

n=\frac{q}{E}

q = 53127.72 J

E = 1.44\times 10^{-24}J

Putting values in above equation, we get:

n=\frac{53157.72J}{1.44\times 10^{-24}J}=3.7\times 10^{28}

Hence, the number of photons are 3.7\times 10^8

4 0
3 years ago
A laser beam is incident on two slits with a separation of 0.195 mm, and a screen is placed 5.30 m from the slits. If the bright
kumpel [21]

Answer:

λ = 596 nm.

Explanation:

Fringe width = λ D / d

λ is wave length , D is screen distance and d is slit separation.

Putting the values

1.62 x 10⁻² =(  λ x 5.3 ) / .195 x 10⁻³

\lambda=\frac{1.62\times10^{-2}\times195\times10^{-6}}{5.3}

λ = 596 nm.

8 0
3 years ago
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