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sveticcg [70]
3 years ago
14

to get a flat uniform cylindrical satellite spinning at the correct rate, scientists fire 4 tangential rockets. suppose that the

satellite has a mass of 3600 kg and a radius of 4m, and that the rockets each add a mass of 250 kg. what is the steady force required of each rocket if the satellite is to reach 32 rpm in 5 min starting from rest.
Physics
1 answer:
Elena-2011 [213]3 years ago
3 0

Answer:

The steady force required of each rocket is 31.36 N

Explanation:

Given that,

Mass of satellite = 3600 Kg

Radius = 4 m

Mass of rocket = 250 kg

Time = 5 min

Angular velocity = 32 rpm

We need to calculate the moment of inertia

Using formula of moment of inertia

I=\dfrac{1}{2}Mr^2+4mr^2

Where, M = mass of satellite

r = radius

m = mass of rocket

Put the value into the formula

I=\dfrac{1}{2}\times3600\times4^2+4\times250\times4^2

I=44800\ kg-m^2

We need to calculate the angular acceleration

Using formula of angular acceleration

\alpha=\dfrac{\omega}{t}

Where, \omega = angular velocity

t = time

Put the value into the formula

\alpha=\dfrac{2\pi\times32}{3600\times5}

\alpha=0.0112\ rad/s^2

We need to calculate the steady force required of each rocket

Using formula of torque

\tau=I\alpha

4rF=I\alpha

F=\dfrac{I\alpha}{4r}

Put the value into the formula

F=\dfrac{44800\times0.0112}{4\times4}

F=31.36\ N

Hence, The steady force required of each rocket is 31.36 N

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Eva8 [605]

Answer:

Total length of spring 0.647 m

Explanation:

We have given mass of the person m = 150 kg

Acceleration due to gravity g=9.8m/sec^2

Spring constant k = 10000 N/m

Nominal length of spring = 0.50

According to hook's law

mg=kx

150\times 9.8=10000\times x

x = 0.147 m

So total length of spring = 0.50+0.147 = 0.647 m

4 0
3 years ago
Pleaseee someone help me TEST TOMORROW
Liono4ka [1.6K]
3.A

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4 0
3 years ago
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A nonconducting container filled with 25 kg of water at 23°C is fitted with a stirrer, which is made to turn by gravity acting o
Paul [167]

Explanation:

Given that,

Weight of water = 25 kg

Temperature = 23°C

Weight of mass = 32 kg

Distance = 5 m

(a). We need to calculate the amount of work done on the water

Using formula of work done

W=mgh

W=32\times9.8\times5

W=1568\ J

The amount of work done on the water is 1568 J.

(b). We need to calculate the internal-energy change of the water

Using formula of internal energy

The change in internal energy of the water equal to the amount of the  work done on the water.

\Delta U=W

\Delta U=1568\ J

The  change in internal energy is 1568 J.

(c). We need to calculate the final temperature of the water

Using formula of the change internal energy

\Delta U=mc_{p}\Delta T

\Delta U=mc_{p}(T_{2}-T_{1})

T_{2}=T_{1}+\dfrac{\Delta U}{mc_{p}}

T_{2}=23+\dfrac{1568}{25\times4.18\times10^{3}}

T_{2}=23.01^{\circ}\ C

The final temperature of the water is 23.01°C.

(d). The amount of heat removed from the water to return it to it initial temperature is the change in internal energy.

The amount of heat is 1568 J.

Hence, This is the required solution.

6 0
3 years ago
You test a moon buggy on Earth. When the buggy hits a bump, it oscillates up and down on its springs with a period of 4 seconds.
Blizzard [7]

Answer:

Remains same

Explanation:

T = Time period of oscillation

m = mass

k = spring constant

Time period of oscillation is given as

T = 2\pi \sqrt{\frac{m}{k} }

we know that as we move from earth to moon, the value of spring constant "k"  and mass "m" remains unchanged because they do not depend on the acceleration due to gravity.

Time period depends on spring constant inversely and directly on the mass.

hence the time period remains the same.

3 0
4 years ago
You are in downtown Chicago (where streets run N-S and E-W). You started from 600 N. Michigan Avenue, and walked 3 blocks toward
pickupchik [31]

Answer:

Displacement is 565.69 m at 45° west of north

Explanation:

Let north represent positive y axis and east represent positive x axis.

We have journey started from 600 N. Michigan Avenue, and walked 3 blocks toward north, 4 blocks toward west, and 1 block toward north to a train station.

3 blocks toward north = 300 j m

4 blocks toward west = -400 i m

1 blocks toward north = 100 j m

Total displacement = -400 i + 400 j m

Magnitude

     s=\sqrt{(-400)^2+400^2}=565.69m

Direction,

     \theta =tan^{-1}\left ( \frac{400}{-400}\right )=135^0

     Direction is 45° west of north.

Displacement is 565.69 m at 45° west of north

4 0
3 years ago
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