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marysya [2.9K]
3 years ago
5

-2 5/8 is bigger than -5/12

Physics
1 answer:
Neko [114]3 years ago
8 0

Answer:

hey...................................

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10. A satellites is in a circular orbit around the earth at a height of 360 km above the earth’s surface. What is its time perio
Afina-wow [57]

Answer:

Orbital speed=8102.39m/s

Time period=2935.98seconds

Explanation:

For the satellite to be in a stable orbit at a height, h, its centripetal acceleration V2R+h must equal the acceleration due to gravity at that distance from the center of the earth g(R2(R+h)2)

V2R+h=g(R2(R+h)2)

V=√g(R2R+h)

V= sqrt(9.8 × (6371000)^2/(6371000+360000)

V= sqrt(9.8× (4.059×10^13/6731000)

V=sqrt(65648789.18)

V= 8102.39m/s

Time period ,T= sqrt(4× pi×R^3)/(G× Mcentral)

T= sqrt(4×3.142×(6.47×10^6)^3/(6.673×10^-11)×(5.98×10^24)

T=sqrt(3.40×10^21)/ (3.99×10^14)

T= sqrt(0.862×10^7)

T= 2935.98seconds

4 0
3 years ago
Can you travel faster by not running forward ?
GrogVix [38]
Yeah i think with a car or a plane:)
5 0
3 years ago
I need helpppppp asap
asambeis [7]
It’s c because it’s not Control so that means that it would be broken and non fix able
5 0
3 years ago
A meter stick is suspended vertically at a pivot point 22 cm from the top end. It is rotated on the pivot until it is horizontal
suter [353]

Answer:

5.82812 rad/s

Explanation:

L = Length of meter stick = 1 m = 100 cm

m_c = The center of mass of the stick = \frac{L}{2}-0.22=0.5-0.22=0.28\ m

\omega = Angular velocity

Moment of inertia of the system is given by

I=I_c+mr^2\\\Rightarrow I=\frac{mL^2}{12}+mr^2\\\Rightarrow I=\frac{m1^2}{12}+m0.28^2\\\Rightarrow I=m(\frac{1}{12}+0.0784)

As the energy in the system is conserved

mgh=I\frac{\omega^2}{2}\\\Rightarrow mgh=m(\frac{1}{12}+0.0784)\frac{\omega^2}{2}\\\Rightarrow gh=(\frac{1}{12}+0.0784)\frac{\omega^2}{2}\\\Rightarrow \omega=\sqrt{\frac{2gh}{\frac{1}{12}+0.0784}}\\\Rightarrow \omega=\sqrt{\frac{2\times 9.81\times 0.28}{\frac{1}{12}+0.0784}}\\\Rightarrow \omega=5.82812\ rad/s

The maximum angular velocity is 5.82812 rad/s

4 0
3 years ago
A fire engine approaches a wall at 5 m/s while the siren emits a tone of 500 Hz frequency. At the time, the speed of sound in ai
Svetach [21]

Answer:

The  values is  f_b  =14.9 \  beats/s

Explanation:

From the question we are told that

   The  speed of the fire engine is  v =  5\ m/s

    The frequency of the tone is  f =  500 \ Hz

    The speed of sound in air is v_s  =  340 \ m/s

The  beat frequency is mathematically represented as

     f_b  =  f_a  -  f

Where  f_a is the frequency of sound heard by the people in the fire engine and is is mathematically evaluated as

   f_a  =  [\frac{v_s + v }{v_s  -v} ]* f

substituting values

  f_a  =  [\frac{340 + 5 }{340  -5} ]* 500

  f_a  = 514.9 \  Hz

Thus  

      f_b  =514.9 -  500

      f_b  =14.9 \  beats/s

8 0
3 years ago
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