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IrinaVladis [17]
3 years ago
5

HELP

Physics
1 answer:
svet-max [94.6K]3 years ago
6 0

Answer:

Cold Front 3  //  Stationary Front 1  //  Warm Front 2  //  Occluded Front 4

Explanation:

It's simple.  Warm front means the warm air is pressing forward,  which is why it's a warm front.  Stationary Front, meaning they're at a standstill, also makes sense because stationary means not moving.  Then since your last option is Occluded Front, since the others already have an answer, you have no choice but to match 4 with it.  I took the quiz and got the answer right.  :D

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A 545-kg satellite is in a circular orbit about Earth at a height above Earth equal to Earth's mean radius. (a) Find the satelli
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Answer

given,

mass of satellite = 545 Kg

R = 6.4 x 10⁶ m

H = 2 x 6.4 x 10⁶ m

Mass of earth = 5.972 x 10²⁴ Kg

height above earth is equal to earth's mean radius

a) satellite's orbital velocity

   centripetal force acting on satellite = \dfrac{mv^2}{r}

     gravitational force = \dfrac{GMm}{r^2}

    equating both the above equation

    \dfrac{mv^2}{r} = \dfrac{GMm}{r^2}

      v = \sqrt{\dfrac{GM}{r}}

      v = \sqrt{\dfrac{6.67 \times 10^{-11}\times 5.972 \times 10^{24}}{2 \times 6.4 \times 10^6}}

          v = 5578.5 m/s

b) T= \dfrac{2\pi\ r}{v}

   T= \dfrac{2\pi\times 2\times 6.4 \times 10^6}{5578.5}

   T= \dfrac{2\pi\times 2\times 6.4 \times 10^6}{5578.5}

          T = 14416.92 s

          T = \dfrac{14416.92}{3600}\ hr

          T = 4 hr

c) gravitational force acting

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     F = 5202 N

4 0
3 years ago
Your electric drill rotates initially at 5.21 rad/s. You slide the speed control and cause the drill to undergo constant angular
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Answer:

The drill's angular displacement during that time interval is 24.17 rad.

Explanation:

Given;

initial angular velocity of the electric drill, \omega _i = 5.21 rad/s

angular acceleration of the electric drill, α = 0.311 rad/s²

time of motion of the electric drill, t = 4.13 s

The angular displacement of the electric drill at the given time interval is calculated as;

\theta = \omega _i t \ + \ \frac{1}{2}\alpha t^2\\\\\theta = (5.21 \ \times \ 4.13) \ + \ \frac{1}{2}(0.311)(4.13)^2\\\\\theta = (21.5173 ) \ + \ (2.6524)\\\\\theta =24.17 \ rad

Therefore, the drill's angular displacement during that time interval is 24.17 rad.

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