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nikdorinn [45]
3 years ago
13

In the real world, orbiting objects such as satellites and stations do not need to be constantly accelerating. What is different

about the real world satellites and the theoretical satellites we are discussing here?
Physics
1 answer:
JulijaS [17]3 years ago
8 0
What do you mean by theoretical satellites? When you mean real world satellites are you talking about the ones in space because I'm not sure what else you talking about.. I wanna help but could you explain plz??
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Romashka [77]

Answer:

it seems that the answer should be 7.5 mph as the average

Explanation:

sorry i did (s x t) when it was (s/t)

3 0
3 years ago
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A stunt driver wants to make his car jump over 8 cars parked side by side below a horizontal ramp.
Genrish500 [490]

<u>Answer:</u>

a) Minimum speed must he drive off the horizontal ramp = 39.78 m/s

b) Minimum speed must he drive off the horizontal ramp with 7° above the horizontal  = 23.93 m/s

<u>Explanation:</u>

a) The height of ramp = 1.5 meter

   Horizontal distance he must clear = 22 meter

   The car is having horizontal motion and vertical motion. In case of vertical motion the acceleration on the car is acceleration due to gravity.

   We have equation of motion, s= ut+\frac{1}{2} at^2, s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

 In case of vertical motion initial velocity = 0 m/s, acceleration = 9.8 m/s^2, we need to calculate time when displacement = 1.5 meter.

 1.5=0*t+\frac{1}{2} *9.8*t^2\\ \\ t = 0.553 seconds

So the car has to cover a distance of 22 meter in 2.119 seconds.

 So minimum speed required = 22/0.553 = 39.78 m/s

 Minimum speed must he drive off the horizontal ramp = 39.78 m/s

b) When the take of angle is 7⁰ the vertical speed of car is not zero = V sin 7 = 0.122 V

 So the in case of vertical motion we have initial velocity = 0.122 V, S = -1.5 meter( below ramp), acceleration = -9.8 m/s^2

Substituting

     -1.5=0.122V*t-\frac{1}{2} *9.8*t^2\\ \\ 4.9t^2-0.122Vt-1.5=0

In case of horizontal motion

    Horizontal speed of car = V cos 7 = 0.993V

    So it has to travel 22 meter in t seconds

            0.993Vt = 22, Vt = 22.155 m

    Substituting in the equation 4.9t^2-0.122Vt-1.5=0

    We will get 4.9t^2-0.122*22.155-1.5=0\\ \\ t = 0.926 seconds

   Speed required = 22.155/0.926 = 23.93 m/s

  Minimum speed must he drive off the horizontal ramp with 7° above the horizontal  = 23.93 m/s

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Lapatulllka [165]
Your answer is C hope i helped
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