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4 years ago

15

When the space shuttle coasts in a circular orbit at constant speed about the earth, is it accelerating? if so, in what directio

n?

2 answers:

pochemuha4 years ago

5 0

Yes. Acceleration means any change in speed or direction of motion. When an object coasts in a circular orbit at constant speed around the Earth, its direction is constantly changing. The acceleration is "CENTRIPETAL", which points toward the center of the circle.

Lesechka [4]4 years ago

4 0

Answer: YES it is accelerating and the direction is towards the centre

Explanation:

Acceletion is defined as the ratio of change in velocity in a direction with time. When a space shuttle moves in a circular orbit, accelerating at a constant speed, it's movement is centripetal. And it accelerates towards the centre.

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Magnesium hydroxide is a common ___?

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Magnesium hydroxide is the inorganic compound with the chemical formula Mg(OH)2. Magnesium hydroxide is a common component of antacids, such as milk of magnesia, as well as laxatives

Explanation:

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3 years ago

A motor running at 2600 rev/min is suddenly switched off and decelerates uniformly to rest after 10 s. Find the angular decelera

Mama L [17]

Angular acceleration = (change in angular speed) / (time for the change)

change in angular speed = (zero - 2,600 RPM) = -2,600 RPM

time for the change = 10 sec

Angular acceleration = -2600 RPM / 10 sec = -260 rev / min-sec

(-260 rev/min-sec) x (1 min / 60 sec) = <em>-(4 1/3) rev / sec²</em>

Since the acceleration is negative, the motor is slowing down.
You might call that a 'deceleration' of (4 1/3) rev/sec² .

The average speed is 1/2(2,600 + 0) = 1,300 rev/min = (21 2/3) rev/sec.

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6 0

4 years ago

An object is pushed from rest across a sheet of ice, accelerating at 8.0 m/s^2 [E] over a displacement of 1.05 m [E]. The object

tatiyna

Answer:

$D_T=18.567m$

Explanation:

From the question we are told that:

Acceleration $a=8.0 m/s^2$

Displacement $d=1.05 m$

Initial time $t_1=6.0s$

Final Time $t_2=2.5s$

Generally the equation for Velocity of 1.05 travel is mathematically given by

Using Newton's Law of Motion

$V^2=2as$

$V=\sqrt{2*6*1.05}$

$V=4.1m/s$

Generally the equation for Distance traveled before stop is mathematically given by

$d_2=v*t_1$

$d_2=3.098*4$

$d_2=12.392$

Generally the equation for Distance to stop is mathematically given by

Since For this Final section

Final velocity $v_3=0 m/s$

Initial velocity $u_3=4.1 m/s$

Therefore

Using Newton's Law of Motion

$-a_3=(4.1)/(2.5)$

$-a_3=1.64m/s^2$

Giving

$v_3^2=u^2-2ad_3$

Therefore

$d_3=\frac{u_3^2}{2ad_3}$

$d_3=\frac{4.1^2}{2*1.64}$

$d_3=5.125m$

Generally the Total Distance Traveled is mathematically given by

$D_T=d_1+d_2+d_3$

$D_T=5.125m+12.392+1.05 m$

$D_T=18.567m$

6 0

3 years ago