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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.

You might be interested in

A car tire rotates with an average angular speed of 29 rad/s. In what time interval will the tire rotate 3.5 times?

meriva

Answer:

"0.758315 sec" is the appropriate choice.

Explanation:

The given values are:

Angular speed,

= 29 rad/s

or,

= $\frac{29}{2 \pi} \ rps$

Tire rotates,

= 3.5 times

Now,

The time interval will be:

⇒ $\Delta t=\frac{3.5}{\frac{29}{2 \pi} }$

⇒ $=\frac{2 \pi\times 3.5}{29}$

⇒ $=\frac{21.99}{29}$

⇒ $=0.758315 \ sec$

8 0

3 years ago

A teacher asks students to make a model of a transform plate boundary the students use blocks to represent tectonic plates and S

nirvana33 [79]

Answer:

Hello your question is incomplete hence I will give you a general answer as regards to tectonic plates sliding past each other in a sideways direction

answer : The Transform boundary is been demonstrated by the students when sliding tectonic plates past each other in sideways directions

Explanation:

The event that can be demonstrated by the students using blocks to represent tectonic plates and sliding the clocks past each other in sideways direction is Transform Boundary of the tectonic plates

4 0

3 years ago

A satellite is in a circular orbit around the Earth at an altitude of 2.80 3 106 m. Find (a) the period of the orbit, (b) the sp

Katen [24]

Explanation:

Given that,

Altitude $h= 2.803\times10^{6}\ m$

We need to calculate the radius

$r=R+h$

Where, R = radius of the earth

h = radius of altitude

Put the value into the formula

$r=(6.38\times10^{6}+2.803\times10^{6})$

$r=9.18\times10^{6}\ m$

(a). We need to calculate the period of the orbit,

Using formula of period

$T^2=\dfrac{4\pi^2}{GM}r^3$

$T^2=\dfrac{4\pi^2}{6.67\times10^{-11}\times5.98\times10^{24}}\times(9.18\times10^{6})^3$

$T^2=76570372.9509\ sec$

$T=8750.45\ sec$

(b). We need to calculate the speed of the satellite

Using formula of speed

$v^2=\dfrac{GM}{r}$

Put the value into the formula

$v^2=\dfrac{6.67\times10^{-11}\times5.98\times10^{24}}{9.18\times10^{6}}$

$v^2=43449455.3377\ m/s$

$v=6.59\times10^{3}\ m/s$

(c). We need to calculate the acceleration of the satellite

Using formula of acceleration

$a_{c}=\dfrac{v^2}{r}$

Put the value into the formula

$a_{c}=\dfrac{(6.59\times10^{3})^2}{9.18\times10^{6}}$

$a_{c}=4.73\ m/s^2$

Hence, This is the required solution.

6 0

4 years ago

A certain car engine delivers enough force to create 630 N⋅m of torque when the engine is operating at 3200 revolutions per minu

jekas [21]

The appropriate expression for the calculation of power by relating the angular energy in a given time.

In other words the instantaneous power of an angular accelerating body is the torque times the angular velocity

$P=\tau\omega$

Where

$\tau =$ Torque

$\omega =$ Angular speed

Our values are given by

$\tau = 630Nm$

$\omega = 3200rev/min$

The angular velocity must be transformed into radians per second then

$\omega = 3200rev/min (\frac{2\pi rad}{60s})$

$\omega = 335.103rad/s$

Replacing,

$P=(630)(335.103)$

$P = 211.11*10^3W$

$P = 211.1kW$

The average power delivered by the engine at this rotation rate is 211.1kW

5 0

3 years ago

Nevermind i dont need it anymore! thank you.

Sidana [21]

it's alright ok fine ! no problem you most welcome!

3 0

2 years ago