Recall: earth applies _____________ on Earth or equivalently

The space shuttle releases a satellite into a circular orbit 630 km above the Earth.

solniwko [45]

Answer:

7,539 m/s

Explanation:

Let's use this equation to find the gravitational acceleration of this space shuttle:

$\displaystyle g=\frac{GM}{r^2}$

We know that G is the gravitational constant: 6.67 * 10^(-11) Nm²/kg².

M is the mass of the planet, which is Earth in this case: 5.972 * 10^24 kg.

r is the distance from the center of Earth to the space shuttle: radius of Earth (6.3781 * 10^6 m) + distance above the Earth (630 km → 630,000 m).

Plug these values into the equation:

$\displaystyle g=\frac{(6.67\cdot 10^-^1^1 \ Nm^2kg^-^2)(5.972\cdot 10^2^4 \ kg)}{[(6.3781\cdot 10^6 \ m)+(630000 \ m)]^2}$

Remove units to make the equation easier to read.

$\displaystyle g=\frac{(6.67\cdot 10^-^1^1 )(5.972\cdot 10^2^4 )}{[(6.3781\cdot 10^6)+(630000 )]^2}$

Multiply the numerator out.

$\displaystyle g=\frac{(3.983324\cdot 10^1^4)}{[(6.3781\cdot 10^6)+(630000 )]^2}$

Add the terms in the denominator.

$\displaystyle g=\frac{(3.983324\cdot 10^1^4)}{[(7008100)]^2}$

Simplify this equation.

$\displaystyle g=8.11045189 \ \frac{m}{s^2}$

The acceleration due to gravity g = 8.11045189 m/s². Now we use the equation for acceleration for an object in circular motion which contains v and r.

$\displaystyle a = \frac{v^2}{r}$

a = g, v is the velocity that the space shuttle should be moving (what we are trying to solve for), and r is the radius we had in the previous equation when solving for g.

Plug these values into the equation and solve for v.

$\displaystyle 8.11045189 \ \frac{m}{s^2} = \frac{v^2}{7008100 \ m}$

Remove units to make the equation easier to read.

$\displaystyle 8.11045189 = \frac{v^2}{7008100}$

Multiply both sides by 7,008,100.

$56838857.89=v^2$

Take the square root of both sides.

$v=7539.154985$

The shuttle should be moving at a velocity of about 7,539 m/s when it is released into the circular orbit above Earth.

5 0

2 years ago

Where do we hear loud sound, at antinode or node?

tester [92]

Answer:

node

Explanation:

on the graph node is higher than antinode

so it can get or hear loud sounds faster

4 0

2 years ago

__ modified the concept by adding an internal combustion engine and marketing hybrids that were part electric and part gas power

navik [9.2K]

Hybrid

<u>Hybrid</u> modified the concept by adding an internal combustion engine and marketing hybrids that were part electric and part gas powered.

The driving wheels of hybrid vehicles receive power from their drivetrains.
A hybrid car has numerous sources of propulsion.
There are numerous hybrid configurations.
A hybrid vehicle might, for instance, get its energy from burning gasoline while alternating between an electric motor and a combustion engine.
Although they have primarily been employed for rail locomotives, electrical vehicles have a long history of integrating internal combustion and electrical transmission, like in a diesel-electric power-train.
Because the electric drive transmission directly substitutes the mechanical gearbox rather than serving as an additional source of motive power, a diesel-electric powertrain does not meet the definition of a hybrid.
Only the electric/ICE hybrid car type was readily accessible on the market as of 2017.
One type used parallel operation to power both motors at the same time.
Another ran in series, using one source to supply power solely and the other to supply electricity.
Either source may act as the main driving force, with the other source serving to strengthen the main.

To learn more about hybrid vehicles visit:

brainly.com/question/14610495

#SPJ4

3 0

1 year ago

A student is pushing a box across the room. To push the box three times farther, the student needs to do how much work?

Travka [436]

Answer:

Removing some of the books reduced the mass of the box, and less force was needed to push it across the floor.

8 0

2 years ago

Suppose a soccer player kicks the ball from a distance 29 m toward the goal. find the initial speed of the ball if it just passe

Rudik [331]

The ball's vertical velocity at the time it just passes over the goal is 0 m/s. Its initial vertical velocity is unknown and we denote it by $v\sin39^\circ$ , where $v$ here is the ball's initial speed. Vertically, the only force acting on the ball is gravity, which attributes a downward acceleration of 9.8 m/s^2. We expect the maximum height achieved by the ball to be 2.4 m, so we can find the initial speed by solving

$\left(0\,\dfrac{\mathrm m}{\mathrm s}\right)^2-\left(v\sin39^\circ\right)^2=2\left(-9.8\,\dfrac{\mathrm m}{\mathrm s^2}\right)(2.4\,\mathrm m)$

$\implies v=11\,\dfrac{\mathrm m}{\mathrm s}$

6 0

2 years ago