Ask question

Ask question

All categories

3 years ago

6

What was the speed of a space shuttle that orbited Earth at an altitude of 1482 km?

2 answers:

lisov135 [29]3 years ago

7 0

Answer:

v = 7121.3 m/s

Explanation:

gregori [183]3 years ago

3 0

Answer:

v = 7121.3 m/s

Explanation:

As we know that the centripetal force for the space shuttle is due to gravitational force of earth due to which it will rotate in circular path with constant speed

so here we will have

$\frac{mv^2}{r} = \frac{GMm}{r^2}$

here we know that

r = orbital radius = 6370 km + 1482 km

$r = 7.852 \times 10^6 m$

also we know that

$M = 5.97 \times 10^{24} kg$

now we will have

$v^2 = \frac{(6.67 \times 10^{-11})(5.97 \times 10^{24})}{7.852 \times 10^6}$

$v^2 = 5.07 \times 10^7$

$v = 7121.3 m/s$

You might be interested in

A 4 kg car with frictionless wheels is on a ramp that makes a 25° angle with the horizontal. The car is connected to a 1 kg mass

stellarik [79]

Refer to the diagram shown below.

W₁ = (4 kg)*(9.8 m/s²) = 39.2 N
W₂ = (1 kg)*(9.8 m/s²) = 9.8 N

The normal reaction on the 4-kg mass is
N = (39.2 N)*cos(25°) = 35.5273 N
The force acting down the inclined plane due to the weight is
F = (39.2 N)*sin(25°) = 16.5666 N

The net force that accelerates the 4-kg mass at a m/²s down the plane is
F - W₂ = (4 kg)*(a m/s²)
4a = 16.5666 - 9.8
a = 1.6917 m/s²

Answer: 1.69 m/s² (nearest hundredth)

7 0

3 years ago

When you ride a bicycle at constant speed, nearly all the energy you expend goes into the work you do against the drag force of

ratelena [41]

Answer:

Power output = 96.506 watts

Explanation:

Drag coefficient (Cd) = 0.9

V = 7.3 m/s

Air density (ρ) = 1.225 kg/m^(3)

Area (A) = 0.45 m^2

Let's find the drag force ;

Fd=(1/2)(Cd)(ρ)(A)(v^(2))

So Fd = (1/2)(0.9)(1.225)(0.45)(7.3^(2)) = 13.22N

Drag power = Drag Force x Drag velocity.

Thus drag power, = 13.22 x 7.3 = 96.506 watts

8 0

2 years ago

A ray of light incident in water strikes the surface separating water from air making an angle of 10 ° with the normal to the su

labwork [276]

Answer:

a

$\theta _2 = 13^o$

b

$\theta _1 =32.94^o$

c

$\theta_c = 53.05^o$

Explanation:

From the question we are told that

The angle of incidence is $\theta_1 = 10^o$

The refractive index of water is $n_1 = 1.3$

Generally Snell's law is mathematically represented as

$n_1 sin(\theta_1) = n_2 sin(\theta_ 2)$

Here $n_2$ is the refractive index of air with value $n_2 = 1$

$\theta_2$ is the angle of refraction

So

$\theta _2 = sin^{-1}[\frac{n_1 * sin(\theta _1)}{n_2} ]$

=> $\theta _2 = sin^{-1}[\frac{1.3 * sin(10)}{1} ]$

=> $\theta _2 = 13^o$

Given that the angle should not be greater than $\theta _2 =45^o$ then the angle of incidence will be

$\theta _1 = sin^{-1}[\frac{n_2 * sin(\theta _2)}{n_1} ]$

=> $\theta _1 = sin^{-1}[\frac{1 * sin(45)}{1.3} ]$

=> $\theta _1 =32.94^o$

Generally for critical angle is mathematically represented as

$\theta_c = sin^{-1}[\frac{n_2}{n_1} ]$

=> $\theta_c = sin^{-1}[\frac{1}{1.3} ]$

=> $\theta_c = 53.05^o$

4 0

2 years ago

A 2-kg book falls off a shelf. it hits a student traveling 2 m/s. how much kinetic energy does the book have?

jekas [21]

Kinetic energy = 1/2 * m * V^2

K = 1/2 * 2kg * 4

5 0

2 years ago

A 0.14 kilogram baseball is throw in a straight line at a velocity of 30 m/sec.what is the momentum of the baseball

Masja [62]

4.2 kg m/s hope this helps

4 0

2 years ago