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White raven [17]

2 years ago

3

A 7600 kg satellite is in a circular orbit around Earth at a height of 2300 km above Earth's surface. What is this satellite's s

peed

1 answer:

Cloud [144]2 years ago

8 0

Answer:

6779.7m/s

Explanation:

Using

GMm/(Re +h)² = mv²/ (Re+h)

So making v subject we have

V= √GM/Re+h

So

V = √ 6.67*10^-11 x 5.97*10^24/(6371+2300)*10^3

V= 6779.7m/s

Note h = height of satellite

Re= radius of the earth

M = mass of the earth

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(d) A beam of white light shines onto a sheet of white paper. An identical beam of light shines onto a mirror. The light is scat

irakobra [83]

QUESTION:-A beam of white light shines onto a sheet of white paper. An identical beam of light shines onto a mirror. The light is scattered from the paper and reflected from the mirror.
Describe how scattering by paper and reflection by a mirror are different from each other.

ANSWER: Scattering sends or reflects light from each point on the object in all directions, whereas reflection sends light from each point on the object in one direction only (or to one point)

5 0

2 years ago

A battleship launches a shell horizontally at 100 m/s from the ship’s deck that’s 50 m above the water. The shell is intended to

Annette [7]

Answer:

The shell will land 10.18m away from the buoy.

Explanation:

In order to solve this problem, we must first do a sketch of what the problem looks like (see attached picture).

Now, there are two cases, one with the tailwind and another with the tailwind. In both cases the shell would have the same vertical initial velocity and acceleration, therefore the shell would hit the water in the same amount of time. So we need to first find the time it takes the shell to hit the water.

In order to do so we can use the following equation:

$y_{f}=y_{0}+V_{0}t+\frac{1}{2}at^{2}$

now, we know that the final height and the initial velocity are to be zero, so we can simplify the equation like this:

$0=y_{0}+\frac{1}{2}at^{2}$

and solve for t:

$t=\sqrt{\frac{-2y_0}{a}}$

now we can substitute the values:

$t=\sqrt{\frac{-2(50m)}{-9.81m/t^2}}$

t=3.19s

Since it takes 3.19s for the shell to hit the water, that's the amount of time it spends flying horizontally.

So we can consider the shell to move at a constant speed if there was no tailwind, so we can find the distance from the ship to point A to be:

$x_{A}=V_{x}t$

$x_{A}=(100m/s)(3.19)$

$x_{A}=319m$

We can now find the distance between the ship to point B, which is the point the ball falls due to the tailwind. Since the movement will be accelerated in this scenario, we can find the distance by using the following formula:

$x_{f}=V_{x0}t+\frac{1}{2}a_{x}t^{2}$

So we can substitute the given values:

$x_{f}=(100m/s)(3.19s)+\frac{1}{2}(2m/s^{2})(3.19s)^{2}$

Which yields:

$x_{f}=329.18m$

so now we can use the A and B points to find by how far the shell missed the buoy:

Distance=329.18m-319m=10.18m

So the shell missed the buoy by 10.18m.

8 0

2 years ago

An activity that is relatively short in time (< 10 seconds) and has few repetitions predominately uses the _____________ ener

tensa zangetsu [6.8K]

An activity that is relatively short in time <10 seconds and has few repetitions predominantly uses the ATP/PC energy system. The cellular respiration procedure that changes food energy into ATP which is a form of energy is largely reliant on oxygen obtainability. During exercise the source and request of oxygen obtainable to muscle is unnatural by period and strength and by the individual’s cardiorespiratory suitability level.
Steps of the ATP-PC system:

1. Primarily, ATP kept in the myosin cross-bridges which is microscopic contractile parts of muscle is broken down to issue energy for muscle shrinkage. This action consents the by-products of ATP breakdown which are the adenosine diphosphate and one single phosphate all on its own.

2. Phosphocreatine is then broken down by the enzyme creatine kinase into creatine and phosphate.

3. The energy free in the breakdown of PC permits ADP and Pi to rejoin creating more ATP. This newly made ATP can now be broken down to issue energy to fuel activity.

5 0

2 years ago

A 10.0 cm3 sample of copper has a mass of 89.6

Romashka-Z-Leto [24]

Density is mass divides by volume, so
89.6g / 10cm^3 =8.96g /cm^3

*cm^3 is a standard unit of volume*

4 0

2 years ago

luggage handler pulls a 20.0 kg suitcase up a ramp inclined at 34.0 ∘ above the horizontal by a force F⃗ of magnitude 165 N that

Andreas93 [3]

Answer:

a) W = 643.5 J, b) W = -427.4 J

Explanation:

a) Work is defined by

W = F. x = F x cos θ

in this case they ask us for the work done by the external force F = 165 N parallel to the ramp, therefore the angle between this force and the displacement is zero

W = F x

let's calculate

W = 165 3.9

W = 643.5 J

b) the work of the gravitational force, which is the weight of the body, in ramp problems the coordinate system is one axis parallel to the plane and the other perpendicular, let's use trigonometry to decompose the weight in these two axes

sin θ = Wₓ / W

cos θ = Wy / W

Wₓ = W sinθ = mg sin θ

Wy = W cos θ

the work carried out by each of these components is even Wₓ, it has to be antiparallel to the displacement, so the angle is zero

W = Wₓ x cos 180

W = - mg sin 34 x

let's calculate

W = -20 9.8 sin 34 3.9

W = -427.4 J

The work done by the component perpendicular to the plane is ero because the angle between the displacement and the weight component is 90º, so the cosine is zero.

3 0

2 years ago