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3 years ago

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The Hubble Space Telescope orbits 600 km above Earth's surface. Earth's radius is about 6370 km. Use the Pythagorean Theorem to

find the distance x from the telescope to Earth's horizon. Round your answer to the nearest ten kilometers.

1 answer:

musickatia [10]3 years ago

7 0

To solve this problem we will apply the concept given by Pythagoras in the description of the lengths of the legs of a rectangular triangle and if equality against the square of the hypotenuse, that is

$a^2+b^2=c^2$

Here,

a, b = Legs of a triangle

c = Hypotenuse

According to the attached chart then we would have to

$a=x\\b = 6370km\\c = 600+6370km = 6970km$

Substituting the given the lengths into the Pythagorean Theorem.

$a^2+b^2 = c^2 \\x^2 +(6370)^2 = (6970)^2\\x = 2829.13km \approx 2830km$

Therefore the distance x is 2830km.

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3 years ago

Which imaging technique uses x-rays to make cross-sectional images of the body? mri ct scan x-ray imaging ultrasound

Vitek1552 [10]

Computed tomography (CT) is an imaging tool that combines x-rays with computer technology to produce a more detailed, cross-sectional image of your body.

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CT is a painless, noninvasive way for your healthcare provider to diagnose conditions.

A CT scan lets your doctor see the size, shape, and position of structures that are deep inside your body, such as organs, tissues, or tumors.

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2 years ago

Which statements are true concerning Newton's law of gravitation? The gravitational force is related to the mass of each object.

OLEGan [10]

Answer:

The gravitational force is related to the mass of each object.

The gravitational force is an attractive force.

Explanation:

Gravitational force is a long range force of attraction between any two masses.

Mathematically given as :

$F=G.\frac{m_1.m_2}{r^2}$

where:

$m_1 & m_2$ are the masses

r= distance between the center of mass of the two objects.

G= gravitational constant = $6.67\times 10^{-11} m^3.kg^{-1}.s^{-2}$

From the above relation of eq. (1) it is clear that,

Gravitational force is inversely proportional to the square of the distance and directly proportional to the masses.

The mass of an object is independent of its size due to the fact that density may vary for different objects.

The force of gravity varies with height as:

$\frac{g}{g_x} =(\frac{r_x}{r} )^2$

where:

$g=9.8\,m.s^{-2}$

$g_x=$ gravity at height $r_x$ of the center of mass of the object from the center of mass of the earth.

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where: m= mass of the object.

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3 years ago

The independent variable:

ValentinkaMS [17]

Answer:

A. Is the one that the experimenter manipulates directly

Explanation:

The independent variable is the one that is manipulated during an experiment by the experimenter.

The dependent variable is the one that is effected by the independent variable in an experiment.

3 0

3 years ago

A cannon fires a 0.2 kg shell with initial velocity vi = 9.2 m/s in the direction θ = 46 ◦ above the horizontal. The shell’s tra

Sedbober [7]

Answer:

∆h = 0.071 m

Explanation:

I rename angle (θ) = angle(α)

First we are going to write two important equations to solve this problem :

Vy(t) and y(t)

We start by decomposing the speed in the direction ''y''

$sin(\alpha) = \frac{Vyi}{Vi}$

$Vyi = Vi.sin(\alpha ) = 9.2 \frac{m}{s} .sin(46) = 6.62 \frac{m}{s}$

Vy in this problem will follow this equation =

$Vy(t) = Vyi -g.t$

where g is the gravity acceleration

$Vy(t) = Vyi - g.t= 6.62 \frac{m}{s} - (9.8\frac{m}{s^{2} }) .t$

This is equation (1)

For Y(t) :

$Y(t)=Yi+Vyi.t-\frac{g.t^{2} }{2}$

We suppose yi = 0

$Y(t) = Yi +Vyi.t-\frac{g.t^{2} }{2} = 6.62 \frac{m}{s} .t- 4.9\frac{m}{s^{2} } .t^{2}$

This is equation (2)

We need the time in which Vy = 0 m/s so we use (1)

$Vy (t) = 0\\0=6.62 \frac{m}{s} - 9.8 \frac{m}{s^{2} } .t\\t= 0.675 s$

So in t = 0.675 s → Vy = 0. Now we calculate the y in which this happen using (2)

$Y(0.675s) = 6.62\frac{m}{s}.(0.675s)-4.9 \frac{m}{s^{2} } .(0.675s)^{2} \\Y(0.675s) =2.236 m$

2.236 m is the maximum height from the shell (in which Vy=0 m/s)

Let's calculate now the height for t = 0.555 s

$Y(0.555s)= 6.62 \frac{m}{s} .(0.555s)-4.9\frac{m}{s^{2} } .(0.555s)^{2} \\Y(0.555s) = 2.165m$

The height asked is

∆h = 2.236 m - 2.165 m = 0.071 m

6 0

3 years ago