What do you think that astronomers mean when they use the term observable universe? (Hint: Thi.nk of the time it takes for light from very distant objects to reach Earth.)

Answers

Answer 1

Answer: When people use the term "observable universe" it means that the light from other parts of the universe hasn't yet had time to reach us, so we cannot see it.

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The space shuttle is located exactly half way between the earth and the moon. Which statement is true regarding the gravitational pull on the shuttle?

Answers

Answer:

The correct answer is option B)

Explanation:

Considering the given question as -

The space shuttle is located exactly half way between the earth and the moon. Which statement is true regarding the gravitational pull on the shuttle? A) The moon pulls more on the shuttle. B) The earth pulls more on the shuttle. C) Both are equal due to equal distances. D) Both are equal due to the mass of the shuttle.

We know that gravitational pull (F) between any two bodies of mass $M_(1)$ and $M_(2)$ is given by -

F = $(GM_(1)M_(2) )/(r^(2) )$ where 'r' is the distance between the two bodies.

Let ,

$M_(e)$ : Mass of the earth

$M_(m)$ : Mass of the moon

m : Mass of the satellite

$r_(e)$ : Distance of satellite from earth

$r_(m)$ : Distance of satellite from moon

Given that $r_(e)$ = $r_(m)$

Let $r_(e)$ = $r_(m)$ =r

Force on satellite by the earth is -

$F_(e)$ = $(GM_(e)m )/(r^(2) )$

Force on satellite by the moon is -

$F_(m)$ = $(GM_(m)m )/(r^(2) )$

∵ Mass of earth ( $M_(e)$ ) > Mass of moon ( $M_(m)$ )

∴ $F_(e)$ > $F_(m)$

∴ The gravitational pull of earth on satellite is more than that of the moon.

An artificial satellite circling the Earth completes each orbit in 129 min(a) Find the altitude of the satellite.

(b) What is the value of g at the location of this satellite?

Answers

(a) $2.09\cdot 10^6 m$ above Earth's surface

The orbital speed of a satellite orbiting the Earth can be found using the equation

$v=\sqrt{(GM)/(r)}$

where

G is the gravitational constant

$M=5.98\cdot 10^(24) kg$ is the Earth's mass

r is the radius of the satellite's orbit

The orbital speed can also be rewritten as the ratio between the circumference of the orbit and the orbital period, T:

$(2\pi r)/(T)$

where

T = 129 min = 7740 s is the period

Combining the two equations,

$(2\pi r)/(T)=\sqrt{(GM)/(r)}$

And solving for r,

$((2\pi)^2 r^2)/(T^2)=(GM)/(r)\nr=\sqrt[3]{(GMT^2)/((2\pi)^2)}=\sqrt[3]{((6.67\cdot 10^(-11))(5.98\cdot 10^(24))(7740)^2)/((2\pi)^2)}=8.46\cdot 10^6 m$

This is, however, the orbital radius: this means we have to subtract the Earth's radius to find the altitude of the satellite, which is

$R=6.37\cdot 10^6 m$

therefore, the altitude of the satellite is

$h=r-R=8.46\cdot 10^6 - 6.37\cdot 10^6 =2.09\cdot 10^6 m$

b) $5.57 m/s^2$

The value of g at the location of the satellite is given by

$g=(GM)/(r^2)$

where:

G is the gravitational constant

$M=5.98\cdot 10^(24) kg$ is the Earth's mass

$r=8.46 \cdot 10^6 m$ is the radius of the satellite's orbit

Substituting into the equation, we find

$g=((6.67\cdot 10^(-11))(5.98\cdot 10^(24)))/((8.46\cdot 10^6)^2)=5.57 m/s^2$

Final answer:

The satellite orbits at an altitude of approximately 800 km. The gravitational constant, 'g', at this location is approximately 8.66 m/s^2.

Explanation:

The orbital period of an artificial satellite can be used to calculate the altitude at which it orbits. For a satellite that completes each orbit in 129 min (or approximately 2.15 hr), we can apply Kepler's third law which states that the square of the period of a satellite is proportional to the cube of its semi-major axis (distance from the center of the Earth to the satellite).

The formula for the altitude is given by: h = [(GMT^2)/(4π^2)]^(1/3) - R, where G is the gravitational constant, M the mass of Earth, T the orbital period, and R the Earth's radius. With the values G=6.67 x 10^-11 N(m/kg)^2, M=5.98 x 10^24 kg, T=2.15 hr = 7740s, and R=6.371 x 10^6 m, we get h approximately equals 800 km.

The value of 'g' at the satellite's location is given by g = GM/(R+h)^2. Substituting the aforementioned values, we get g to be approximately 8.66 m/s^2. This is less than the 9.81 m/s^2 at Earth's surface due to the increased distance from the Earth's center.

Learn more about Artificial Satellite Orbit here:

brainly.com/question/31204877

#SPJ11

A flashlight has a resistance of 30 Q and is connectedby a wire to a 90 V source of voltage. What is the
current in the flashlight?

Answers

Answer:

3 A

Explanation:

The relationship between voltage, current and resistance in a conductor is expressed by Ohm's law, which states that:

$V=RI$

where:

V is the potential difference across the conductor

R is the resistance of the conductor

I is the current flowing through it

In this problem, for this flashlight we have:

V = 90 V is the potential difference

$R=30 \Omega$ is the resistance of the flashlight

Solving for I, we find the current:

$I=(V)/(R)=(90)/(30)=3 A$

Use mass in a sentence

Answers

the mass of a house is up to 120 meters

Is this piece of chalk made out of mass?

A 60kg bicyclist (including the bicycle) is pedaling to theright, causing her speed to increase at a rate of 3.1 m/s2,
despite experiencing a 60N drag. Neglect any friction
impeding her motion.

How many forces are acting on the bicyclist?

What is the magnitude of the net force on the bicyclist?

How much force is the bicyclist generating through her
pedaling?

Answers

a) 4 forces

b) 186 N

c) 246 N

Explanation:

Let's count the forces acting on the bicylist:

1) Weight ( $W=mg$ ): this is the gravitational force exerted on the bicyclist by the Earth, which pulls the bicyclist towards the Earth's centre; so, this force acts downward (m = mass of the bicyclist, g = acceleration due to gravity)

2) Normal reaction (N): this is the reaction force exerted by the road on the bicyclist. This force acts vertically upward, and it balances the weight, so its magnitude is equal to the weight of the bicyclist, and its direction is opposite

3) Applied force ( $F_A$ ): this is the force exerted by the bicylicist to push the bike forward. Its direction is forward

4) Air drag ( $R$ ): this is the force exerted by the air on the bicyclist and resisting the motion of the bike; its direction is opposite to the motion of the bike, so it is in the backward direction

So, we have 4 forces in total.

Here we can find the net force on the bicyclist by using Newton's second law of motion, which states that the net force acting on a body is equal to the product between the mass of the body and its acceleration:

$F_(net)=ma$

where

$F_(net)$ is the net force

m is the mass of the body

a is its acceleration

In this problem we have:

m = 60 kg is the mass of the bicyclist

$a=3.1 m/s^2$ is its acceleration

Substituting, we find the net force on the bicyclist:

$F_(net)=(60)(3.1)=186 N$

We can write the net force acting on the bicyclist in the horizontal direction as the resultant of the two forces acting along this direction, so:

$F_(net)=F_a-R$

where:

$F_(net)$ is the net force

$F_a$ is the applied force (forward)

$R$ is the air drag (backward)

In this problem we have:

$F_(net)=186 N$ is the net force (found in part b)

$R=60 N$ is the magnitude of the air drag

Solving for $F_a$ , we find the force produced by the bicyclist while pedaling:

$F_a=F_(net)+R=186+60=246 N$

How did Orion become a constellation?

Answers

While the Orion constellation is named after the hunter in Greek mythology, it is anything but stealthy. Orion, which is located on the celestial equator , is one of the most prominent and recognizable constellations in the sky and can be seen throughout the world.

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