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gizmo_the_mogwai [7]
3 years ago
11

The table below shows the right ascensions of two stars at a location.

Physics
1 answer:
zlopas [31]3 years ago
5 0
Star 1 - 4 hours right ascension
 Star 2 - 3 hours right ascension
 Subtracting hours right ascension
 4 hours right ascension - 3 hours right ascension = 1 hours right ascension.
Thus,
 1) star 1 will rise 1 hour before star 2
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9 m/s to a total stop. what is the objects acceleration
MAVERICK [17]
We would need to know the time it took to slow to a stop.
5 0
3 years ago
Light takes 8 minutes to reach the Earth, and the speed of light is 3.0×10^8 m/s. a) What is the orbital speed of the Earth arou
spin [16.1K]

Answer:

(a) 28690 m/s (b) 2.46x10^{33}J

Explanation:

The orbital speed is define as:

v = \frac{2 \pi r}{T}   (1)

Where r is the radius of the trajectory and T is the orbital period.

To determine the orbital speed of the Earth it is necessary to know the orbital period and the radius of the trajectory. That can be done by means of the Kepler's third law and average velocity equation.

The average velocity in a Uniform Rectilinear Motion is defined as:

v = \frac{d}{t}   (2)

Where v is the velocity, d is the covered distance and t is the time.

Equation 2 can be rewritten for d to get:

d = vt   (3)

In this case, v will be the speed of light and t, the 8 minutes that takes to reach the Earth.

The time will be converted to seconds so the units in equation 3 can match:

8min . \frac{60s}{1min} ⇒  480s

t = 480s

Replacing all those values in equation 3 it is gotten:

d = (3.0x10^{8}m/s)(480s)

d = 1.44x10^{11}m

Kepler’s third law is defined as:

T^{2} = r^{3}

Where T is orbital period and r is the radius of the trajectory.

T = \sqrt{r^{3}}

T = \sqrt{(1.44x10^{11}m)^{3}}

It is necessary to pass from meters to astronomical unit (AU), 1 AU is defined as the distance between the Earth and the Sun.

T = \sqrt{1AU}

T = 1AU

That can be expressed in units of years.

T = 1AU . \frac{1year}{1AU}

T = 1year

But there are 31536000 seconds in one year:

T = 1year . \frac{31536000s}{1year}

T = 31536000s

Finally, equation  1 can be used:

v = \frac{2 \pi (1.44x10^{11}m)}{(31536000s)}

v = 28690 m/s

<u>So Earth orbital speed around the Sun is 28690 m/s.</u>

<em>b) What is its kinetic energy?</em>

The kinetic energy is defined as:

E = \frac{1}{2}mv^{2}  (4)

Notice that it is necessary to found the mass of the Earth, that can be done combining the Universal law of gravity and Newton's second law:

F = \frac{GMm}{r^{2}}

ma = \frac{GMm}{r^{2}}  (5)

M will be isolated in equation 5:

M = \frac{r^{2}a}{G}

Where r is the radius of the Earth (6.38x10^{6}m)

M = \frac{(6.38x10^{6}m)^{2}(9.8m/s^{2})}{6.67x10^{-11}kg.m/s^{2}.m^{2}/Kg^{2}})

M = 5.98x10^{24} Kg

E = \frac{1}{2}( 5.98x10^{24} Kg))(28.690m/s)^{2}

E = 2.46x10^{33}Kg.m^{2}/s^{2}

E = 2.46x10^{33}J

<u>Hence, the kinetic energy of Earth is 2.46x10^{33}J.</u>

8 0
3 years ago
A horizontal force of magnitude 356 N is employed to push a 78.0-kg block a distance of 18.6 m on a rough horizontal surface. If
icang [17]

Answer:

6621.6 Joule

0.46525

Explanation:

F = Force = 356 N

s = Displacement = 18.6 m

m = Mass of block = 78 kg

Work done

W=F\times s\\\Rightarrow W=356\times 18.6\\\Rightarrow W=6621.6\ J

The work done by the horizontal force is 6621.6 Joule

F=\mu mg\\\Rightarrow \mu=\frac{F}{mg}\\\Rightarrow \mu=\frac{356}{78\times 9.81}\\\Rightarrow \mu=0.46525

The coefficient of friction between the block and the rough surface is 0.46525

5 0
3 years ago
A total Δν of 15 km/s is required to achieve an interplanetary mission. The proposed rocket has two stages. The first stage alon
AlexFokin [52]

Answer:

102000 kg

Explanation:

Given:

A total Δν = 15 km/s

first stage mass = 1000 tonnes

specific impulse of liquid rocket =  300 s

Mass flow rate of liquid fuel = 1500 kg/s

specific impulse of solid fuel = 250 s

Mass flow of solid fuel = 200 kg/s

First stage burn time = 1 minute = 1 × 60 seconds = 60 seconds

Now,

Mass flow of liquid fuel in 1 minute = Mass flow rate × Burn time

or

Mass flow of liquid fuel in 1 minute = 1500 × 60 = 90000 kg

Also,

Mass flow of solid fuel in 1 minute = Mass flow rate × Burn time

or

Mass flow of solid fuel in 1 minute = 200 × 60 = 12000 kg

Therefore,

The total jettisoned mass flow of the fuel in first stage

= 90000 kg +  12000 kg

= 102000 kg

3 0
3 years ago
Oceans receive freshwater from precipitation and rivers. Yet ocean levels do not change very much from these actions. Why are oc
Neko [114]
Ocean levels are not greatly affected because when water is deposited or received from the ocean, at the same time water is being evaporated from the ocean to condense in the clouds and them form into precipitation.
5 0
3 years ago
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