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3 years ago

Please help me answer this question I will be given the brainliest as well

Physics

1 answer:

Sophie [7]3 years ago

3 0

draw a horizontal straight line from 25 to 75 and connects the line drawn at 75 to the bottom of the graph (0) at 90 with a straight line

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A train travels a distance of 2000 km at an average speed of 120 km per hour. How long did the trip take? SHOW WORK

Igoryamba

S = d/t, s = 120, d = 2000, t = ?
Input the values,
120 = 2000/t
Make t the subject of the formula by cross multiplying, Therefore,
120t = 2000
Divide both sides by 120
t = 16.7hrs to 1 decimal place.
If you're asked to convert it, you can.

6 0

3 years ago

PLZ HELP

Nadya [2.5K]

1) The mass of the continent is $3.3\cdot 10^{21}kg$

2) The kinetic energy of the continent is 1683 J

3) The speed of the jogger must be 6.57 m/s

Explanation:

The continent can be represented as a slab of size

$d=5850 km = 5.85\cdot 10^6 m$

and depth

$t = 35 km = 3.5\cdot 10^4 m$

So its volume is

$V=d^2 t = (5.85\cdot 10^6)^2(3.5\cdot 10^4)=1.20\cdot 10^{18} m^3$

We also know that the density of the continent is

$\rho = 2750 kg/m^3$

Therefore, we can calculate its mass as:

$m=\rho V=(2750)(1.20\cdot 10^{18})=3.3\cdot 10^{21} kg$

The kinetic energy of the continent is given by

$K=\frac{1}{2}mv^2$

where

m is its mass

v is its speed

We have already calculate its mass, while the speed is

v = 3.2 cm/year

We have to convert into SI units first, as follows:

$v=3.2 \frac{cm}{year} \cdot \frac{1}{100 cm/m} \cdot \frac{1}{(365 d/y)(24h/d)(60min/h)(60 s/min)}=1.01\cdot 10^{-9} m/s$

The mass is

$m=3.3\cdot 10^{21} kg$

So, the kinetic energy of the continent is

$K=\frac{1}{2}(3.3\cdot 10^{21})(1.01\cdot 10^{-9})^2=1683 J$

Here we have a jogger having the same kinetic energy of the continent, so

$K=1683 J$

And the kinetic energy of the jogger can be expressed as

$K=\frac{1}{2}mv^2$

where

m = 78 kg is the mass of the jogger

v is his speed

We can therefore re-arrange the equation to find the speed of the man, and we get:

$v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(1683)}{78}}=6.57 m/s$

Learn more about kinetic energy:

brainly.com/question/6536722

#LearnwithBrainly

8 0

3 years ago

Two satellites, X and Y, are orbiting Earth. Satellite X is 1.2 × 106 m from Earth, and Satellite Y is 1.9 × 105 m from Earth. W

jenyasd209 [6]

Answer: Satellite X has a greater period and a slower tangential speed than Satellite Y

Explanation:

According to Kepler’s Third Law of Planetary motion “The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”.

$T^{2}=\frac{4\pi^{2}}{GM}r^{3}$ (1)

Where;

$G=6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}$ is the Gravitational Constant

$M=5.972(10)^{24}kg$ is the mass of the Earth

$r$ is the semimajor axis of the orbit each satellite describes around Earth (assuming it is a circular orbit, the semimajor axis is equal to the radius of the orbit)

So for satellite X, the orbital period $T_{X}$ is:

$T_{X}^{2}=\frac{4\pi^{2}}{GM}r_{X}^{3}$ (2)

Where $r_{X}=1.2(10)^{6}m$

$T_{X}^{2}=\frac{4\pi^{2}}{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.972(10)^{24}kg)}(1.2(10)^{6}m)^{3}$ (3)

$T_{X}=413.712 s$ (4)

For satellite Y, the orbital period $T_{Y}$ is:

$T_{Y}^{2}=\frac{4\pi^{2}}{GM}r_{Y}^{3}$ (5)

Where $r_{Y}=1.9(10)^{5}m$

$T_{Y}^{2}=\frac{4\pi^{2}}{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.972(10)^{24}kg)}(1.9(10)^{5}m)^{3}$ (6)

$T_{Y}=26.064 s$ (7)

This means $T_{X}>T_{Y}$

Now let's calculate the tangential speed for both satellites:

<u>For Satellite X:</u>

$V_{X}=\sqrt{\frac{GM}{r_{X}}}$ (8)

$V_{X}=\sqrt{\frac{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.972(10)^{24}kg)}{1.2(10)^{6}m}}$

$V_{X}=18224.783 m/s$ (9)

<u>For Satellite Y:</u>

$V_{Y}=\sqrt{\frac{GM}{r_{Y}}}$ (10)

$V_{Y}=\sqrt{\frac{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.972(10)^{24}kg)}{1.9(10)^{5}m}}$

$V_{Y}= 45801.13 m/s$ (11)

This means $V_{Y}>V_{X}$

Therefore:

Satellite X has a greater period and a slower tangential speed than Satellite Y

4 0

3 years ago

Read 2 more answers

Name and describe the three types of binary systems.

Pachacha [2.7K]

Answer and Explanation:

The binary system is a star system about half of the star are the part of the binary system

The important binary systems are

Visual binary : in visual binary system the two stars resolved visually with nay sort of optical device the orbit of binary orbit is very large about center of mass.
Spectroscopic binary : in this binary system the stars are detected by close analysis of light and after detection it is found that there are two seller spectrum present instead of one.
Eclipsing binary : eclipsing binary system is a several types of variable stars this binary system is very important in astrophysics

5 0

3 years ago

For a storm to be a blizzard, the wind must be at least

Anton [14]

For a storm to be a blizzard, the wind must be at least 35 miles per hour. This is just one criteria for considering a storm to be a blizzard. The wind speed of 35 miles per hour should reduce the visibility to less than 400 meters. The last criteria is that the storm must continue for a time frame of at least 3 hours.

7 0

3 years ago