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3 years ago

Terry can ride 30 miles in 2 hours. If his riding speed is

Physics

2 answers:

Nastasia [14]3 years ago

6 0

Answer: 20.4 miles

Explanation:

Here we need to use the equation:

Velocity = Distance/Time.

Initially we have that he can travel 30 miles in 2 hours, so the velocity is:

V = 30mi/2h = 15mph

Now, we reduce the velocity by 3 mph, so the new velocity is 15mph - 3 mph = 12mph.

Now we want to know the distance traveled in 1.7 hours with this velocity, this is.

Velocity*Time = Distance

12mi/h*1.7h = 20.4 miles

marishachu [46]3 years ago

5 0

Answer:

Explanation:

Terry can ride at a speed of

V = 30 miles in 2hours

Speed = distance / time

V = 30 /2

V = 15 mile/hour

So, we want to know the distance traveled in 1.7hours

Then,

Speed = distane / time

Distance = speed × time

Distance =15 × 1.7

Distance = 25.5 miles

So, the distance traveled in 1.7hours is 25.5 miles

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A red blood cell is 7 x 10-6 m in size. This means it is 7 __________ in size. What unit completes the sentence?

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If a red blood cell is 7 x $10^-^6$ m in size, it means that such a cell is 7 microns or micrometer in size. The unit that completes the sentence will, thus, be microns or micrometer.

According to international system of measurement:

$10^{-1}$ m = decimeter

$10^{-2}$ m = centimeter

$10^{-3}$ m = millimeter
$10^{-6}$ m = micrometer or micron
$10^{-9}$ m = nanometer

Thus, a measurement of 7 x $10^-^6$ m is the same is 7 micons or 7 micrometer.

More on metric conversion can be found here: brainly.com/question/17767575

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2 years ago

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3 years ago

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tatyana61 [14]

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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

A capacitor is made from two hollow, coaxial, iron cylinders, one inside the other. The inner cylinder is negatively charged and

Musya8 [376]

Answer: 3.12 * 10^12 F ( 3.12 pF)

Explanation: To calculate this capacitor of two hollow, coaxial, iron cylinders, we have to determine the potental differente between them and afeter that to use C=Q/ΔV

The electric field in th eregion rinner<r<router

By using the Gaussian law

∫E*ds=Q inside/εo

E*2*π*rinner^2*L= Q /εo

E=Q/(2*π*εo*r^2)

[Vab]=\int\limits^a_b {E} \, dr

where a and b are the inner and outer radii.

Then we have:

ΔV= 2*k*(Q/L)* ln (b/a)

replacing the values and using that C=Q/ΔV

we have:

C= L/(2*k*ln(b/a)=0.17/(2*9*10^9*3.023)=3.12 pF

5 0

3 years ago