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2 years ago

8

A cyclist on a training ride records the distance she travels away from home. The data only shows the first150 minutes of the ri

de before her cycling computer ran out of battery.

1 answer:

Varvara68 [4.7K]2 years ago

3 0

Answer:

A) 58 km

B) 30 mins

Explanation:

In pic details

graph in pic

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Check my work please

katrin [286]

We can use the ideal gas equation which is expressed as PV = nRT. At a constant volume and number of moles of the gas the ratio of T and P is equal to some constant. At another set of condition, the constant is still the same. Calculations are as follows:

T1/P1 = T2/P2

P2 = T2 x P1 / T1

P2 = 273 x 340 / 713

<span>P2 = 130 kPa</span>

6 0

3 years ago

The Hubble Space Telescope orbits the Earth at approximately 612,000m altitude. Its mass is 11,100 kg and the mass of earth is 5

nexus9112 [7]

Answer:

7.55 km/s

Explanation:

The force of gravity between the Earth and the Hubble Telescope corresponds to the centripetal force that keeps the telescope in uniform circular motion around the Earth:

$G\frac{mM}{R^2}=m\frac{v^2}{R}$

where

$G=6.67\cdot 10^{-11} m^3 kg^{-1} s^{-2}$ is the gravitational constant

$m=11,100 kg$ is the mass of the telescope

$M=5.97\cdot 10^{24} kg$ is the mass of the Earth

$R=r+h=6.38\cdot 10^6 m+612,000 m=6.99\cdot 10^6 m$ is the distance between the telescope and the Earth's centre (given by the sum of the Earth's radius, r, and the telescope altitude, h)

v = ? is the orbital velocity of the Hubble telescope

Re-arranging the equation and substituting numbers, we find the orbital velocity:

$v=\sqrt{\frac{GM}{R}}=\sqrt{\frac{(6.67\cdot 10^{-11})(5.97\cdot 10^{24} kg)}{6.99\cdot 10^6 m}}=7548 m/s=7.55 km/s$

6 0

3 years ago

Where is a trench most likely to occur?

Dafna1 [17]

D. convergent plate boundary involving an oceanic plate

7 0

3 years ago

Read 2 more answers

Jake drops a book out of a window from a height of 10 meters. At what velocity does the book hit the ground?

user100 [1]

G = 9.81 m/sec^2) g = 9.81 $\frac{meters}{ second^{2} }$

<span>Solving for velocity : </span>

$velocity^{2}$ <span> = 2gh </span>
<span>v = </span> $2gh^{ \frac{1}{2} }$
<span>v = (2 x 9.81 x 10)^1/2 </span>
<span>v = 196.2 m/sec (answer)</span>

8 0

3 years ago

A beam of monochromatic light with a wavelength of 400 nm in air travels into water. what is the wavelength of the light in wate

slava [35]

The refractive index of water is $n=1.33$ . This means that the speed of the light in the water is:
$v= \frac{c}{n}= \frac{3 \cdot 10^8 m/s}{1.33 }=2.26 \cdot 10^8 m/s$

The relationship between frequency f and wavelength $\lambda$ of a wave is given by:
$\lambda= \frac{v}{f}$
where v is the speed of the wave in the medium. The frequency of the light does not change when it moves from one medium to the other one, so we can compute the ratio between the wavelength of the light in water $\lambda_w$ to that in air $\lambda$ as
$\frac{\lambda_w}{\lambda}= \frac{ \frac{v}{f} }{ \frac{c}{f} } = \frac{v}{c}$
where v is the speed of light in water and c is the speed of light in air. Re-arranging this formula and by using $\lambda=400 nm$ , we find
$\lambda_w = \lambda \frac{v}{c}=(400 nm) \frac{2.26 \cdot 10^8 m/s}{3 \cdot 10^8 m/s}=301 nm$
which is the wavelength of light in water.

5 0

3 years ago