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matrenka [14]
3 years ago
9

Irina rode her bike to work at an average speed of 16 miles per hour. It started to rain, so she got a ride home along the same

route in her coworker’s car at an average speed of 27 miles per hour. If Irina’s ride home in the car took 24 minutes (0.4 of an hour), how many hours was her bike ride to work, to the nearest tenth of an hour?
Physics
2 answers:
motikmotik3 years ago
7 0
Average speed of Irina riding home in her coworkers car = 27 miles per hour
Time taken by Irina to reach home = 0.4 hour.
Then
In 1 hour the car traveled for a distance = 27 miles
So in 0.4 hour, the distance traveled by the car = 27 * 0.4 miles
                                                                           = 10.8 miles
Now
While traveling to office, Irina travels 16 miles in = 1 hour
So
Irina travels 10.8 miles in = (16/10.8) hours
                                         = 1.48 hours.]
So Irina takes 1.48 hours of bike riding to reach her place of work.
KengaRu [80]3 years ago
6 0

Answer:

0.7 hours

Explanation:

From the way back, we can calculate the distance between Irina's work and Irina's home. In fact, we know that the car takes 0.4 hourse traveling at 27 mph, so the distance covered should be

d=vt=(27 mph)(0.4 h)=10.8 mi

When Irina rides to work with her bike, she travels at a speed of 16 mph. So we can find the time she takes by dividing the total distance (10.8 miles) by her speed:

t=\frac{d}{v}=\frac{10.8 mi}{16 mph}=0.7 h

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3 0
3 years ago
The International Space Station has a mass of 1.8 × 105 kg. A 70.0-kg astronaut inside the station pushes off one wall of the st
Aleonysh [2.5K]

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a = 5.83 \times 10^{-4} m/s^2

Explanation:

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Monochromatic light from a distant source is incident on a slit 0.750 mm wide. On a screen 2.00 m away, the distance from the ce
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Answer:

\lambda= 506.25 nm

Explanation:

Diffraction is observed when a wave is distorted by an obstacle whose dimensions are comparable to the wavelength. The simplest case corresponds to the Fraunhofer diffraction, in which the obstacle is a long, narrow slit, so we can ignore the effects of extremes.

This is a simple case, in which we can use the Fraunhofer single slit diffraction equation:

y=\frac{m \lambda D}{a}

Where:

y=Displacement\hspace{3}from\hspace{3} the\hspace{3} centerline \hspace{3}for \hspace{3}minimum\hspace{3} intensity =1.35mm\\\lambda=Light\hspace{3} wavelength \\D=Distance\hspace{3}between\hspace{3}the\hspace{3}screen\hspace{3}and\hspace{3}the\hspace{3}slit=2m\\a=width\hspace{3}of\hspace{3}the\hspace{3}slit=0.750mm\\m=Order\hspace{3}number=1

Solving for λ:

\lambda=\frac{y*a}{mD}

Replacing the data provided by the problem:

\lambda=\frac{(1.35\times 10^{-3})*(0.750\times 10^{-3})}{1*2} =5.0625\times 10^{-7}m =506.25nm

7 0
3 years ago
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