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Lemur [1.5K]
3 years ago
12

A bicycle racer sprints at the end of a race to clinch a victory. The racer has an initial velocity of 12.0 m/s and accelerates

at the rate of 0.650 m/s2 for 7.00 s. (a) What is his final velocity (in m/s)?
Physics
1 answer:
Sveta_85 [38]3 years ago
7 0

Answer:

Final speed of the racer, v = 16.55m/s

Explanation:

It is given that,

Initial velocity of the racer, u = 12 m/s

Acceleration, a=0.65\ m/s^2

time taken, t = 7 s

We need to find the final velocity of the racer. Let it is given by v. It can be calculated using first equation of motion as :

v=u+at

v=12+0.65\times 7

v = 16.55 m/s

So, the final speed of the racer is 16.55 m/s. Hence, this is the required solution.

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Answer:

D. You speed up

Explanation:

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This involves convection currents.
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6 0
3 years ago
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Galois drove 60.0 kilometers due west in 5.00 hours and then drove 43.0 kilometers due north in 3.00 hours.
IceJOKER [234]

Answer:

Explanation:

See the attachment for the details.  A right triangle is formed to find the hypotenuse of the two legs consisting of the actual driving distances and times.  The hypotenuse gives the vector information for the displacement at the end of 8 hours of driving.  

The individual driving times and distances are summed to provide:

(<u>a) How far did he travel?</u>

103 km

<u>(b) What was his average speed?</u>

12.88 km/h

<u>(c) What was his displacement?</u>

73.82 km

<u>(d) What was his average velocity?</u>

9.228 km/h

8 0
2 years ago
Balanced forces acting on an object cause the object to accelerate. Please select the best answer from the choices provided T F
Pani-rosa [81]
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7 0
3 years ago
If the distance between slits on a diffraction grating is 0.50 mm and one of the angles of diffraction is 0.25°, how large is th
Trava [24]

Answer:

-  path differnce = 2.18*10^-6

-  1538 lines

Explanation:

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path\ difference\ = dsin\theta           (1)

d: separation between slits = 0.50mm = 0.50*10^-3 m

θ: angle of a diffraction = 0.25°

Then, the path difference is:

path\ difference\ =(0.50*10^{-3}m)sin(0.25\°)=2.18*10^{-6}m

- The maximum number of bright lines are calculated by using the following formula:

m\lambda = dsin\theta           (2)

m: order of the bright

λ: wavelength = 650nm

The maximum bright is calculated for an angle of 90°:

m=\frac{(0.50*10^{-3}m)sin90\°}{650*10^{-9}m} \approx 769

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8 0
3 years ago
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