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stich3 [128]
3 years ago
7

"Ryan drove to the mountains last weekend. There was heavy traffic on the way there, and the trip took 7 hours. When Ryan drove

home, there was no traffic and the trip only took 4 hours. If his average rate was 27 miles per hour faster on the trip home, how far away does Ryan live from the mountains?"
Physics
1 answer:
Flura [38]3 years ago
8 0

Answer:

Distance between Ryan home and mountain will be 252 miles

Explanation:

Let the distance from home to mountains is d

It is given that it took 7 hours to go mountains from home

Let the speed in going to mountain is v

Then according to question speed in coming from mountain is v+27

And time taken in come from mountain to home is hour

As distance will be same in both case

Distance will be equal to distance=speed\times time

So v\times 7=(v+27)\times 4

3v=108

v = 36 miles per hour

So distance will be equal to d=36\times 7=252miles

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A car of mass 1167 kg accelerates on a flat highway from 10 m/s to 28.0 m/s. How much work does the car's engine do on the car?
denis23 [38]

Answer:

Workdone = 465766038 Joules.

Explanation:

<u>Given the following data;</u>

Mass = 1167

Initial velocity = 10m/s

Final velocity =28m/s

To find the workdone;

We know that from the workdone theorem, the workdone by an object or a body is directly proportional to the kinetic energy possessed by the object due to its motion.

Mathematically, it is given by the equation;

W = Kf - Ki

W = ½MVf² - ½MVi²

Substituting into the equation

W = ½(1167)*28² - ½(1167)*10²

W = ½ * 1361889* 784 - ½ * 1361889 * 100

W = 533860488 - 68094450

Workdone = 465766038 Joules.

7 0
2 years ago
PLSS HELP ITS TIMED <br> WILL MARK BRAINLIEST
dlinn [17]

Answer:

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Explanation:

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4 0
3 years ago
Monochromatic light falling on two very narrow slits 0.048mm apart. Successive fringes on a screen 5.00m away are 6.5cm apart ne
tino4ka555 [31]

Answer:

λ = 5.85 x 10⁻⁷ m = 585 nm

f = 5.13 x 10¹⁴ Hz

Explanation:

We will use Young's Double Slit Experiment's Formula here:

Y = \frac{\lambda L}{d}\\\\\lambda = \frac{Yd}{L}

where,

λ = wavelength = ?

Y = Fringe Spacing = 6.5 cm = 0.065 m

d = slit separation = 0.048 mm = 4.8 x 10⁻⁵ m

L = screen distance = 5 m

Therefore,

\lambda = \frac{(0.065\ m)(4.8\ x\ 10^{-5}\ m)}{5\ m}

<u>λ = 5.85 x 10⁻⁷ m = 585 nm</u>

Now, the frequency can be given as:

f = \frac{c}{\lambda}

where,

f = frequency = ?

c = speed of light = 3 x 10⁸ m/s

Therefore,

f = \frac{3\ x\ 10^8\ m/s}{5.85\ x\ 10^{-7}\ m}\\\\

<u>f = 5.13 x 10¹⁴ Hz</u>

5 0
2 years ago
A race car has a centripetal acceleration of 13.33 m/s^2 as it goes around a curve. if the curve is a circle with a radius 30 m
anzhelika [568]

Answer:

The speed of the car, v = 19.997 m/s

Explanation:

Given,

The centripetal acceleration of the car, a = 13.33 m/s²

The radius of the curve, r = 30 m

The centripetal force acting on the car is given by the formula

                                   F = mv²/r

Where    v²/r is the acceleration component of the force

                                       a = v²/r

Substituting the values in the above equation

                                        13.33 = v²/30

                                         v² = 13.33 x 30

                                         v² = 399.9

                                         v = 19.997 m/s

Hence, the speed of the car, v = 19.997 m/s

3 0
2 years ago
A car speedometer has a 4% uncertainty. What is the range of possible speeds (in km/h) when it reads 110 km/h?
Kruka [31]
4% of 110 is 4.4. So the possible range of speeds is the interval from 110-4.4 till 110+4.4.
105.6 till 114.4
4 0
2 years ago
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