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Arisa [49]
4 years ago
12

The formula v = √ 2.3 r models the maximum safe speed, v , in miles per hour, at which a car can travel on a curved road with ra

dius of curvature r , in feet. A highway crew measures the radius of curvature at an exit ramp on a highway as 370 feet. What is the maximum safe speed? FThe formula v = √ 2.3 r models the maximum safe speed, v , in miles per hour, at which a car can travel on a curved road with radius of curvature r , in feet. A highway crew measures the radius of curvature at an exit ramp on a highway as 370 feet. What is the maximum safe speed? F
Physics
1 answer:
Archy [21]4 years ago
6 0

Answer:

562 miles per hour.

Explanation:

As given in the question, the formula for the maximum speed on a curved road is

v=\sqrt{2.3} r

Given value of r=370 feet

So the maximum safe speed will be

v=\sqrt{2.3} \times 370 = 1.52\times 370 = 562.4 miles per hour.

Rounding off to the nearest whole number we get the maximum safe speed at the curved road is 562 miles per hour.

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A 31 cm tall object is placed in front of a concave mirror with a radius of 34 cm. The distance of the object to the mirror is 9
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Focal length is the distance between the center of a convex lens or a concave mirror and the focal point of the lens or mirror — the point where parallel rays of light meet, or converge. From the optics the focal length of the mirror can be defined as the radius of the mirror divided between two, or in other words, half the radius of the mirror.

f = \frac{R}{2}

f = \frac{34}{2}

f = 17cm

Therefore the focal length of the mirror is 17cm

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3 years ago
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Kepler's first law states that planetary orbits are elliptical in shape. (C)

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As soon as a traffic light turns green, a car speeds up from rest to its cruising speed 22.3 m/sec with constant acceleration 4.
Ganezh [65]

Answer:  (a) The bicycle is ahead of the car for 4 s.

               (b) The bicycle leads the car by the maximum distance of 55 m.

Explanation:

(a)

Use the equation of the motion to calculate the time taken by the car.

v=u+at  

As it is given in the problem, a car speeds up from rest to its cruising speed 22.3 m/sec with constant acceleration 4.02 m/sec2 .

Put u=0, v=22.3 m/sec and a=4.02 m/sec^2.

22.3=0+4.02t

t=\frac{22.3}{4.02}

t= 5.5 s

Use the equation of the motion to calculate the time taken by the  bicycle.

v=u+at_{1}

As it is given in the problem, a cyclist speeds up from rest to its cruising speed 8.94 m/sec with constant acceleration 5.81 m/sec^2.

Put u=0, v=8.94 m/sec and a=5.81 m/sec^2.

8.94=0+5.81t_{1}

t_{1}=\frac{8.94}{5.81}[tex]t_{1}=1.5 s

Calculate the time interval for which the bicycle is ahead of the car.

t-t_{1}= 5.5 s - 1.5s

t-t_{1}= 4s

Therefore, the bicycle is ahead of the car for 4 s.

(b)

Use the equation motion to calculate the distance covered by the car.

S=ut+\frac{1}{2}at^{2}

As it is given in the problem, a car speeds up from rest to its cruising speed 22.3 m/sec with constant acceleration 4.02 m/sec^2 .

Put t= 5.5 s, u=0 s and a=4.02 m/sec^2.

S=(0)t+\frac{1}{2}(4.02)(5.5)^{2}

S= 60.8 m

Use the equation motion to calculate the distance covered by the bicycle.

S_{1}=ut+\frac{1}{2}at^{2}

As it is given in the problem, a cyclist speeds up from rest to its cruising speed 8.94 m/sec with constant acceleration 5.81 m/sec^2.

Put t= 1.5 s, u=0 s and a=5.81 m/sec^2.

S_{1}=(0)t+\frac{1}{2}(5.18)(1.5)^{2}

S_{1}= 5.8 m

Calculate the maximum distance covered by the bicycle to lead the car.

S-S_{1}=60.8-5.8=55m

Therefore, the bicycle leads the car by the maximum distance of 55 m.

6 0
3 years ago
On a global annual average basis, what is the most important process in cooling earth’s surface?
const2013 [10]
I believe that the answer to the question provided above is that <span>the most important process in cooling earth’s surface is the production of oxygen by trees , since it has cooling effect and CO2 is reduced.</span>
Hope my answer would be a great help for you.    If you have more questions feel free to ask here at Brainly.
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Anything that occupies space and has mass is called
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Answer:

is called <em>matter</em> :)

Explanation:

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