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Ronch [10]
3 years ago
11

On a trip to the Colorado Rockies, you notice that when the freeway goes steeply down a hill, there are emergency exits every fe

w miles. These emergency exits are straight ramps which leave the freeway and are sloped uphill. They are designed to stop runaway trucks and cars that lose their brakes on downhill stretches of the freeway even if the road is covered with ice. You are curious, so you stop at the next emergency exit to take some measurements. You determine that the exit rises at an angle of 10o from the horizontal and is 100m long. What is the maximum speed of a truck that you are sure will be stopped by this road, even if the frictional force of the road surface is negligible?
Physics
1 answer:
Zanzabum3 years ago
4 0

Answer:

The maximum speed that the truck can have and still be stopped by the 100m road is the speed that it can go and be stopped at exactly 100m. Since there is no friction, this problem is similar to a projectile problem. You can think of the problem as being a ball tossed into the air except here you know the highest point and you are looking for the initial velocity needed to reach that point. Also, in this problem, because there is an incline, the value of the acceleration due to gravity is not simply g; it is the component of gravity acting parallel to the incline. Since we are working parallel to the plane, also keep in mind that the highest point is given in the problem as 100m. Solving for the initial velocity needed to have the truck stop after 100m, you should find that the maximum velocity the truck can have and be stopped by the road is 18.5 m/s.

Explanation:

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EastWind [94]

The Speed of a moving vehicle can be measured. The average time it will taken too cover both distance are shown below;

The average travel times traveled are :

0.25 meters = 2.23 seconds

0.50 meters = 3.13 seconds

Therefore, the Average travel time to cover 0.25 meters is:

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(2.24 + 2.21 + 2.23) / 3 = 6.68 = 2.2266 = 2.23 seconds ( approximated to 2 decimal places).

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(Trial 1 + Trial 2 + Trial 3) ÷ 3

(3.16 + 3.08 + 3.15) / 3 = 9.39 ÷ 3 = 3.13 seconds

Conclusively, the average time taken to cover 0.25 and 0.50 meters is said to be 2.23 and 3.13 seconds for the both..

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What is the most interesting or surprising thing you learned about contact and non-contact forces?
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<h2>Density of the unknown liquid is 771.93 kg/m³</h2>

Explanation:

An empty graduated cylinder weighs 55.26 g

Weight of empty cylinder = 55.26 g = 0.05526 kg

Volume of liquid filled = 48.1 mL = 48.1 x 10⁻⁶ m³

Weight of cylinder plus liquid = 92.39 g = 0.09239 kg

Weight of liquid = 0.09239 - 0.05526

Weight of liquid = 0.03713 kg

We have

        Mass = Volume x Density

        0.03713 = 48.1 x 10⁻⁶ x Density

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balandron [24]
Hey

Potential Difference given is : 2V

Resistance is : 2 ohms

By Ohm's Law, one can easily utilize the relation :

v = ir

Where, { v , i , r } are the potential difference, current and Resistance Respectively.


Hence,
i =  \frac{v}{r}  =  \frac{2}{2}  = 1a

Hence, the Current is 1 Ampere
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