Steep safety ramps are built beside mountain highway to enable vehichles with fedective brakes to stop safely. a truck enters a
1000 ft ramps at a high speed vo and travels 600ft in 7 s at constant deceleration before its speed is reduced to uo/2. Assuming the same constant deceleration.
Determine:
a. The additional time required for the truck to stop.
b. The additional distance traveled by the truck.
Determine:
a. The additional time required for the truck to stop.
b. The additional distance traveled by the truck.
1 answer:
Answer:
a. 6 seconds
b. 180 feet
Explanation:
Images attached to show working.
a. You have the position of the truck so you integrate twice. Use the formula and plug in the time t = 7 sec. Check out uniform acceleration. The time at which the truck's velocity is zero is when it stops.
b. Determine the initial speed. Plug in the time calculated in the previous step. From this we can observe that the truck comes to a stop before the end of the ramp.
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Answer:
Yes, the flow is turbulent.
Explanation:
Reynolds number gives the nature of flow. If he Reynolds number is less than 2000 then the flow is laminar else turbulent.
Given:
Diameter of pipe is 10mm.
Velocity of the pipe is 1m/s.
Temperature of water is 200°C.
The kinematic viscosity at temperature 200°C is m2/s.
Calculation:
Step1
Expression for Reynolds number is given as follows:
Here, v is velocity, is kinematic viscosity, d is diameter and Re is Reynolds number.
Substitute the values in the above equation as follows:
Re=64226.07579
Thus, the Reynolds number is 64226.07579. This is greater than 2000.
Hence, the given flow is turbulent flow.
Answer:
4m/s
Explanation:
We know that power supplied by the motor should be equal to the rate at which energy is increased of the mass that is to be hoisted
Mathematically
\
We also know that Power = force x velocity ..................(i)
The force supplied by the motor should be equal to the weight (mg) of the block since we lift the against a force equal to weight of load
=> power = mg x Velocity........(ii)
While hoisting the load at at constant speed only the potential energy of the mass increases
Thus Potential energy = Mass x g x H...................(iii)
where
g = accleration due to gravity (9.81m/s2)
H = Height to which the load is hoisted
Equating equations (ii) and (iii) we get
m x g x v =
thus we get v = H/t
Applying values we get
v = 6/1.5 = 4m/s