The formula v = √2.5r models the maximum safe speed, v, in miles per hour, at which a car can travel on a curved road with radiu
1 answer:
Answer:
The maximum safe speed of the car is 30.82 m/s.
Explanation:
It is given that,
The formula that 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 r, is in feet is given by :
.........(1)
A highway crew measures the radius of curvature at an exit ramp on a highway as 380 feet, r = 380 feet
Put the value of r in equation (1) as :
v = 30.82 m/s
So, the maximum safe speed of the car is 30.82 m/s. Hence, this is the required solution.
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Answer:
x = 4.32 [m]
Explanation:
We must divide this problem into three parts, in the first part we must use Newton's second law which tells us that the force is equal to the product of mass by acceleration.
∑F = m*a
where:
F = force = 700 [N]
m = mass = 2030 [kg]
a = acceleration [m/s²]
Now replacing:
Then we can determine the final speed using the principle of conservation of momentum and amount of movement.
where:
m₁ = mass of the car = 2030 [kg]
v₁ = velocity at the initial moment = 0 (the car starts from rest)
Imp₁₋₂ = The impulse or momentum (force by the time)
v₂ = final velocity after the impulse [m/s]
Now using the following equation of kinematics, we can determine the distance traveled.
where:
v₂ = final velocity = 1.72 [m/s]
v₁ = initial velocity = 0
a = acceleration = 0.344 [m/s²]
x = distance [m]
Answer:
1) 10.1 s 2) 909 m 3) 90.0 m/s 4) -99m/s 5) just over the bomb.
Explanation:
1)
- In the vertical direction, as the bomb is dropped, its initial velocity is 0.
- So, we can find the time required for the bomb to reach the earth, applying the following kinematic equation for displacement:
- where Δy = -500 m (taking the upward direction as positive).
- a=-g=-9.8 m/s²
- Replacing these values in (1), and solving for t, we have:
- The time required for the bomb to reach the earth is 10.1 s.
2)
- In the horizontal direction, once released from the helicopter, no external influence acts on the bomb, so it will continue moving forward at the same speed. that it had, equal to the helicopter.
- As the time must be the same for both movements, we can find the horizontal displacement just as the product of this speed times the time, as follows:
3)
- The horizontal component of the bomb's velocity is the same that it had when left the helicopter. i.e. 90 m/s.
4)
- In order to find the vertical component of the bomb's velocity just before it strikes the earth, we can apply the definition of acceleration, remembering that v₀ = 0, as follows:
5)
- If the helicopter keeps flying horizontally at the same speed, it will be always over the bomb, as both travel horizontally at the same speed.
- So, when the bomb hits the ground, the helicopter will be exactly over it.