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Dmitrij [34]
3 years ago
14

A car is traveling at 120 km/h (75 mph). When applied the braking system can stop the car with a deceleration rate of 9.0 m/s2.

The typical reaction time for an alert driver is 0.8 s versus 3 s for a sleepy driver. Assuming a typical car length of 5 m, calculate the number of additional car lengths approximately it takes the sleepy driver to stop compared to the alert driver. Group of answer choices
Physics
1 answer:
Bumek [7]3 years ago
8 0

Answer:

the number of additional car lengths approximately it takes the sleepy driver to stop compared to the alert driver is 15

Explanation:

Given that;

speed of car V  = 120 km/h = 33.3333 m/s

Reaction time of an alert driver = 0.8 sec

Reaction time of an alert driver = 3 sec

extra time taken by sleepy driver over an alert driver = 3 - 0.8 = 2.2 sec

now, extra distance that car will travel in case of sleepy driver  will be'

S_d = V × 2.2 sec

S_d = 33.3333 m/s × 2.2 sec

S_d = 73.3333 m

hence, number of car of additional car length  n will be;

n = S_n / car length

n = 73.3333 m / 5m

n = 14.666 ≈ 15

Therefore, the number of additional car lengths approximately it takes the sleepy driver to stop compared to the alert driver is 15

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C. middle layer, density increases with depth as pressure increases

Explanation:

The mantle is the middle layer of the earth. It's density increases with depth as pressure increases.

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3 years ago
One end of a thin rod is attached to a pivot, about which it can rotate without friction. Air resistance is absent. The rod has
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Answer:

5.24 m/s

Explanation:

So for the rod to be able to rise upward to the straight up position, the kinetic energy caused by linear speed v0 must be just enough to convert into the potential energy.

Since the rod is uniform in mass, we can treat the body as 1 point, at its center of mass, or geometric center, aka 0.35 / 2 = 0.175 m from the pivot.

For the rod to swing from bottom to top, the center must have moved a distance of h = 0.175 * 2 = 0.35 m, vertically speaking.

Since we neglect friction and air resistance, according to the law of energy conservation then:

E_k = E_p

mv^2/2 = mgh

Where v is the speed at the center of mass, g = 9.81 m/s2 is the gravitational acceleration, and m is the mass. We can divide both sides by m

v^2 = 2gh = 2*9.81*0.35 = 6.867

v = \sqrt{6.867} = 2.62 m/s

As this is only the speed at the center of mass, the speed at the bottom end would be different, to calculate this, we need to find the common angular speed:

\omega = v / r = 2.62 / 0.175 = 14.97 rad/s

Where r is the rotation radius, or the distance from pivot point to the center of mass

v_0 = \omega R = 14.97*0.35 = 5.24 m/s

Where R is the distance from the pivot to the bottom end of the rod

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Calculate the potentia energy of a car with a mass of 3800kg that is on a hill 110 meters above sea level? USE 10 instead of 9.8
Arada [10]

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Where:

m = mass = 3800 kg

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7 0
1 year ago
Show that the optimal launch angle for a projectile subject to gravity is 45o by carrying out the following steps: 6. Write down
Lorico [155]

Answer:

sin  2θ = 1    θ=45

Explanation:

They ask us to prove that the optimal launch angle is 45º, for this by reviewing the parabolic launch equations we have the scope equation

            R = Vo² sin 2θ / g

Where R is the horizontal range, Vo is the initial velocity, g the acceleration of gravity and θ the launch angle. From this equation we see that the sine function is maximum 2θ = 90 since sin 90 = 1 which implies that θ = 45º; This proves that this is the optimum angle to have the maximum range.

We calculate the distance traveled for different angle

          R = vo² Sin (2 15) /9.8

          R = Vo² 0.051 m

In the table are all values ​​in two ways

Angle (θ)                  distance R (x)

 0                 0                     0

15                 0.051 Vo²        0.5 Vo²/g

30                0.088 vo²        0.866   Vo²/g

45                0.102 Vo²        1   Vo²/g

60                0.088 Vo²      0.866   Vo²/g

75                0.051 vo²        0.5   Vo²/g

90                0                     0

See graphic ( R Vs θ)  in the attached ¡, it can be done with any program, for example EXCEL

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