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a_sh-v [17]
3 years ago
8

Before 1960, people believed that the maximum attainable coefficient of static friction for an automobile tire on a roadway was

μs = 1. around 1962, three companies independently developed racing tires with coefficients of 1.6. this problem shows that tires have improved further since then. the shortest time interval in which a piston-engine car initially at rest has covered a distance of one-quarter mile is about 4.43 s. ]
Physics
1 answer:
True [87]3 years ago
5 0

This problem is looking for the minimum value of μs that is necessary to achieve the record time. To solve this problem:


Assuming the front wheels are off the ground for the entire ¼ mile = 402.3 m, the acceleration a = µs·9.8 m/s².


For a constant acceleration, distance = 402.3


m = 1/2at^2 = 804.6 m / (4.43 s)^2 = a = µs·9.8 m/s^2



µs = 804.6 m / (4.43s)^2 / 9.8 m/s^2 = 4.18

You might be interested in
which type of landform develops at plate boundaries were one oceanic plate desends beneth another oceanic plate
viktelen [127]
Umm I think the ocean floor but I don't think so
3 0
3 years ago
A basketball player throws a ball at 60° to the horizontal at a net 4.0m away. The height of the net is 3.3m. The ball leaves hi
Hoochie [10]
Excellent work!
Your calculations are correct, but near the end, you have forgotten to cancel u.  If you cancel the u, the linear term becomes a constant, and the resultant equation becomes a simple quadratic which is much easier to solve.
I get u=7.468 m/s using g=9.81 (as you did)
You will rework and should get 7.468 m/s as well.

Congrats for the good work!

4 0
3 years ago
When starting a foot race, a 70.0kg sprinter exerts an average force of 650 N backward on the ground for 0.800 s. How far does h
melomori [17]

Answer:

Answer is option b) 2.97m

Explanation:

With the relationship between the force exerted by the runner and the mass that it has, I can determine the acceleration it will have:

F= m × a ⇒ a= (650 kg ×(m/s^2)) / (70kg)= 9.286 (m/s^2)

With the acceleration that prints the force exerted and the time I can determine the distance traveled in the interval:

Distance= (1/2) × a × t^2 = (1/2) × 9.286 (m/s^2) × ((0.8s)^2)= 2.97m

8 0
3 years ago
An oak block had the following dimension, 2 cm by 2 cm and 5 cm in height. A copper block had the same dimensions. The copper bl
antiseptic1488 [7]

Answer:

ρ_body = 1000 kg / m³

Explanation:

This is an exercise in fluid mechanics, specifically we must use the Archimedean principle, which states that the thrust is equal to the weight of the dislodged liquid.

In this case let's start by finding the volume of our body

oak block

      v = l to h

       v = 0.02 0.02 0.05

       V = 2 10⁻⁵ m³

cooper block indicate that it has the same dimensions so its volume is the same, the total volume of the body is

        V_total = 4 10⁻⁵ m³

as they indicate that the body is fully submerged there is a balance between weight and thrust

          B - W = 0

the push is

          B = ρ_fluid g V_total

the body weight is

         ρ_body = M / V_total

          M = ρ_body V_total

         W = Mg

          W = ρ_body V_total g

we substitute

          ρ_fluid g V_total = ρ_body V_total g

        ρ_body = ρ_fluid

in this case the body is in equilibrium in the fluid, in case the density of the body is greater than that of the fluid, the body sinks

Therefore the average density is equal to the density of the fluid, since since it is water the density is

              ρ_body = 1000 kg / m³

8 0
3 years ago
Consider a motor that exerts a constant torque of 25.0 N⋅m to a horizontal platform whose moment of inertia is 50.0 kg⋅m2 . Assu
Step2247 [10]

To solve this exercise it is necessary to apply the concepts related to Work and Kinetic Energy. Work from the rotational movement is described as

W=\tau \Delta\theta

In the case of rotational kinetic energy we know that

KE = \frac{1}{2}I\omega^2

PART A) \theta is given in revolutions and needs to be in radians therefore

\theta = 12rev(\frac{2\pi rad}{1rev})

\theta = 24\pi rad

Replacing in the work equation we have to

W=\tau \Delta\theta

W= (25)(24\pi)

W = 1884.95J

PART B) From the torque and moment of inertia it is possible to calculate the angular acceleration and the final speed, with which the kinetic energy can be determined.

\tau = I \alpha

Rearrange for the angular acceleration,

\alpha = \frac{\tau}{I}

\alpha = \frac{25}{50}

\alpha = 0.5rad/s

From the kinematic equations of angular motion we have,

\omega_f^2=\omega_i^2+2\alpha\theta

\omega_f^2=0+2*0.5*24\pi

\omega_f=\sqrt{0+2*0.5*24\pi}

\omega_f = 8.68rad/s

In this way the rotational kinetic energy would be given by

KE = \frac{1}{2}I\omega_f^2

KE = \frac{1}{2}(50)(8.68)^2

KE = 1883.56J

3 0
3 years ago
Read 2 more answers
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