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Aliun [14]
2 years ago
13

For typical rubber-on-concrete friction, what is the shortest time in which a car could accelerate from 0 to 70 mph? suppose tha

t μs=1. 00μs=1. 00 and μk=0. 80μk=0. 80
Physics
1 answer:
algol132 years ago
3 0

The shortest time in which a car could accelerate from 0 to 70 mph is 3.193 seconds.

To find the answer, we have to know more about the friction.

<h3>How to find the shortest time?</h3>
  • We have the expression for final velocity as,

                          v=u+at

  • It is given in the question that,

                        v=70mph=31.29m/s\\1mile=1609.3m,1hour=3600s\\u=0\\k_s=1\\k_k=0.8

  • It is given that; the car could accelerate from 0 to 70mph. Thus, acceleration will be,

                    a=g*k_s=9.8*1=9.8m/s^2

  • Thus, the shortest time will be,

                    t=\frac{v-u}{a} \\t=\frac{31.29}{9.8}=3.193s

Thus, we can conclude that, the shortest time in which a car could accelerate from 0 to 70 mph is 3.193 seconds.

Learn more about the shortest time here:

brainly.com/question/3017271

#SPJ4

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Vectors A and B lie in the xy-plane. Vector A has a magnitude of 19.1 and is at an angle of 125.5º counterclockwise from the +x-
Nady [450]

Answer:

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Explanation:

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8 0
3 years ago
A 1100 kg sports car accelerates from 0m/s to 32m/s in 10 seconds. What is the average power of the engine?
timurjin [86]

Answer:

The average power of the engine of the sports car is 56.32 kW

Explanation:

Given;

mass of the sports car, m = 1100 kg

initial velocity of the sports car, u = 0 m/s

final velocity of the sports car, v = 32 m/s

time of motion, t = 10 s

The kinetic energy of the car is given by;

K.E = ¹/₂m(v² - u²)

K.E = ¹/₂mv²

K.E = ¹/₂ x 1100 x 32²

K.E = 563200 J

The average power of the engine of the sports car is given by;

Pavg = Energy / time

Pavg = 563200 / 10

Pavg = 56320 W

Pavg = 56.32 kW

Therefore, the average power of the engine of the sports car is 56.32 kW

8 0
3 years ago
A 2.8 kg grinding wheel is in the form of a solid cylinder of radius 0.1 m. a) What constant torque will bring it from rest to a
kogti [31]

Answer:

a) τ =  0.672 N m , b) θ = 150 rad , c) W = 100.8 J

Explanation:

a) for this part let's start by finding angular acceleration, when the angular velocity stops it is zero (w = 0)

       w = w₀ + α t

       α = -w₀ / t

       α = 120 / 2.5

       α = 48 rad / s²

The moment of inertia of a cylinder is

       I = ½ M R²

Let's calculate the torque

      τ = I α

      τ = ½ M R² α

      τ = ½ 2.8 0.1² 48

      τ =  0.672 N m

b) we look for the angle by kinematics

      θ = w₀ t + ½ α t2

      θ = ½ α t²

      θ = ½ 48 2.5²

      θ = 150 rad

c) work in angular movement

      W = τ θ

      W = 0.672 150

      W = 100.8 J

7 0
3 years ago
5. A cheetah with a mass of 70 kg was clocked running at 72 mph (32 m's). How many joules of
blsea [12.9K]

Answer:

Kinetic energy = 35840 Joules

Explanation:

Given the following data;

Mass = 70kg

Velocity = 32m/s

To find the kinetic energy;

Kinetic energy can be defined as an energy possessed by an object or body due to its motion.

Mathematically, kinetic energy is given by the formula;

K.E = \frac{1}{2}MV^{2}

Where, K.E represents kinetic energy measured in Joules.

M represents mass measured in kilograms.

V represents velocity measured in metres per seconds square.

K.E = \frac{1}{2}MV^{2}

Substituting into the equation, we have;

K.E = \frac{1}{2}*70*32^{2}

K.E = \frac{1}{2}*70*1024

K.E = 35 * 1024

K.E = 35840 Joules.

Therefore, the kinetic energy possessed by the cheetah is 35840 Joules.

7 0
3 years ago
In a controlled experiment why is it important to change just one variable parameter at a time
kiruha [24]
Think of it this way: If you test only one variable then you know that the difference in the experimental and control setup is that one independent variable. If you test more than one you will not know which one made the difference. Hope this helps :)
3 0
3 years ago
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