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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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A rock is thrown vertically upward from some height above the ground. It rises to some maximum height and falls back to the grou
True [87]

Answer:

At the highest point the velocity is zero, the acceleration is directed downward.

Explanation:

This is a free-fall problem, in the case of something being thrown or dropped, the acceleration is equal to -gravity, so -9.80m/s^2. So, the acceleration is never 0 here.

I attached an image from my lecture today, I find it to be helpful. You can see that because of gravity the acceleration is pulled downwards.

At the highest point the velocity is 0, but it's changing direction and that's why there's still an acceleration there.

6 0
3 years ago
A stuntman is being pulled along a rough road at a constant velocity by a cable attached to a moving truck. The cable is paralle
Alex73 [517]

Answer:

715 N

Explanation:

Since the system is moving at a constant velocity, the net force must be 0. The tension on the road is equal and opposite direction with the kinetic friction force created by the road and the stuntman.

Let g = 9.8 m/s2

Gravity and equalized normal force is:

N = P = mg = 107*9.8 = 1048.6 N

Kinetic friction force and equalized tension force on the rope is

T = F_{\mu} = N\mu = 1048.6 * 0.682 = 715.1452 N

6 0
3 years ago
A 180-ohm resistor has 0.10 A of current in it. what is the potential difference across the resistor
Firlakuza [10]
We know V=IR (Ohm's law).

We are given R=180Ω and I=0.1A, then V=(0.1AΩ)(180Ω). Therefore

V=18V
5 0
3 years ago
An engineer is designing a runway. She knows that a plane, starting at rest, needs to reach a speed of 180mph at take-off. If th
kvv77 [185]

Answer:

The plane would need to travel at least 8,\!580\; {\rm ft} (8.58 \times 10^{3}\; {\rm ft}.)

The 10,\!000\; {\rm ft} runway should be sufficient.

Explanation:

Convert unit of the the take-off velocity of this plane to \rm ft\cdot s^{-1}:

\begin{aligned}v &= 180\; {\rm mph} \\ &= 180\; {\rm mi \cdot hrs^{-1}} \times \frac{1\; {\rm hrs}}{3600\; {\rm s}} \times \frac{5280\; {\rm ft}}{1\; {\rm mi}} \\ &= 264\; {\rm ft \cdot s^{-1}}\end{aligned}.

Initial velocity of the plane: u = 0\; {\rm ft \cdot s^{-1}}.

Take-off velocity of the plane v =264\; {\rm ft\cdot s^{-1}}.

Let x denote the distance that the plane travelled along the runway. Since acceleration is constant but unknown, make use of the SUVAT equation x = ((u + v) / 2) \, t.

Notice that this equation does not require the value of acceleration. Rather, this equation make use of the fact that the distance travelled (under constant acceleration) is equal to duration t times average velocity (u + v) / 2.

The distance that the plane need to cover would be:

\begin{aligned}x &= \left(\frac{u + v}{2}\right)\, t \\ &= \frac{0\; {\rm ft \cdot s^{-1}} + 264\; {\rm ft \cdot s^{-1}}}{2} \times 65.0\; {\rm s} \\ &= 8.58\times 10^{3}\; {\rm ft}\end{aligned}.

4 0
2 years ago
Which particle is most likely to interact with your hand? Select one: a. Alpha particle b. Beta particle c. Gamma particle d. Ne
IgorLugansk [536]

Answer:

The correct option is a

Explanation:

The alpha particle has the lowest penetrating power of the trio of alpha, beta and gamma particles and can be stopped by a sheet of paper and hence cannot penetrate a human skin. Beta particle has a higher penetrating power than alpha particle (some of it penetrates the human skin and some do not) while the gamma particle has the highest penetrating power (with all of it penetrating the human skin).

From the above description, it can be deduced that the alpha particle will stay and interact with the hand (because of its low penetrating power) as the remaining particles move through the skin.

4 0
3 years ago
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