All categories

3 years ago

A porsche 911 ca go from 0 mph to 60 mph in 3.4secs whats ur acceleration

Physics

1 answer:

lawyer [7]3 years ago

8 0

The acceleration is $7.9 m/s^2$

Explanation:

The acceleration of a body is given by

$a=\frac{v-u}{t}$

where

a is the acceleration

v is the final velocity

u is the initial velocity

t is the time taken for the velocity to change from u to v

Here we first have to convert the velocities into SI units. Keeping in mind that

1 mile = 1609 m

1 h = 3600 s

We have:

$u=0 mph = 0 m/s$

$v=60 mph \cdot \frac{1609 m/mi}{3600 s/h}=26.8 m/s$

And the time interval is

t = 3.4 s

Therefore the acceleration is

$a=\frac{26.8-0}{3.4}=7.9 m/s^2$

Learn more about acceleration:

brainly.com/question/9527152

brainly.com/question/11181826

brainly.com/question/2506873

brainly.com/question/2562700

#LearnwithBrainly

You might be interested in

In some states, the highway patrol uses planes carrying radar devices to monitor traffic. As the plane flies over traffic, the r

amid [387]

Speed/ velocity/ how fast the car is moving!
Hope this assists you!

5 0

3 years ago

Read 2 more answers

A drag racer starts from rest and accelerates at 7.4 m/s2. How far will he travel in 2.0 seconds?

Leto [7]

Using the kinematic equation below we can determine the distance traveled if t=2, a=7.4m/s^2. First we must determine the final velocity:

$v_{final}=v_{initial}+\frac{1}{2}at\\\\v_{final}=0+(7.4m/s^2)(2s)=34.8m/s$

Now we will determine the distance traveled:

$v_{final}^2=v_{initial}^2+2a \Delta x\\\\\Delta x = \frac{v_{final}^2}{2a} =\frac{(34.8)^2}{(2)(7.4)}=81.83 m$

Therefore, the drag racer traveled 81.83 meters in 2 seconds.

3 0

3 years ago

What is the ratio of escape speed from earth to circular orbital speed? ignore air resistance.

klio [65]

About 40 000 km/h
Here you go:

8 0

3 years ago

Read 2 more answers

Firecracker A is 300 m from you. Firecracker B is 600 m from you in the same direction. You see both explode at the same time. D

ss7ja [257]

Answer:

e see that the distances are different, the only way that the two beams of light approach simultaneously is that event 2 (farthest) occurs first than event 1

Explanation:

This is an ejercise in special relativity, where the speed of light is constant.

Let's carefully analyze the approach, we see the two events at the same time.

The closest event time is

c = (x₁-300) / t

t = (x₁-300) / c

The time for the other event is

t = (x₂- 600) / c

since they tell us that we see the events simultaneously, we can equalize

(x₁ -300) / c = (x₂ -600) / c

x₁ = x₂ - 300

We see that the distances are different, the only way that the two beams of light approach simultaneously is that event 2 (farthest) occurs first than event 1

3 0

3 years ago

Suppose the microphone on an AM radio is not working properly. Which best explains how this will affect the process of amplitude

Anna007 [38]

You can have it your own definition of

3 0

3 years ago

Read 2 more answers