Complete question
A 2700 kg car accelerates from rest under the action of two forces. one is a forward force of 1157 newtons provided by traction between the wheels and the road. the other is a 902 newton resistive force due to various frictional forces. how far must the car travel for its speed to reach 3.6 meters per second? answer in units of meters.
Answer:
The car must travel 68.94 meters.
Explanation:
First, we are going to find the acceleration of the car using Newton's second Law:
(1)
with m the mass , a the acceleration and
the net force forces that is:
(2)
with F the force provided by traction and f the resistive force:
(2) on (1):

solving for a:

Now let's use the Galileo’s kinematic equation
(3)
With Vo te initial velocity that's zero because it started from rest, Vf the final velocity (3.6) and
the time took to achieve that velocity, solving (3) for
:


Answer:
Explanation:
Givens
vi = 10 m/s
a = 1.5 m/s^2
d = 600 m
vf = ?
Formula
vf^2 = vi^2 + 2*a*d
Solution
vf^2 = 10^2 + 2*1.5 * 600
vf^2 = 100 + 1800
vf^2 = 1900
sqrt(vf^2) = sqrt(1900)
vf = 43.59 m/s
Im not so sure but it should be the
instantaneous speed
Answer:
You know the saying "Opposites attract" well that is how you can remember that South and North Magnetic Poles connect.
Explanation:
Hope this helps ya
1. electrons
2. positive to negative
3. insulator
4. TRUE
5. closed circuit
6. TRUE
7. series
8. TRUE
9. v=ir
10. TRUE
Hope this helps! :)