2. A 5.2 x 10 kg car accelerates from rest under the

actions of two forces. One is a forward force of 3600

N provided by traction between the wheels and the

road. The other is 3350 N resistive force due to

various frictional forces. Determine how far the car

must travel for its speed to reach 7 m/s.

5.095m

Step-by-step explanation:

We first calculate the net force acting on the car

FN = Forward Force - Resistive Force

FN = 3 600 - 3350 = 250N

ACCORDING TO NEWTON'S SECOND LAW ACCELERATION IS GIVEN BY:

a = Fnet/m

=250N/5.2*10kg = 4.8077m/s^2

Since V = at

t = v/a 1

Equation of motion for the car;

S =ut + 1/2at^2

Where u = initial velocity for the car is zero

U=0

S=1/2at^2 2

Substitute to value of t in equation 1 into equation 2

S = 1/2 a(v/a)^2 = v^2/2a

= 7^2/2*4.8077=49/9.6154 = 5.095m

The car will travel for 5.095m

this shows a reflection over the x-axis and is translated one unit down and 6 units to the right (x + 6, y - 1)

hope this !

~just a girl in love with shawn mendes

there are 70 more legs than eyes at the dog show (196 – 126 = 70). dogs have two more legs than eyes, and humans have the same number of legs as eyes. thus 70 is twice the number of dogs at the show, so the number of dogs is half of 70, which is 35. to get the number of humans, subtract 35 from 63: 63 – 35 = 28.

step-by-step explanation: