Answer: The question is incomplete or some details are missing. Here is the complete question ; (a) The driver of a car slams on the brakes when he sees a tree blocking the road. The car slows uniformly with acceleration of −5.55 m/s2 for 4.05 s, making straight skid marks 63.0 m long, all the way to the tree. With what speed (in m/s) does the car then strike the tree? m/s
(b) What If? If the car has the same initial velocity, and if the driver slams on the brakes at the same distance from the tree, then what would the acceleration need to be (in m/s2) so that the car narrowly avoids a collision? m/s2
a ) With what speed (in m/s) does the car then strike the tree? m/s = 4.3125m/s
b) then what would the acceleration need to be (in m/s2) so that the car narrowly avoids a collision? m/s2 = -5.696m/s2
Explanation:
The detailed steps and calculation is as shown in the attached file.
The correct answer is
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force per unit charge.
In fact, the electric field strength is defined as the electric force per unit charge experienced by a positive test charge located in the electric field. In formula:
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where
E is the electric field strength
F is the electric force experienced by the charge
q is the positive test charge.
I believe it’s A, i could be wrong tho 3
A bowling ball because of the force it has one everyone on
The problem about describes a perfectly inelastic collision. We are tasked to find the initial velocity of an object having a mass of 6 kg moving due west. It is given in the problem that after collision the cart sticks together and it stops. Thus, the final mass is the sum of the two cart and the final velocity is zero. For a perfectly inelastic collision,
m1v1-m2v2=vf(m1+m2)
By Substitution,
3(4)-6(v2)=0
6v2=12
v2=2
Therefor, the initial velocity if a 6 kg cart is 2 m/s
