Answer:
e. Object X has traveled four times as far as object Y.
Explanation:
The distance covered by an object in uniform accelerated motion is given by:

where
u is the initial velocity
t is the time
a is the acceleration
The two objects in the problem have same initial velocity, u = 0 (since they start from rest), so we can rewrite the equation as

We see that the distance covered is proportional to the square of the time. In this problem, the two objects X and Y have same acceleration, but object X accelerates for twice the time: since
, this means that the distance covered by X will be
times higher that the distance covered by object Y.
mass (m) = 8kg
velocity (v) = 3 m/s²
Ke = ½mv²
ke = ½ mv²
½ × 8 × 3²
½ × 24²
12
Answer:
A . Are the masses repelling or attracting
answer:
<em><u>So, positive energy densities ("masses") attract each other by gravitational interaction. This is the general idea.</u></em>
<em><u>B</u></em>. What is the magnitude of the electrical force between the objects<em><u>?</u></em>
<em><u>answer</u></em><em><u>:</u></em>
<em><u> </u></em><em><u>The magnitude of the electrostatic force F between two point charges q1 and q2 is directly proportional to the product of the magnitudes of charges and inversely proportional to the square of the distance between them.</u></em>
C. What would the magnitude of the force be if one of the charges was 1/3 the amount?
answer:
<em><u>0.45</u></em>
D. ) What would the force be if the distance between the charges was only 1 meter (with the original charges)?
answer;
<em><u>Fmin=2.3×10^−28</u></em><em><u>N</u></em>
<em><u>E</u></em><em><u>:</u></em><em><u> </u></em>What would the force be if one of the charges was 1/3 the amount AND the distance was 1 meter?
answer;
<em><u>Electrostatic force is directly related to the charge of each object. So if the charge of one object is doubled, then the force will become two times greater.</u></em>
<em><u>#</u></em><em><u>CARRYONLEARNING</u></em><em><u>:</u></em><em><u>)</u></em>
<em><u>LOVEUALL</u></em>
Answer:
a) t = 19.6 s, b) fr = 1.274 10⁴ N
Explanation:
This is a Newton's second law problem
Y Axis
for the cabin
N₁-W₁ = 0
N₁ = W₁
for the trailer
N₂- W₂ = 0
N₂ = W₂
X axis
for the cabin plus trailer, where friction is only in the cabin
fr = (m₁ + m₂) a
the friction force equation is
fr = μ N
we substitute
μ N₁ = (m₁ + m₂) a
μ m₁ g = (m₁ + m₂) a
a = μ g 
let's calculate
a = 0.65 9.8
a = 1,274 m / s²
a) to find the stopping distance we can use kinematics
Let's slow down the sI system
v₀ = 90 km / h (1000 m / 1km) (1h / 3600s) = 25 m / s
v = v₀ - a t
when it is stopped its speed is zero
0 = v₀ - at
t = v₀ / a
t = 25 / 1.274
t = 19.6 s
b) the friction force is
fr = 0.65 2000 9.8
fr = 1.274 10⁴ N
This is the braking force and also the forces that couple the cars.
Glaciers cuase they was there now its not and a glacier carved the great lakes in the us