Answer:
It will take 15.55s for the police car to pass the SUV
Explanation:
We first have to establish that both the police car and the SUV will travel the same distance in the same amount of time. The police car is moving at constant velocity and the SUV is experiencing a deceleration. Thus we will use two distance fromulas (for constant and accelerated motions) with the same variable for t and x:
1. 
2. 
Since both cars will travel the same distance x, we can equal both formulas and solve for t:
We simplify the fraction present and rearrange for our formula so that it equals 0:
In the very last step we factored a common factor t. There is two possible solutions to the equation at 
and:
What this means is that during the displacement of the police car and SUV, there will be two moments in time where they will be next to each other; at 
(when the SUV passed the police car) and 
(when the police car catches up to the SUV)
Work needed: 720 J
Explanation:
The work needed to stretch a spring is equal to the elastic potential energy stored in the spring when it is stretched, which is given by
where
k is the spring constant
x is the stretching of the spring from the equilibrium position
In this problem, we have
E = 90 J (work done to stretch the spring)
x = 0.2 m (stretching)
Therefore, the spring constant is
Now we can find what is the work done to stretch the spring by an additional 0.4 m, that means to a total displacement of
x = 0.2 + 0.4 = 0.6 m
Substituting,
Therefore, the additional work needed is
Learn more about work:
brainly.com/question/6763771
brainly.com/question/6443626
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It’s will be B because the circuit had a open or close so if that doesn’t work than it’s because it’s open
The acceleration of the crate after it begins to move is 0.5 m/s²
We'll begin by calculating the the frictional force
Mass (m) = 50 Kg
Coefficient of kinetic friction (μ) = 0.15
Acceleration due to gravity (g) = 10 m/s²
Normal reaction (N) = mg = 50 × 10 = 500 N
<h3>Frictional force (Fբ) =?</h3>
Fբ = μN
Fբ = 0.15 × 500
<h3>Fբ = 75 N</h3>
- Next, we shall determine the net force acting on the crate
Frictional force (Fբ) = 75 N
Force (F) = 100 N
<h3>Net force (Fₙ) =?</h3>
Fₙ = F – Fբ
Fₙ = 100 – 75
<h3>Fₙ = 25 N</h3>
- Finally, we shall determine the acceleration of the crate
Mass (m) = 50 Kg
Net force (Fₙ) = 25 N
<h3>Acceleration (a) =?</h3>
a = Fₙ / m
a = 25 / 50
<h3>a = 0.5 m/s²</h3>
Therefore, the acceleration of the crate is 0.5 m/s²
Learn more on friction: brainly.com/question/364384
<h2>Answer</h2>
option D)
2.4 seconds
<h2>Explanation</h2>
Given in the question,
mass of car = 1200kg
speed of car = 19m/s
Force due to direction of travel
F = ma
= 12000(a)
Force to due frictional force in reverse direction
-F = mg(friction coefficient)
= -12000(9.81)(0.8)
<h2>
-mg(friction coefficient) = ma </h2>
(cancelling mass from both side of equation)
g(0.8) = a
(9.81)(0.8) = a
a = 7.848 m/s²
<h2>Use Newton Law of motion</h2><h3>vf - vo = a • t</h3>
where vf = final velocity
vo = initial velocity
a = acceleration
t = time
0 - 19 = 7.8(t)
t = 19/7.8
= 2.436 s
≈ 2.4s
