Answer:
<h2>18150 J</h2>Explanation:
The kinetic energy of the car can be found by using the formula
m is the Mass
v is the velocity
From the question we have
We have the final answer as
<h3>18150 J</h3>Hope this helps you
Why are simple machines important?
2 answers:
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a) 4 forces
b) 186 N
c) 246 N
Explanation:
a)
Let's count the forces acting on the bicylist:
1) Weight (): this is the gravitational force exerted on the bicyclist by the Earth, which pulls the bicyclist towards the Earth's centre; so, this force acts downward (m = mass of the bicyclist, g = acceleration due to gravity)
2) Normal reaction (N): this is the reaction force exerted by the road on the bicyclist. This force acts vertically upward, and it balances the weight, so its magnitude is equal to the weight of the bicyclist, and its direction is opposite
3) Applied force (): this is the force exerted by the bicylicist to push the bike forward. Its direction is forward
4) Air drag (): this is the force exerted by the air on the bicyclist and resisting the motion of the bike; its direction is opposite to the motion of the bike, so it is in the backward direction
So, we have 4 forces in total.
b)
Here we can find the net force on the bicyclist by using Newton's second law of motion, which states that the net force acting on a body is equal to the product between the mass of the body and its acceleration:
where
is the net force
m is the mass of the body
a is its acceleration
In this problem we have:
m = 60 kg is the mass of the bicyclist
is its acceleration
Substituting, we find the net force on the bicyclist:
c)
We can write the net force acting on the bicyclist in the horizontal direction as the resultant of the two forces acting along this direction, so:
where:
is the net force
is the applied force (forward)
is the air drag (backward)
In this problem we have:
is the net force (found in part b)
is the magnitude of the air drag
Solving for , we find the force produced by the bicyclist while pedaling:
Missing question in the text:
"A.What are the magnitude and direction of the electric field at the point in question?
"A.What are the magnitude and direction of the electric field at the point in question?
B.<span>What would be the magnitude and direction of the force acting on a proton placed at this same point in the electric field?"</span>
<span>Solution:
A) A charge q </span>under an electric field of intensity E will experience a force F equal to:
In our problem we have and
, so we can find the magnitude of the electric field:
The charge is negative, therefore it moves against the direction of the field lines. If the force is pushing down the charge, then the electric field lines go upward.
B) The proton charge is equal to
Therefore, the magnitude of the force acting on the proton will be
And since the proton has positive charge, the verse of the force is the same as the verse of the field, so upward.