A proton is released such that its initical velocity is from right to left across this page. THe proton's path, however, is defl
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Answer:
Action force: Would be the force of your feet against the Earth given by the weight defined as:
Where g is a constat who represent the gravity in Earth
Reaction force: Would be the force of the Earth pushing against your feet. And on this case is represented by the normal force defined as:
Where represent the friction coefficient between the ground and the object.
And the friction force.
If we don't have any other forces involved in the y axis we can conclude that:
And as we can see we have that Action force = Reaction force
So then the third Law of Newton is satisfied.
Explanation:
For this case we have this:
Action force: Would be the force of your feet against the Earth given by the weight defined as:
Where g is a constat who represent the gravity in Earth
Reaction force: Would be the force of the Earth pushing against your feet. And on this case is represented by the normal force defined as:
Where represent the friction coefficient between the ground and the object.
And the friction force.
If we don't have any other forces involved in the y axis we can conclude that:
And as we can see we have that Action force = Reaction force
So then the third Law of Newton is satisfied.
<h2>Answer:</h2>
The rate of deceleration is -0.14
Using one of the equations of motion;
v = u + at
where;
v = final velocity of the boat = 0m/s (since the boat decelerates to a stop)
u = initial velocity of the boat = 25m/s
a = acceleration of the boat
t = time taken for the boat to accelerate/decelerate from u to v = 3 minutes
<em>Convert the time t = 3 minutes to seconds;</em>
=> 3 minutes = 3 x 60 seconds = 180seconds.
<em>Substitute the values of v, u, t into the equation above. We have;</em>
v = u + at
=> 0 = 25 + a(180)
=> 0 = 25 + 180a
<em>Make a the subject of the formula;</em>
=> 180a = 0 - 25
=> 180a = -25
=> a = -25/180
=> a = -0.14
The negative value of a shows that the boat is decelerating.
Therefore, the rate of deceleration of the speed boat is 0.14