Answer:
Approximately
. (Assuming that
, and that the tabletop is level.)
Explanation:
Weight of the book:
.
If the tabletop is level, the normal force on the book will be equal (in magnitude) to weight of the book. Hence,
.
As a side note, the
and
on this book are not equal- these two forces are equal in size but point in the opposite directions.
When the book is moving, the friction
on it will be equal to
, the coefficient of kinetic friction, times
, the normal force that's acting on it.
That is:
.
Friction acts in the opposite direction of the object's motion. The friction here should act in the opposite direction of that
applied force. The net force on the book shall be:
.
Apply Newton's Second Law to find the acceleration of this book:
.
<span>A free body diagram is a representation of how the forces that are acting on a point or particle interact. You place your point at the origin and then draw your forces.
</span>
Answer:
81.3ohms
Explanation:
Resistance is known to provide opposition to the flow of electric current in an electric circuit.
Power dissipated by the computer is expressed as;
Power = current (I) × Voltage(V)
P = IV... (1)
Note that from ohms law, V = IR
I = V/R ... (2)
Substituting equation 2 into 1, we will have;
P = (V/R)×V
P = V²/R.. (3)
Given source voltage = 100V, Power dissipated = 123W
To get resistance R of the computer, we will substitute the given value into equation 3 to have
123 = 100²/R
R = 100²/123
R = 10,000/123
R = 81.3ohms
The resistance of the computer is 81.3ohms
Momentum should be conserved. The momentum of both
objects must balance with their initial and final momentum.
Let m1 and v1 be the mass and velocity of the
bowling ball
And m2 and v2 be the mass and velocity of the
bowling pin
(m1v1)i + (m2v2)i = (m1v1)f + (m2v2)f
30 kg m/s + (1.5 kg)(0 m/s) = 13kg m/s + 1.5v2f
V2f = 11.33 m/s
<span>So the momentum = 1.5 kg(11.33 m/s) = 17 kg m/s</span>
Answer:
We live in a very adaptable environment. If we cannot adapt to the environment, we will not be able to survive.
Explanation: