Answer:
4.9 m/s²
Explanation:
Draw a free body diagram. There are two forces on the object:
Weight force mg pulling straight down,
and normal force N pushing perpendicular to the plane.
Sum the forces in the parallel direction.
∑F = ma
mg sin θ = ma
a = g sin θ
a = (9.8 m/s²) (sin 30°)
a = 4.9 m/s²
Answer:
The tank is losing
Explanation:
According to the Bernoulli’s equation:
We are being informed that both the tank and the hole is being exposed to air :
∴ P₁ = P₂
Also as the tank is voluminous ; we take the initial volume ≅ 0 ;
then can be determined as:
h₁ = 5 + 15 = 20 m;
h₂ = 15 m
as it leaves the hole at the base.
radius r = d/2 = 4/2 = 2.0 mm
(a) From the law of continuity; its equation can be expressed as:
J =
J = πr²
J =
J =
b)
How fast is the water from the hole moving just as it reaches the ground?
In order to determine that; we use the relation of the velocity from the equation of motion which says:
v² = u² + 2gh ₂
v² = 9.9² + 2×9.81×15
v² = 392.31
The velocity of how fast the water from the hole is moving just as it reaches the ground is :
Answer:
The work is -67.76 J
Explanation:
The law of conservation of energy is considered one of one of the fundamental laws of physics and states that the total energy of an isolated system remains constant. except when it is transformed into other types of energy.
This is summed up in the principle that energy can neither be created nor destroyed in the universe, only transformed into other forms of energy.
In this case you must calculate the loss of kinetic energy. This loss is actually the work done against the resistive force in the air. Friction is the only force other than gravity that acts on the ball.
So, the loss of kinetic energy is
You know:
Final velocity of the ball:
Replacing:
= -67.76 J
Friction work is always negative because friction is always against displacement.
<u><em>The work is -67.76 J</em></u>