Answer:
toward the center
Explanation:
Before answering, let's remind the first two Newton Laws:
1) An object at rest tends to stay at rest and an object moving at constant velocity tends to continue its motion at constant velocity, unless acted upon a net force
2) An object acted upon a net force F experiences an acceleration a according to the equation
where m is the mass of the object.
In this problem, we have an object travelling at constant speed in a circular path. The fact that the trajectory of the object is circular means that the direction of motion of the object is constantly changing: this means that its velocity is changing, so it has an acceleration. And therefore, a net force is acting on it. The force that keeps the object travelling in the circular path is called centripetal force, and it is directed towards the center of the circle (because it prevents the object from continuing its motion straight away).
So, the correct answer is
toward the center
Answer:
V₁ = 1.75 m³
Explanation:
Assuming the gas to be an ideal gas. At constant temperature, the relationship between the volume and temperature of an ideal gas is given by Boyle's Law as follows:
where,
P₁ = Initial Pressure of the Gas = 4 KPa
V₁ = Initial Volume of the Gas = ?
P₂ = Final Pressure of the Gas = 2 KPa
V₂ = Final Volume of the Gas = 3.5 m³
Therefore,
<u>V₁ = 1.75 m³</u>
Answer:
the correct answer is reduce friction
Answer:
Revolutions made before attaining angular velocity of 30 rad/s:
θ = 3.92 revolutions
Explanation:
Given that:
L(final) = 10.7 kgm²/s
L(initial) = 0
time = 8s
<h3>
Find Torque:</h3>
Torque is the rate of change of angular momentum:
<h3>Find Angular Acceleration:</h3>
We know that
T = Iα
α = T/I
where I = moment of inertia = 2.2kgm²
α = 1.34/2.2
α = 0.61 rad/s²
<h3>
Find Time 't'</h3>
We know that angular equation of motion is:
ω²(final) = ω²(initial) +2αθ
(30 rad/s)² = 0 + 2(0.61 rad/s²)θ
θ = (30 rad/s)²/ 2(0.61 rad/s²)
θ = 24.6 radians
Convert it into revolutions:
θ = 24.6/ 2π
θ = 3.92 revolutions
The cross section is the little tiny circle you see when you cut a wire
and look at the flat, cut end.
The cross-sectional area of the wire is the area of that little circle.
It's equal to
Area = (pi) x (1/4) x (Diameter of the wire)²
