Answer:
1.63 N
Explanation:
F = GMm/r^2
= (6.67x10^-11)(10x10^5)(3x10^5) / 3.5^2
= 1.63 N ( 3 sig. fig.)
So, the angular frequency of the blades approximately <u>36.43π rad/s</u>.
<h3>Introduction</h3>
Hi ! Here I will discuss about the angular frequency or what is also often called the angular velocity because it has the same unit dimensions. <u>Angular frequency occurs, when an object vibrates (either moving harmoniously / oscillating or moving in a circle)</u>. Angular frequency can be roughly interpreted as the magnitude of the change in angle (in units of rad) per unit time. So, based on this understanding, the angular frequency can be calculated using the equation :
With the following condition :
= angular frequency (rad/s)
= change of angle value (rad)- t = interval of the time (s)
<h3>Problem Solving</h3>
We know that :
- = change of angle value = 1,000 revolution = 1,000 × 2π rad = 2,000π rad/s >> Remember 1 rev = 2π rad/s.
- t = interval of the time = 54.9 s.
What was asked :
- = angular frequency = ... rad/s
Step by step :
<h3>Conclusion :</h3>
So, the angular frequency of the blades approximately 36.43π rad/s.
Kinetic energy is the energy possessed by an object when that object is moving in space. The higher the mass of an object or higher the speed of an object the higher the kinetic energy will be.
So to calculate the Kinetic Energy we can use the following formula
K.E=(1/2)*m*v^2
Inserting the values in formula gives:
K.E=1/2*7.26*2^2
14.52J
This is the final answer which gives the kinetic energy of the ball.
The correct answer is the reverse wave I took the test
Answer:
18.1 × 10⁻⁶ A = 18.1 μA
Explanation:
The current I in the wire is I = ∫∫J(r)rdrdθ
Since J(r) = Br, in the cylindrical wire. With width of 10.0 μm, dr = 10.0 μm. r = 1.20 mm. We have a differential current dI. We integrate first by integrating dθ from θ = 0 to θ = 2π.
So, dI = J(r)rdrdθ
dI/dr = ∫J(r)rdθ = ∫Br²dθ = Br²∫dθ = 2πBr²
Now I = (dI/dr)dr at r = 1.20 mm = 1.20 × 10⁻³ m and dr = 10.0 μm = 0.010 mm = 0.010 × 10⁻³ m
I = (2πBr²)dr = 2π × 2.00 × 10⁵ A/m³ × (1.20 × 10⁻³ m)² × 0.010 × 10⁻³ m = 0.181 × 10⁻⁴ A = 18.1 × 10⁻⁶ A = 18.1 μA
