Answer:
a) correct answer is C
, b) 14º from the west to the north, c) v_{1g} = 300.79 km / h
Explanation:
This is a relative speed exercise using the addition of speeds.
1) when it is not specified regarding what is being measured, the medicine is carried out with respect to the Z Earth, therefore the correct answer is C
2 and 3) In this case we must compose the speed using the Pythagorean Theorem.
² =
² +
²
where v_{1a} is the speed of the airplane with respect to the air, v_{1g} airplane speed with respect to the Earth, v_{ag} air speed with respect to the Earth
in this case let's clear the speed of the airplane with respect to the Earth
v_{1g} = √(v_{1a}² - v_{ag}²)
v_{1g} = √ (310² - 75²)
v_{1g} = 300.79 km / h
we find the direction of the airplane using trigonometry
sin θ = v_{ag} / v_{1a}
θ = sin⁻¹ (v_{ag} /v_{1a})
θ = sin⁻¹ (75/310)
θ= 14º
the pilot must direct the aircraft at an angle of 14º from the west to the north
Answer:
Torque = –207.4 Nm
Explanation:
Given M = 3.2kg, r = 5.4m, α = –12rad/s² (it is slowing down)
Torque = I × α
α = angular acceleration
I = moment of inertia
I = MR² for a circular hoop
Torque = 3.2×5.4×(– 12)
Torque = –207.4 Nm
Answer:
Explanation:
We Often solve the the integral neutron transport equation using the collision probability (CP) method which usually requires flat flux (FF) approach. In this research, it has been carried out in the cylindrical nuclear fuel cell with the spatial of mesh with quadratic flux approach. This simply means that the neutron flux at any region of the nuclear fuel cell is forced to follow the pattern of a quadratic function.
Furthermore The mechanism may be referred to as the process of non-flat flux (NFF) approach. The parameters that calculated in this study are the k-eff and the distribution of neutron flux. The result shows that all parameters are in accordance with the result of SRAC.
Answer:
0.84
Explanation:
m = Massa balok
g = Percepatan gravitasi
= Sudut kemiringan
= Koefisien gesekan statik antara balok dan bidang miring
Gaya balok karena beratnya diberikan oleh
![F=mg\sin\theta](https://tex.z-dn.net/?f=F%3Dmg%5Csin%5Ctheta)
Gaya gesekan diberikan oleh
![f=\mu mg\cos\theta](https://tex.z-dn.net/?f=f%3D%5Cmu%20mg%5Ccos%5Ctheta)
Kondisi dimana balok mulai bergerak adalah ketika gaya balok akibat beratnya sama dengan gaya gesek pada balok.
![mg\sin\theta=\mu mg\cos\theta\\\Rightarrow \mu=\dfrac{mg\sin\theta}{mg\cos\theta}\\\Rightarrow \mu=\tan\theta\\\Rightarrow \mu=\tan40^{\circ}\\\Rightarrow \mu=0.84](https://tex.z-dn.net/?f=mg%5Csin%5Ctheta%3D%5Cmu%20mg%5Ccos%5Ctheta%5C%5C%5CRightarrow%20%5Cmu%3D%5Cdfrac%7Bmg%5Csin%5Ctheta%7D%7Bmg%5Ccos%5Ctheta%7D%5C%5C%5CRightarrow%20%5Cmu%3D%5Ctan%5Ctheta%5C%5C%5CRightarrow%20%5Cmu%3D%5Ctan40%5E%7B%5Ccirc%7D%5C%5C%5CRightarrow%20%5Cmu%3D0.84)
Koefisien gesekan statik antara balok dan bidang miring adalah 0.84.
The net force is 12 N to the left.