<span> Beryllium has an exclusive </span>+2<span> oxidation state in all of its compounds</span>
Answer:
r = 2161.9 m
Explanation:
Aerodynamic lift(L) is perpendicular to the wing, which is tilted 40 degrees to the horizontal.
Since the plane is moving in a horizontal circle, the vertical component of the lift must cancel the weight W of the airplane, but the horizontal component is the centripetal force that keeps it in a circle.
L is perpendicular to wing at angle θ with respect to horizontal
Thus,
Vertical component of lift is:
L cosθ = W = mg
Thus, m = L cosθ / g - - - - (eq1)
Horizontal component of lift is:
L sinθ = centripetal force = mv² / r - - - - (eq2)
Combining equations 1 and 2,we have;
L sinθ = (L cosθ / g)(v² / r)
L cancels out on both sides to give;
tanθ = v²/ rg
r = v² / (g tanθ)
We are given;
velocity; v = 480 km/hr = 480 x 10/36 = 133.33 m/s
r = 133.33²/[(9.8) tan(40)] = 2161.9 m
If we have the angle and magnitude of a vector A we can find its Cartesian components using the following formula
Where | A | is the magnitude of the vector and 
is the angle that it forms with the x axis in the opposite direction to the hands of the clock.
In this problem we know the value of Ax and Ay and we need the angle .
Vector A is in the 4th quadrant
So:
So:
So:
= -47.28 ° +360° = 313 °
= 313 °
Option 4.
Hello There!
Well, technically no. It has to be one or the other, it can't be both.
Hope This Helps You!
Good Luck :)
- Hannah ❤
Given Information:
Current = I = 20 A
Diameter = d = 0.205 cm = 0.00205 m
Length of wire = L = 1 m
Required Information:
Energy produced = P = ?
Answer:
P = 2.03 J/s
Explanation:
We know that power required in a wire is
P = I²R
and R = ρL/A
Where ρ is the resistivity of the copper wire 1.68x10⁻⁸ Ω.m
L is the length of the wire and A is the area of the cross-section and is given by
A = πr²
A = π(d/2)²
A = π(0.00205/2)²
A = 3.3x10⁻⁶ m²
R = ρL/A
R = 1.68x10⁻⁸*(1)/3.3x10⁻⁶
R = 5.09x10⁻³ Ω
P = I²R
P = (20)²*5.09x10⁻³
P = 2.03 Watts or P = 2.03 J/s
Therefore, 2.03 J/s of energy is produced in 1.00 m of 12-gauge copper wire carrying a current of 20 A
