Answer:
V = ∫∫∫rdrdθdz integrating from z = 2 to z = 4, r = 0 to √(16 - z²) and θ = 0 to 2π
Step-by-step explanation:
Since we have the radius of the sphere R = 4, we have R² = r² + z² where r = radius of cylinder in z-plane and z = height² of cylinder.
So, r = √(R² - z²)
r = √(4² - z²)
r = √(16 - z²)
Since the region is above the plane z = 2, we integrate z from z = 2 to z = R = 4
Our volume integral in cylindrical coordinates is thus
V = ∫∫∫rdrdθdz integrating from z = 2 to z = 4, r = 0 to √(16 - z²) and θ = 0 to 2π
5)
a. The equation that describes the forces which act in the x-direction:
<span> Fx = 200 * cos 30 </span>
<span>
b. The equation which describes the forces which act in the y-direction: </span>
<span> Fy = 200 * sin 30 </span>
<span>c. The x and y components of the force of tension: </span>
<span> Tx = Fx = 200 * cos 30 </span>
<span> Ty = Fy = 200 * sin 30 </span>
d.<span>Since desk does not budge, </span><span>frictional force = Fx
= 200 * cos 30 </span>
<span> Normal force </span><span>= 50 * g - Fy
= 50 g - 200 * sin 30
</span>____________________________________________________________
6)<span> Let F_net = 0</span>
a. The equation that describes the forces which act in the x-direction:
(200N)cos(30) - F_s = 0
b. The equation that describes the forces which act in the y-direction:
F_N - (200N)sin(30) - mg = 0
c. The values of friction and normal forces will be:
Friction force= (200N)cos(30),
The Normal force is not 490N in either case...
Case 1 (pulling up)
F_N = mg - (200N)sin(30) = 50g - 100N = 390N
Case 2 (pushing down)
F_N = mg + (200N)sin(30) = 50g + 100N = 590N
Answer:
82+25=107 yan po ang answer
Answer: He got to be elected president of the Republic of Texas.
