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:
![r = \sqrt[3]{\dfrac{3V}{4 \pi}}](https://tex.z-dn.net/?f=%20r%20%3D%20%5Csqrt%5B3%5D%7B%5Cdfrac%7B3V%7D%7B4%20%5Cpi%7D%7D%20)
Step-by-step explanation:




![r = \sqrt[3]{\dfrac{3V}{4 \pi}}](https://tex.z-dn.net/?f=%20r%20%3D%20%5Csqrt%5B3%5D%7B%5Cdfrac%7B3V%7D%7B4%20%5Cpi%7D%7D%20)
The area of the triangle is 4 units²
<h3>What is the area of a triangle?</h3>
The area of a triangle is the half the base multiplied with the height of that triangle
Area = 1/ 2 × b × h
From the figure given,
base = 7 - 3 = 4 units
height = 4 - 2 = 2 units
Area = 1/ 2 × 4 × 2
Area = 1/ 2 × 8
Area = 4 units²
Thus, the area of the triangle is 4 units²
Learn more about a triangle here:
brainly.com/question/1475130
#SPJ1
Answer:
C (2, -5/2)
Step-by-step explanation:
To find the midpoint of two points
mid = (x1+x2)/2, (y1+y2)/2
= (2+2)/2, (4+-9)/2
= 4/2, -5/2
=2,-5/2