7,14,21,28,35,42,49,56,63,70,77,84,91,98,105,112,119,126
18,36,54,72,90,108,126
So the first person to get both of them is the 126th person
Answer:
325 and 375 mph
Step-by-step explanation:
Given :
Distance between plane A and B = 2800
Recall :
Distance = speed * time
Let speed of A = v1
Speed of B = v2
v1 = v2 + 50
Time taken = 4 hours
Distance = speed * time
Total distance = 2800
2800 = v1 * 4 + v2 * 4
2800 = (4v1) + (4v2)
2800 = 4(v2+50) + 4v2
2800 = 4v2 + 200 + 4v2
2800 = 8v2 + 200
2800 - 200 = 8v2
2600 = 8v2
v2 = 2600/8
v2 = 325 mph
v1 = v2 + 50
v1 = 325 + 50
v1 = 375 mph
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
