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: 0.0228
Step-by-step explanation:
Given : The mean and the standard deviation of finish times (in minutes) for this event are respectively as :-
If the distribution of finish times is approximately bell-shaped and symmetric, then it must be normally distributed.
Let X be the random variable that represents the finish times for this event.
z score : 
Now, the probability of runners who finish in under 19 minutes by using standard normal distribution table :-
Hence, the approximate proportion of runners who finish in under 19 minutes = 0.0228
Answer:
You have to have the graph to answer it.
Step-by-step explanation:
Domain: (infinity, 4]
range: [-6, infinity)
