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:
4/10=1
Step-by-step explanation:
Divide both fractions by 10 and you get 4/10=1
Answer:
Rachel
Step-by-step explanation:
We need to measure how far (towards the left) are the students from the mean in<em> “standard deviations units”</em>.
That is to say, if t is the time the student ran the mile and s is the standard deviation of the class, we must find an x such that
mean - x*s = t
For Rachel we have
11 - x*3 = 8, so x = 1.
Rachel is <em>1 standard deviation far (to the left) from the mean</em> of her class
For Kenji we have
9 - x*2 = 8.5, so x = 0.25
Kenji is <em>0.25 standard deviations far (to the left) from the mean</em> of his class
For Nedda we have
7 - x*4 = 8, so x = 0.25
Nedda is also 0.25 standard deviations far (to the left) from the mean of his class.
As Rachel is the farthest from the mean of her class in term of standard deviations, Rachel is the fastest runner with respect to her class.
