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Here's what I got to question A and B
a) Fs(max)=(0.560)(88.9N)=49.8N Fy=88.9N-49.8N=39.1N Fx=(39.1N)/(cos(40))=51.0N sqrt{39.1^2+51.0^2}=F
F=64.3N
b) 88.9N +49.8 N=138.7N=Fy Fx=(138.7N)/(cos(40))=181.06N sqrt{138.7^2+181.06^2}=F
F=228N
The answer:
the full question is as follow:
<span>A Texas rancher wants to fence off his four-sided plot of flat land. He measures the first three sides, shown as A, B, and C in Figure below , where A = 4.90 km and θC = 15°. He then correctly calculates the length and orientation of the fourth side D. What is the magnitude and direction of vector D?
As shown in the figure,
A + B + C + D = 0, so to find the </span>magnitude and direction of vector D, we should follow the following method:
D = 0 - (A + B + C) ,
let W = - (A + B + C), so the magnitude and direction of vector D is the same of the vector W characteristics
Magnitude
A + B + C = <span> (4.90cos7.5 - 2.48sin16 - 3.02cos15)I</span>
<span>+ (-4.9sin7.5 + 2.48cos16 + 3.02sin15)J
</span>= 1.25I +2.53J
the magnitude of W= abs value of (A + B + C) = sqrt (1.25² + 2.53²)
= 2.82
the direction of D can be found by using Dx and Dy value
we know that tan<span>θo = Dx / Dy = 1.25 / 2.53 =0.49
</span>tanθo =0.49 it implies θo = arctan 0.49 = 26.02°
direction is 26.02°
Answer:
I =
(K+5)
Explanation:
Given :
J = k+5
Now selecting a thin ring in the wire of radius "r" and thickness dr.
Current through the thin ring is
dI = J X 2πrdr
dI = (K+5) x 2πrdr
Now integrating we get
I = 
I = (K+5) 2π
I = (K+5) 2π 
I =
(K+5)
Answer:
reduce passive stiffness and increase range of movement during exercise.
Explanation:
stretching performed as part of a warm up prior to exercise is thought to reduce passive stiffness and increase range of movement during exercise. in general it appears that is static stretching is most beneficial for athletes requiring flexibility for their sports.
<span>d
The mass is doubled which means that both the momentum and kinetic energy are also doubled. Also the normal force that's acting along with the coefficient of kinetic friction is also doubled. So the friction that's working to slow down the crate is doubled. So the crate will have double the kinetic energy that needs to be dissipated, but the rate of dissipation is also doubled, so the total time required to dissipate the kinetic energy is the same. And since both crates start out with the same velocity and since they'll lose energy (and velocity) at the same proportional rate, they'll take the same distance to slide to a stop.</span>