Answer:
dog sleding ----> they do the work
Explanation:
First make sure you draw a force diagram. You should have Fn going up, Fg going down, Ff going left and another Fn going diagonally down to the right. The angle of the diagonal Fn (we'll call it Fn2) is 35° and Fn2 itself is 80N. Fn2 can be divided into two forces: Fn2x which is horizontal, and Fn2y which is vertical. Right now we only care about Fn2y.
To solve for Fn2y we use what we're given and some trig. Drawing out the actual force of Fn2 along with Fn2x and Fn2y we can see it makes a right triangle, with 80 as the hypotenuse. We want to solve for Fn2y which is the opposite side, so Sin(35)=y/80. Fn2y= 80sin35 = 45.89N
Next we solve for Fg. To do this we use Fg= 9.8 * m. Mass = 30kg, so Fg = 9.8 * 30 = 294N.
Since the chair isn't moving up or down, we can set our equation equal to zero. The net force equation in the vertical direction will be Fn + Fn2y -Fg = 0. If we plug in what we know, we get Fn + 45.89 -294 = 0. Then solve this algebraically.
Fn +45.89 -294 = 0
Fn +45.89 = 294
Fn = 248.11 N
You'll get a more accurate answer if you don't round Fn2y when solving for it, it would be something along the lines of 45.88611 etc
Let <em>F₁ </em>and <em>F₂</em> denote the two forces, and <em>R</em> the resultant force.
<em>F₁ </em>and <em>F₂</em> point perpendicularly to one another, so their dot product is
<em>F₁ </em>• <em>F₂</em> = 0
<em />
We're given that one of these vectors, say <em>F₁</em>, makes an angle with <em>R</em> of 30°, so that
<em>F₁</em> • <em>R</em> = ||<em>F₁</em>|| ||<em>R</em>|| cos(30°)
But we also have
<em>F₁</em> • <em>R</em> = <em>F₁ </em>• (<em>F₁ </em>+ <em>F₂</em>) = (<em>F₁ </em>• <em>F₁</em>) + (<em>F₁ </em>• <em>F₂</em>) = <em>F₁ </em>• <em>F₁ </em>=<em> </em>||<em>F₁</em>||²
So, knowing that ||<em>R</em>|| = 100 N, we get that
(100 N) ||<em>F₁</em>|| cos(30°) = ||<em>F₁</em>||²
(100 N) cos(30°) = ||<em>F₁</em>||
||<em>F₁</em>|| ≈ 86.6 N
(And the same would be true for <em>F₂</em>.)
Answer:
The shortest transverse distance between a maximum and a minimum of the wave is 0.1638 m.
Explanation:
Given that,
Amplitude = 0.08190 m
Frequency = 2.29 Hz
Wavelength = 1.87 m
(a). We need to calculate the shortest transverse distance between a maximum and a minimum of the wave
Using formula of distance
Where, d = distance
A = amplitude
Put the value into the formula
Hence, The shortest transverse distance between a maximum and a minimum of the wave is 0.1638 m.