Answer:
The distance from Witless to Machmer is 438.63 m.
Explanation:
Given that,
Machmer Hall is 400 m North and 180 m West of Witless.
We need to calculate the distance
Using Pythagorean theorem

Where,
=distance of Machmer Hall
=distance of Witless
Put the value into the formula


Hence, The distance from Witless to Machmer is 438.63 m.
Answer:
Option B, Fix the piston in place so the volume of the pas remains constant
Explanation:
As we know

The effect on variable due to another variable can be studied by keeping the third variable constant.
Hence, in order the study the variation of temperature with pressure or vice versa, the volume needs to fixed at a certain value.
Hence, option B is correct
Answer:
a) L=0. b) L = 262 k ^ Kg m²/s and c) L = 1020.7 k^ kg m²/s
Explanation:
It is angular momentum given by
L = r x p
Bold are vectors; where L is the angular momentum, r the position of the particle and p its linear momentum
One of the easiest ways to make this vector product is with the use of determinants
![{array}\right] \left[\begin{array}{ccc}i&j&k\\x&y&z\\px&py&pz\end{array}\right]](https://tex.z-dn.net/?f=%7Barray%7D%5Cright%5D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Di%26j%26k%5C%5Cx%26y%26z%5C%5Cpx%26py%26pz%5Cend%7Barray%7D%5Cright%5D)
Let's apply this relationship to our case
Let's start by breaking down the speed
v₀ₓ = v₀ cosn 45
voy =v₀ sin 45
v₀ₓ = 9 cos 45
voy = 9 without 45
v₀ₓ = 6.36 m / s
voy = 6.36 m / s
a) at launch point r = 0 whereby L = 0
. b) let's find the position for maximum height, we can use kinematics, at this point the vertical speed is zero
vfy² = voy²- 2 g y
y = voy² / 2g
y = (6.36)²/2 9.8
y = 2.06 m
Let's calculate the angular momentum
L= ![\left[\begin{array}{ccc}i&j&k\\x&y&0\\px&0&0\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Di%26j%26k%5C%5Cx%26y%260%5C%5Cpx%260%260%5Cend%7Barray%7D%5Cright%5D)
L = -px y k ^
L = - (m vox) (2.06) k ^
L = - 20 6.36 2.06 k ^
L = 262 k ^ Kg m² / s
The angular momentum is on the z axis
c) At the point of impact, at this point the height is zero and the position on the x-axis is the range
R = vo² sin 2θ / g
R = 9² sin (2 45) /9.8
R = 8.26 m
L =
L = - x py k ^
L = - x m voy
L = - 8.26 20 6.36 k ^
L = 1020.7 k^ kg m² /s
I'm in physics, but I think the answer is 22. The angle of reflection is reflecting whatever the angle was so, in this case, it must be 22