The first rule of vectors is that the horizontal and vertical components are separate. Disregarding air resistance, the only thing we have to worry about is gravity.
The appropriate suvat to use for the vertical component is v = u +at
I will take a to be -9.81, you may have to change it to be 10 if your qualification likes g to be 10.
v = 30 + (-9.81x2)
v = 30 - 19.62
=10.38m/s
Therefore we know that after 2.0 s the vertical component will be 10.38ms^-1, ie 10m/s as the answers given are all to 2sf.
The horizontal component is completely separate to the vertical component and since there is no air resistance, it will remain constant throughout the projectiles trajectory. Therefore it will remain at 40ms^-1.
Combining this together we get:
(1) vx=40m/s and vy=10m/s
Mass always stays the same therefore it can’t be c and d. Volume stays the same because she is not removing anything. Therefore it is a
On sources it says it would just be the super giant star
Answer:
Q = 40.1 degrees
Explanation:
Given:
- The weight of the timber W = 670 N
- Water surface level from pivot y = 2.1 m
- The specific density of water Y = 9810 N / m^3
- Dimension of timber = (0.15 x 0.15 x 0.0036) m
Find:
- The angle of inclination Q that the timber makes with the horizontal.
Solution:
- Calculate the Flamboyant Force F_b acting upwards at a distance x along the timber, which is unknown:
F_b = Y * V_timber
F_b = 9810*0.15*0.15*x
F_b = 226.7*x N
- Take static equilibrium conditions for the timber, and take moments about the pivot:
(M)_p = 0
W*0.5*3.6*cos(Q) - x/2 * F_b*cos(Q) = 0
- Plug values in:
670*0.5*3.6 - x^2 * 0.5*226.7 = 0
x^2 = 1206 / 113.35
x = 3.26 m
- Now use the value of x and vertical height y to compute the angle of inclination to be:
sin(Q) = y / x
sin(Q) = 2.1 / 3.26
Q = sin^-1 (0.6441718)
Q = 40.1 degrees