Answer:
xf = 5.68 × 10³ m
yf = 8.57 × 10³ m
Explanation:
given data
vi = 290 m/s
θ = 57.0°
t = 36.0 s
solution
firsa we get here origin (0,0) to where the shell is launched
xi = 0 yi = 0
xf = ? yf = ?
vxi = vicosθ vyi = visinθ
ax = 0 ay = −9.8 m/s
now we solve x motion: that is
xf = xi + vxi × t + 0.5 × ax × t² ............1
simplfy it we get
xf = 0 + vicosθ × t + 0
put here value and we get
xf = 0 + (290 m/s) cos(57) (36.0 s)
xf = 5.68 × 10³ m
and
now we solve for y motion: that is
yf = yi + vyi × t + 0.5 × ay × t
² ............2
put here value and we get
yf = 0 + (290 m/s) × sin(57) × (36.0 s) + 0.5 × (−9.8 m/s2) × (36.0 s) ²
yf = 8.57 × 10³ m
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For part a we can use the kinematic formula distance= velocity(initial)*time+1/2*acceleration*time^2
we know there is only velocity in the horizontal and none in the vertical so v initial in this case is 0, acceleration is gravity =9.81m/s^2
plug in .95=(1/2)*9.81*t^2 solve for t
t=.44seconds
Now for part b we know there is no horizontal acceleration so distance= velcoity* time
we will use horrizontial velocity and the time we just found
distance= .8m/s*.44seconds
=.35 meters
Answer:
B
Explanation:
We know the inital velocity is 16 m/s
Acclereation is -0.5 m/s^2
And that final velocity is 0
We need to find time
The answer is B
Answer:
The work done is -209.42 J.
Explanation:
F(x) = (- 20 - 3 x ) N
x = 0 to x = 6.9 m
Here, the force is variable in nature, so the work done by the variable force is given by
![W =\int F dx\\\\W =\int_{0}^{6.9}(-20- 3x dx )\\\\W= \left [ - 20 x - 1.5 x^2 \right ]_{0}^{6.9}\\\\W = - 20 (6.9 - 0) - 1.5(6.9\times 6.9 - 0)\\\\W =- 138 - 71.42\\\\W = - 209.42 J](https://tex.z-dn.net/?f=W%20%3D%5Cint%20F%20dx%5C%5C%5C%5CW%20%3D%5Cint_%7B0%7D%5E%7B6.9%7D%28-20-%203x%20dx%20%29%5C%5C%5C%5CW%3D%20%5Cleft%20%5B%20-%2020%20x%20-%201.5%20x%5E2%20%5Cright%20%5D_%7B0%7D%5E%7B6.9%7D%5C%5C%5C%5CW%20%3D%20-%2020%20%286.9%20-%200%29%20-%201.5%286.9%5Ctimes%206.9%20-%200%29%5C%5C%5C%5CW%20%3D-%20138%20-%2071.42%5C%5C%5C%5CW%20%3D%20-%20209.42%20%20J)
