Answer:
Acceleration = 9.8 m/sec2
Explanation:
We know that the acceleration due to gravity is 9.8 m/sec2 near the surface of earth.
Remember that the gravity is always affecting everything on earth at every instant, even the rock at the very top of it's trajectory. It doesn't mind if the velocity is zero for an instant or if there's air resistance, the gravity force is there, and always aims towards the center of the earth.
So for all the trajectory of the rock, gravity is 9.8 m/sec2
Answer:
Reflection is when light bounces off an object. If the surface is smooth and shiny, like glass, water or polished metal, the light will reflect at the same angle as it hit the surface. This is called specular reflection. Light reflects from a smooth surface at the same angle as it hits the surface.
Explanation:
Answer:
![\displaystyle y_m=3.65m](https://tex.z-dn.net/?f=%5Cdisplaystyle%20y_m%3D3.65m)
Explanation:
<u>Motion in The Plane</u>
When an object is launched in free air with some angle respect to the horizontal, it describes a known parabolic path, comes to a maximum height and finally drops back to the ground level at a certain distance from the launching place.
The movement is split into two components: the horizontal component with constant speed and the vertical component with variable speed, modified by the acceleration of gravity. If we are given the values of
and
as the initial speed and angle, then we have
![\displaystyle v_x=v_o\ cos\theta](https://tex.z-dn.net/?f=%5Cdisplaystyle%20v_x%3Dv_o%5C%20cos%5Ctheta)
![\displaystyle v_y=v_o\ sin\theta-gt](https://tex.z-dn.net/?f=%5Cdisplaystyle%20v_y%3Dv_o%5C%20sin%5Ctheta-gt)
![\displaystyle x=v_o\ cos\theta\ t](https://tex.z-dn.net/?f=%5Cdisplaystyle%20x%3Dv_o%5C%20cos%5Ctheta%5C%20t)
![\displaystyle y=v_o\ sin\theta\ t -\frac{gt^2}{2}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20y%3Dv_o%5C%20sin%5Ctheta%5C%20t%20-%5Cfrac%7Bgt%5E2%7D%7B2%7D)
If we want to know the maximum height reached by the object, we find the value of t when
becomes zero, because the object stops going up and starts going down
![\displaystyle v_y=o\Rightarrow v_o\ sin\theta =gt](https://tex.z-dn.net/?f=%5Cdisplaystyle%20v_y%3Do%5CRightarrow%20v_o%5C%20sin%5Ctheta%20%3Dgt)
Solving for t
![\displaystyle t=\frac{v_o\ sin\theta }{g}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20t%3D%5Cfrac%7Bv_o%5C%20sin%5Ctheta%20%7D%7Bg%7D)
Then we replace that value into y, to find the maximum height
![\displaystyle y_m=v_o\ sin\theta \ \frac{v_o\ sin\theta }{g}-\frac{g}{2}\left (\frac{v_o\ sin\theta }{g}\right )^2](https://tex.z-dn.net/?f=%5Cdisplaystyle%20y_m%3Dv_o%5C%20sin%5Ctheta%20%5C%20%5Cfrac%7Bv_o%5C%20sin%5Ctheta%20%7D%7Bg%7D-%5Cfrac%7Bg%7D%7B2%7D%5Cleft%20%28%5Cfrac%7Bv_o%5C%20sin%5Ctheta%20%7D%7Bg%7D%5Cright%20%29%5E2)
Operating and simplifying
![\displaystyle y_m=\frac{v_o^2\ sin^2\theta }{2g}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20y_m%3D%5Cfrac%7Bv_o%5E2%5C%20sin%5E2%5Ctheta%20%7D%7B2g%7D)
We have
![\displaystyle v_o=20\ m/s,\ \theta=25^o](https://tex.z-dn.net/?f=%5Cdisplaystyle%20v_o%3D20%5C%20m%2Fs%2C%5C%20%5Ctheta%3D25%5Eo)
The maximum height is
![\displaystyle y_m=\frac{(20)^2(sin25^o)^2}{2(9.8)}=\frac{71.44}{19.6}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20y_m%3D%5Cfrac%7B%2820%29%5E2%28sin25%5Eo%29%5E2%7D%7B2%289.8%29%7D%3D%5Cfrac%7B71.44%7D%7B19.6%7D)
![\displaystyle y_m=3.65m](https://tex.z-dn.net/?f=%5Cdisplaystyle%20y_m%3D3.65m)
Answer:
The magnitude of the tension in the cable, T is 1,064.315 N
Explanation:
Here we have
Length of beam = 4.0 m
Weight = 200 N
Center of mass of uniform beam = mid-span = 2.0 m
Point of attachment of cable = Beam end = 4.0 m
Angle of cable = 53° with the horizontal
Tension in cable = T
Point at which person stands = 1.50 m from wall
Weight of person = 350 N
Therefore,
Taking moment about the wall, we have
∑Clockwise moments = ∑Anticlockwise moments
T×sin(53) = 350×1.5 + 200×2
T = 850/sin(53) = 1,064.315 N.