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Rashid [163]
3 years ago
5

Reducing waste and recycling things are ways to conserve energy. True false

Physics
2 answers:
Leya [2.2K]3 years ago
6 0

I think the correct answer is true :)

nydimaria [60]3 years ago
5 0
True b/c we conserve energy by not using energy to make new materials because we have those available because we recycled
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3 years ago
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3 years ago
A ball is tossed from an upper-story window of a building. the ball is given an initial velocity of 8.00 m/s at an angle of 20.0
otez555 [7]
A) The motion of the ball consists of two indipendent motions on the horizontal (x) and vertical (y) axis. The laws of motion in the two directions are:
x(t)=v_0 \cos \alpha t
y(t)=h-v_0 \sin \alpha t - \frac{1}{2}gt^2
where
- the horizontal motion is a uniform motion, with constant speed v_0 \cos \alpha, where v_0 = 8.00 m/s and \alpha=20.0^{\circ}
- the vertical motion is an uniformly accelerated motion, with constant acceleration g=9.81 m/s^2, initial position h (the height of the building) and initial vertical velocity v_0 \sin \alpha (with a negative sign, since it points downwards)

The ball strikes the ground after a time t=3.00 s, so we can find the distance covered horizontally by the ball by substituting t=3.00 s into the equation of x(t):
x(3.00 s)=v_0 \cos \alpha t=(8 m/s)(\cos 20^{\circ})(3.0 s)=22.6 m

b) To find the height from which the ball was thrown, h, we must substitute t=3.00 s into the equation of y(t), and requiring that y(3.00 s)=0 (in fact, after 3 seconds the ball reaches the ground, so its vertical position y(t) is zero). Therefore, we have:
0=h-v_0 \sin \alpha t -  \frac{1}{2}gt^2
which becomes
h=(8 m/s)(\sin 20^{\circ})(3.0 s)+ \frac{1}{2}(9.81 m/s^2)(3.0 s)^2=52.3 m

c) We want the ball to reach a point 10.0 meters below the level of launching, so we want to find the time t such that 
y(t)=h-10
If we substitute this into the equation of y(t), we have
h-10 = h-v_0 \sin \alpha t-  \frac{1}{2}gt^2
\frac{1}{2}gt^2+v_0 \sin \alpha t -10 =0
4.9 t^2 +2.74 t-10 =0
whose solution is t=1.18 s (the other solution is negative, so it has no physical meaning). Therefore, the ball reaches a point 10 meters below the level of launching after 1.18 s.

4 0
3 years ago
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