The horizontal movement of the rocket is 11m/s, with an acceleration of 1.6m/s². The vertical movement will be downward, with an initial velocity of zero (it was shot horizontally) and a negative acceleration of g (-9.8m/s²)
To see how far the rocket traveled before hitting the ground, let's first figure out the time t at which the rocket hit the ground:
The formula for distance is d= vt + (1/2)at² ,
Where v=initial velocity, d=distance traveled, a=acceleration, and t=time
We want to find how long it took to travel 40 meters (height above the ground), given an initial velocity of 0 and negative acceleration of 9.8
Plugging into the equation:
40 = 0(t) + (1/2) (9.8) (t²) Multiply both sides by (2/9.8)
8.16 = t² Square root of both sides
t= 2.85
The rocket traveled for 2.85 seconds before hitting the ground. Plug this number into our distance formula to find horizontal distance
d= vt + (1/2)at²
d = 11 (2.85) + (1/2) (1.6) (2.85²)
Remember that initial horizonal velocity is 11m/s and horizontal acceleration is 1.6m/s²
Simplify:
d= 31.35 + .8 * 8.16
d = 37.87
The object traveled 37.87 meters before hitting the ground.
Vectors have magnitude and direction, scalars only have magnitude, t<span>here are some </span>quantities<span>, like speed, which have very special definitions</span>
Answer:
338 K
Explanation:
= 2.5 kg, c = 4189 J/(kg K), = 13.5 , = 22.5
= 0.5 kg, = 20
Heat loss by hotter water = heat gained by cooler water
cΔT = cΔT
c( - ) = c( - )
2.5 x 4189 x (22.5 - 13.5) = 0.5 x 4189 x ( - 20)
2.5 x 4189 x 9 = 2094.5 ( - 20)
94252.5 = 2094.5 - 41890
94252.5 + 41890 = 2094.5
136142.5 = 2094.5
=
= 65
= 65
But,
θ K = 273 + θ
= 273 + 65
= 338
The final temperature of the water is 338 K.
Answer:
Displacement from the starting position is 103.21m
Explanation:
If you draw these directions, it will create the two legs of a triangle.
Using this method, you can visualize why your displacement is what it is.
Using the pythagorean theorem
Plug in both values
c = 103.2085
c= 103.21
Answer:
The net effect on the laboratory is cooling of the space.
Explanation:
Given:
power used by the air conditioner,
coefficient of performance,
We know that:
<u>Therefore the cooling power capacity:</u>
The net effect on the laboratory is cooling of the space.