The speed of sound at sea level is 340.29 m/s (meters per seconds).
Answer:
H = 45 m
Explanation:
First we find the launch velocity of the ball by using the following formula:
v₀ = √(v₀ₓ² + v₀y²)
where,
v₀ = launching velocity = ?
v₀ₓ = Horizontal Component of Launch Velocity = 15 m/s
v₀y = Vertical Component of Launch Velocity = 30 m/s
Therefore,
v₀ = √[(15 m/s)² + (30 m/s)²]
v₀ = 33.54 m/s
Now, we find the launch angle of the ball by using the following formula:
θ = tan⁻¹ (v₀y/v₀ₓ)
θ = tan⁻¹ (30/15)
θ = tan⁻¹ (2)
θ = 63.43°
Now, the maximum height attained by the ball is given by the formula:
H = (v₀² Sin² θ)/2g
H = (33.54 m/s)² (Sin² 63.43°)/2(10 m/s²)
<u>H = 45 m</u>
Answer:
1.5 m
Explanation:
Length. L = 12 m
Width, W = 16 m
Area, A = 12 x 16 = 192 m^2
Let the width of pavement be d.
The new length, L' = 12 + 2d
the new width, W' = 16 + 2d
New Area, A' = L' x W' = (12 + 2d)(16 + 2d) = 192 + 56 d + 4d^2
Difference in area = A' - A
285 = 192 + 56 d + 4d^2 - 192
93 = 56 d + 4d^2
4d^2 + 56 d - 93 = 0
![d = \frac{-56\pm \sqrt{56^{2}+4\times 4\times 93}}{8}](https://tex.z-dn.net/?f=d%20%3D%20%5Cfrac%7B-56%5Cpm%20%5Csqrt%7B56%5E%7B2%7D%2B4%5Ctimes%204%5Ctimes%2093%7D%7D%7B8%7D)
\
d = 1.5 m
Thus, the width of the pavement is 1.5 m.
The thermal efficiency of an engine is
![\eta= \frac{W}{Q}](https://tex.z-dn.net/?f=%5Ceta%3D%20%5Cfrac%7BW%7D%7BQ%7D%20)
where
W is the work done by the engine
Q is the heat absorbed by the engine to do the work
In this problem, the work done by the engine is W=200 J, while the heat exhausted is Q=600 J, so the efficiency of the machine is
Answer: Gamma rays, x-rays, ultraviolet rays, visible light, and infrared rays.