<span>The fahrenheit temperature is 927965. It is calculated using the formula 515515 Degree Cx1.8+32=927965. The degree celcius and fahrenheit are two units two measure temperature. If the value is given in celcius it can be converted into fahrenheit using the above formula.</span>
Answer:
y = 17 m
Explanation:
For this projectile launch exercise, let's write the equation of position
x = v₀ₓ t
y =
t - ½ g t²
let's substitute
45 = v₀ cos θ t
10 = v₀ sin θ t - ½ 9.8 t²
the maximum height the ball can reach where the vertical velocity is zero
v_{y} = v_{oy} - gt
0 = v₀ sin θ - gt
0 = v₀ sin θ - 9.8 t
Let's write our system of equations
45 = v₀ cos θ t
10 = v₀ sin θ t - ½ 9.8 t²
0 = v₀ sin θ - 9.8 t
We have a system of three equations with three unknowns for which it can be solved.
Let's use the last two
v₀ sin θ = 9.8 t
we substitute
10 = (9.8 t) t - ½ 9.8 t2
10 = ½ 9.8 t2
10 = 4.9 t2
t = √ (10 / 4.9)
t = 1,429 s
Now let's use the first equation and the last one
45 = v₀ cos θ t
0 = v₀ sin θ - 9.8 t
9.8 t = v₀ sin θ
45 / t = v₀ cos θ
we divide
9.8t / (45 / t) = tan θ
tan θ = 9.8 t² / 45
θ = tan⁻¹ ( 9.8 t² / 45
)
θ = tan⁻¹ (0.4447)
θ = 24º
Now we can calculate the maximum height
v_y² =
- 2 g y
vy = 0
y = v_{oy}^2 / 2g
y = (20 sin 24)²/2 9.8
y = 3,376 m
the other angle that gives the same result is
θ‘= 90 - θ
θ' = 90 -24
θ'= 66'
for this angle the maximum height is
y = v_{oy}^2 / 2g
y = (20 sin 66)²/2 9.8
y = 17 m
thisis the correct
Answer:
Option D 3.9
Explanation:
First, you need to use the correct equation which is the following:
COP = Q/W
Where:
Q = heat absorbed
W = work done by the pump
COP = coefficient of perfomance
We have all the data, so, all you need to do is replace in the above expression and you shoould get the correct result:
COP = 30 / 7.7
COP = 3.896
This result you can round it to 3.9. option D.
When is at the end of the runway the velocity of the plane is given by the equation

where s=1800 m is the runway length. Thus
At half runway the velocity of the plane is

Therefore at midpoint of runway the percentage of takeoff velocity is
‰
Answer:
v =25 m/s
time= 50 s
Velocity =Displacement/Time
Displacement = Velocity × Time
S = 25×50
s=1250m
Explanation:
v =25 m/s
time= 50 s
Velocity =Displacement/Time
Displacement = Velocity × Time