a yoyo in someones hand is an example of potential energy
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:
The study and analysis of light according to its component wavelengths is called spectroscopy.
EXPLANATION:
Spectroscopy is the branch of science that is concerned with the investigation and measurement of spectrum produced when matter interacts with or emits electromagnetic radiation.It helps us to identify atoms and molecules in the object.Spectroscopy is used to find out Dopplers effect (the red shift and blue shift),which tells how fast the object is comming towards earth or moving away from the earth.
Answer:
0.25 L
Explanation:
= Initial pressure = 1 atm
= Initial Temperature = 20 °C
= Initial volume = 4.91 L
= Final pressure = 5.2 atm
= Final Temperature = -196 °C
= Final volume
From ideal gas law we have
![\dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}\\\Rightarrow V_2=\dfrac{P_1V_1T_2}{T_1P_2}\\\Rightarrow P_2=\dfrac{1\times 4.91(273.15-196)}{(20+273.15)\times 5.2}\\\Rightarrow V_2=0.24849\ L\approx 0.25\ L](https://tex.z-dn.net/?f=%5Cdfrac%7BP_1V_1%7D%7BT_1%7D%3D%5Cdfrac%7BP_2V_2%7D%7BT_2%7D%5C%5C%5CRightarrow%20V_2%3D%5Cdfrac%7BP_1V_1T_2%7D%7BT_1P_2%7D%5C%5C%5CRightarrow%20P_2%3D%5Cdfrac%7B1%5Ctimes%204.91%28273.15-196%29%7D%7B%2820%2B273.15%29%5Ctimes%205.2%7D%5C%5C%5CRightarrow%20V_2%3D0.24849%5C%20L%5Capprox%200.25%5C%20L)
The pressure experienced by the balloon is 0.25 L
Complete Question
A gas gun uses high pressure gas tp accelerate projectile through the gun barrel.
If the acceleration of the projective is : a = c/s m/s2
Where c is a constant that depends on the initial gas pressure behind the projectile. The initial position of the projectile is s= 1.5m and the projectile is initially at rest. The projectile accelerates until it reaches the end of the barrel at s=3m. What is the value of the constant c such that the projectile leaves the barrel with velocity of 200m/s?
Answer:
The value of the constant is ![c = 28853.78 \ m^2 /s^2](https://tex.z-dn.net/?f=c%20%3D%2028853.78%20%5C%20m%5E2%20%2Fs%5E2)
Explanation:
From the question we are told that
The acceleration is ![a = \frac{c}{s}\ m/s^2](https://tex.z-dn.net/?f=a%20%3D%20%20%5Cfrac%7Bc%7D%7Bs%7D%5C%20%20%20m%2Fs%5E2)
The initial position of the projectile is s= 1.5m
The final position of the projectile is ![s_f = 3 \ m](https://tex.z-dn.net/?f=s_f%20%3D%20%203%20%5C%20m)
The velocity is ![v = 200 \ m/s](https://tex.z-dn.net/?f=v%20%3D%20200%20%5C%20m%2Fs)
Generally ![time = \frac{ds}{dv}](https://tex.z-dn.net/?f=time%20%20%3D%20%20%5Cfrac%7Bds%7D%7Bdv%7D)
and acceleration is ![a = \frac{v}{time }](https://tex.z-dn.net/?f=a%20%3D%20%20%5Cfrac%7Bv%7D%7Btime%20%7D)
so
![a = v \frac{dv}{ds}](https://tex.z-dn.net/?f=a%20%3D%20v%20%20%5Cfrac%7Bdv%7D%7Bds%7D)
=> ![vdv = a ds](https://tex.z-dn.net/?f=vdv%20%20%3D%20%20a%20ds)
![vdv = \frac{c}{s} ds](https://tex.z-dn.net/?f=vdv%20%20%3D%20%5Cfrac%7Bc%7D%7Bs%7D%20%20ds)
integrating both sides
![\int\limits^a_b vdv = \int\limits^c_d \frac{c}{s} ds](https://tex.z-dn.net/?f=%5Cint%5Climits%5Ea_b%20%20vdv%20%20%3D%20%5Cint%5Climits%5Ec_d%20%5Cfrac%7Bc%7D%7Bs%7D%20%20ds)
Now for the limit
a = 200 m/s
b = 0 m/s
c = s= 3 m
d =
= 1.5 m
So we have
![\int\limits^{200}_{0} vdv = \int\limits^{3}_{1.5} \frac{c}{s} ds](https://tex.z-dn.net/?f=%5Cint%5Climits%5E%7B200%7D_%7B0%7D%20%20vdv%20%20%3D%20%5Cint%5Climits%5E%7B3%7D_%7B1.5%7D%20%5Cfrac%7Bc%7D%7Bs%7D%20%20ds)
![[\frac{v^2}{2} ] \left | 200} \atop {0}} \right. = c [ln s]\left | 3} \atop {1.5}} \right.](https://tex.z-dn.net/?f=%5B%5Cfrac%7Bv%5E2%7D%7B2%7D%20%5D%20%5Cleft%20%7C%20200%7D%20%5Catop%20%7B0%7D%7D%20%5Cright.%20%20%3D%20c%20%5Bln%20s%5D%5Cleft%20%7C%203%7D%20%5Catop%20%7B1.5%7D%7D%20%5Cright.)
![\frac{200^2}{2} = c ln[\frac{3}{1.5} ]](https://tex.z-dn.net/?f=%5Cfrac%7B200%5E2%7D%7B2%7D%20%20%3D%20%20c%20ln%5B%5Cfrac%7B3%7D%7B1.5%7D%20%5D)
=> ![c = \frac{20000}{0.69315}](https://tex.z-dn.net/?f=c%20%3D%20%5Cfrac%7B20000%7D%7B0.69315%7D)
![c = 28853.78 \ m^2 /s^2](https://tex.z-dn.net/?f=c%20%3D%2028853.78%20%5C%20m%5E2%20%2Fs%5E2)