Answer:
i) E = 269 [MJ] ii)v = 116 [m/s]
Explanation:
This is a problem that encompasses the work and principle of energy conservation.
In this way, we establish the equation for the principle of conservation and energy.
i)
![W_{1-2}= (F*d) - (m*g*h)\\W_{1-2}=(500000*2.5*10^3)-(40000*9.81*2.5*10^3)\\W_{1-2}= 269*10^6[J] or 269 [MJ]](https://tex.z-dn.net/?f=W_%7B1-2%7D%3D%20%28F%2Ad%29%20-%20%28m%2Ag%2Ah%29%5C%5CW_%7B1-2%7D%3D%28500000%2A2.5%2A10%5E3%29-%2840000%2A9.81%2A2.5%2A10%5E3%29%5C%5CW_%7B1-2%7D%3D%20269%2A10%5E6%5BJ%5D%20or%20269%20%5BMJ%5D)
At that point the speed 1 is equal to zero, since the maximum height achieved was 2.5 [km]. So this calculated work corresponds to the energy of the rocket.
Er = 269*10^6[J]
ii ) With the energy calculated at the previous point, we can calculate the speed developed.
![E_{k2}=0.5*m*v^2\\269*10^6=0.5*40000*v^2\\v=\sqrt{\frac{269*10^6}{0.5*40000} }\\ v=116[m/s]](https://tex.z-dn.net/?f=E_%7Bk2%7D%3D0.5%2Am%2Av%5E2%5C%5C269%2A10%5E6%3D0.5%2A40000%2Av%5E2%5C%5Cv%3D%5Csqrt%7B%5Cfrac%7B269%2A10%5E6%7D%7B0.5%2A40000%7D%20%7D%5C%5C%20v%3D116%5Bm%2Fs%5D)
Now
- Lower mass=Higher acceleration
- Lower Force=Lower Acceleration
Option B has lowest mass and highest force hence its correct
Answer:
v = 344.1 m / s
d = 1720.5 m
Explanation:
For this problem we must calculate the speed of sound in air at 22ºC
v = 331 RA (1+ T / 273)
we calculate
v = 331 RA (1 + 22/273)
v = 344.1 m / s
the speed of the wave is constant,
v = d / t
d = v t
we calculate
d = 344.1 5
d = 1720.5 m
1. Vpa = 180m/s. @ 0 deg.
Vag = 40m/s @ 120 deg,CCW.
<span>
Vpg = Vpa + Vag,
Vpg = (180 + 40cos120) + i40sin120,
Vpg = 160 + i34.64,
Vpg=sqrt((160)^2 + (34.64)^2)=163.7m/s.
</span>
<span>2. tanA = Y / X = 34.64 / 160 = 0.2165,
A = 12.2 deg,CCW. = 12.2deg. North of
East. </span>
3. 1 hr = 3600s. <span>d = Vt = 163.7m/s * 3600s = 589,320m.
hope this helps</span>
Answer:
796.18 Hz
Explanation:
Applying,
Maximum velocity = Amplitude×Angular velocity
Therefore,
V' = A(2πf)............... Equation 1
Where V' = maximum velocity of the eardrum, A = Amplitude of vibration of the eardrum, f = frequency of the eardrum vibration, π = pie
make f the subject of the equation
f = V'/2πA................ Equation 2
From the question,
Given: V' = 3.6×10⁻³ m/s, A' = 7.2×10⁻⁷ m,
Constant: 3.14.
Substitute these values into equation 2
f = 3.6×10⁻³/( 7.2×10⁻⁷×2×3.14)
f = 796.18 Hz
