Given a = 10 cm/s²
u = 0 cm/s
v = 50 cm/s
we know that
v²=u²+2aS
2500=2×10×S
2500÷20 = S
S= 125 cm
The ramp is 125 cm
Answer: A
Explanation: STEP BY STEP
Answer:
Explanation:
extension in the spring = 40.4 - 31.8 = 8.6 cm = 8.6 x 10⁻² m .
kx = mg
k is spring constant , x is extension , m is mass
k x 8.6 x 10⁻² = 7.52 x 9.8
k = 856.93 N/m
= 857 x 10⁻³ KN /m
b ) Both side is pulled by force of 188 N .
Tension in spring = 188N
kx = T
856.93 x = 188
x = .219.38 m
= 21.938 cm
= 21.9 cm .
length of spring = 31.8 + 21.9
= 53.7 cm .
I belive it would power I’m not 100% sure. If it’s not power then force
Answer:
Psm = 30.66 [Psig]
Explanation:
To solve this problem we will use the ideal gas equation, recall that the ideal gas state equation is always worked with absolute values.
P * v = R * T
where:
P = pressure [Pa]
v = specific volume [m^3/kg]
R = gas constant for air = 0.287 [kJ/kg*K]
T = temperature [K]
<u>For the initial state</u>
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P1 = 24 [Psi] + 14.7 = 165.47[kPa] + 101.325 = 266.8 [kPa] (absolute pressure)
T1 = -2.6 [°C] = - 2.6 + 273 = 270.4 [K] (absolute Temperature)
Therefore we can calculate the specific volume:
v1 = R*T1 / P1
v1 = (0.287 * 270.4) / 266.8
v1 = 0.29 [m^3/kg]
As there are no leaks, the mass and volume are conserved, so the volume in the initial state is equal to the volume in the final state.
V2 = 0.29 [m^3/kg], with this volume and the new temperature, we can calculate the new pressure.
T2 = 43 + 273 = 316 [K]
P2 = R*T2 / V2
P2 = (0.287 * 316) / 0.29
P2 = 312.73 [kPa]
Now calculating the manometric pressure
Psm = 312.73 -101.325 = 211.4 [kPa]
And converting this value to Psig
Psm = 30.66 [Psig]