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3 years ago

A car accelerates uniformly from rest to 60 km/h in 30 s. What is its displacement during this time?

Engineering

1 answer:

Rina8888 [55]3 years ago

5 0

Answer:

252.62 m

Explanation:

Data provided :

Initial speed of the car, u = 0 Km/h (since, the car starts from the rest)

Final speed of the car, v = 60 km/h = 60 × (5/18) = 16.67 m/s

Time, t = 30 s

now,

from the Newton's equation of motion, we have

v = u + at

where,

a is the acceleration of the car

on substituting the values, we have

16.67 = 0 + a × 30

a = 0.55 m/s²

also,

v² - u² = 2as

where, s is the distance traveled

on substituting the values, we get

16.67² - 0² = 2 × 0.55 × s

s = 252.62 m

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Answer:

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Explanation:

Given that

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If the plane is moving upward with acceleration 6 g's then the for F

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$1\ ft/s^2= 0.304\ m/s^2$

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3 years ago

Which of the following is not a relationship set between elements in a sketch

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2 years ago

Oil with a density of 800 kg/m3 is pumped from a pressure of 0.6 bar to a pressure of 1.4 bar, and the outlet is 3 m above the i

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Answer:

23.3808 kW

20.7088 kW

Explanation:

ρ = Density of oil = 800 kg/m³

P₁ = Initial Pressure = 0.6 bar

P₂ = Final Pressure = 1.4 bar

Q = Volumetric flow rate = 0.2 m³/s

A₁ = Area of inlet = 0.06 m²

A₂ = Area of outlet = 0.03 m²

Velocity through inlet = V₁ = Q/A₁ = 0.2/0.06 = 3.33 m/s

Velocity through outlet = V₂ = Q/A₂ = 0.2/0.03 = 6.67 m/s

Height between inlet and outlet = z₂ - z₁ = 3m

Temperature to remains constant and neglecting any heat transfer we use Bernoulli's equation

$\frac {P_1}{\rho g}+\frac{V_1^2}{2g}+z_1+h=\frac {P_2}{\rho g}+\frac{V_2^2}{2g}+z_2\\\Rightarrow h=\frac{P_2-P_1}{\rho g}+\frac{V_2^2-V_1^2}{2g}+z_2-z_1\\\Rightarrow h=\frac{(1.4-0.6)\times 10^5}{800\times 9.81}+\frac{6.67_2^2-3.33^2}{2\times 9.81}+3\\\Rightarrow h=14.896\ m$

Work done by pump

$W_{p}=\rho gQh\\\Rightarrow W_{p}=800\times 9.81\times 0.2\times 14.896\\\Rightarrow W_{p}=23380.8\ W$

∴ Power input to the pump 23.3808 kW

Now neglecting kinetic energy

$h=\frac{P_2-P_1}{\rho g}+z_2-z_1\\\Righarrow h=\frac{(1.4-0.6)\times 10^5}{800\times 9.81}+3\\\Righarrow h=13.19\ m\\$

Work done by pump

$W_{p}=\rho gQh\\\Rightarrow W_{p}=800\times 9.81\times 0.2\times 13.193\\\Rightarrow W_{p}=20708.8\ W$

∴ Power input to the pump 20.7088 kW

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