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lisabon 2012 [21]
4 years ago
10

In a car lift, compressed air exerts a force on a piston with a radius of 2.62 cm. This pressure is transmitted to a second pist

on with a radius of 10.8 cm. a) How large a force must the compressed air exert to lift a 1.44 × 104 N car
Physics
1 answer:
marin [14]3 years ago
8 0

Answer:

847.45 N

Explanation:

F₁=force of exerted by smaller piston

F₂=force of exerted by larger piston=1.44×10⁴ N

A₁=Area of smaller piston= 2.62 cm =0.0265 m

A₂=Area of larger piston= 10.8 cm =0.108 m

Pressure exerted by both the pistons will be equal

P_1=P_2\\\Rightarrow \frac{F_1}{A_1}=\frac{F_2}{A_2}\\\Rightarrow F_1=\frac{F_2}{A_2} A_1\\\Rightarrow F_1=\frac{14400}{\pi\times  0.108^2}\pi\times  0.0262^2\\\Rightarrow F_1=847.45\ N

Hence, force exerted to lift a 14400 N car is 847.45 N

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Answer:

A) At point 1, local acceleration = 0.5 m/s²

At point 2, local acceleration = 1.0 m/s²

B) Average Eulerian convective acceleration over the two points in the cross section shown = 0.5 m/s²

This value is positive indicating an increase in velocity and acceleration kf the fluid as the cross sectional Area of flow reduces.

Explanation:

Local acceleration at those points is the instantaneous acceleration at those points and it is given as

a = dv/dt

At point 1, v₁ = 0.5 t

a₁ =dv₁/dt = 0.5 m/s²

At point 2, v₂ = 1.0 t

a₂ = dv₂/dt = 1.0 m/s²

b) Average Eulerian convective acceleration over the two points in the cross section shown = (change of velocity between the two points)/time

Change of velocity between the two points = v₂ - v₁ = 1.0t - 0.5t = 0.5 t

Time = t

Average acceleration = 0.5t/t = 0.5 m/s²

This value is positive indicating an increase in velocity and acceleration kf the fluid as the cross sectional Area of flow reduces.

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3 years ago
A truck is driving north at 35 miles per hour and passes a car going south at 40 miles per hour. What is the speed of the car fr
inna [77]

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Which one of the following temperatures is equal to 5°C?
natali 33 [55]

Answer : The correct option is, (D) 278 K

Explanation :

We are given temperature 5^oC.

Now the conversion factor used for the temperature is,

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Now put the value of temperature, we get

K=5^oC+273=278K

Therefore, the temperature 278 K is equal to the 5^oC


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3 years ago
A wind-tunnel experimentis performed on a 1/25scale model of a supersonic aircraft. The prototype aircraft flies at 450 m/s in c
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Calculate the change in length of a 90.5 mm aluminum bar that has increased in temperature by from -14.4 oC to 154.6 oC
nignag [31]

Answer:

 ΔL = 3.82 10⁻⁴ m

Explanation:

This is a thermal expansion exercise

          ΔL = α L₀ ΔT

          ΔT = T_f - T₀

where ΔL is the change in length and ΔT is the change in temperature

Let's reduce the length to SI units

          L₀ = 90.5 mm (1m / 1000 mm) = 0.0905 m

let's calculate

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          ΔL = 3.8236 10⁻⁴ m

     

using the criterion of three significant figures

          ΔL = 3.82 10⁻⁴ m

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3 years ago
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