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3 years ago

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Car 1 of mass 1,715kg is sitting at an intersection. Car 2 of mass 2,780kg read ends Car 1, knocking it forward at 8 m/s. If Car

2 is at rest after the collision, how fast was it going when it hit Car 1?

1 answer:

SVEN [57.7K]3 years ago

8 0

Answer:8 miles

Explanation:

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A Venturi tube may be used as a fluid flowmeter. Suppose the device is used at a service station to measure the flow rate of gas

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Answer

given,

flow rate = p = 660 kg/m³

outer radius = 2.8 cm

P₁ - P₂ = 1.20 k Pa

inlet radius = 1.40 cm

using continuity equation

A₁ v₁ = A₂ v₂

π r₁² v₁ = π r₁² v₂

$v_1= \dfrac{r_1^2}{r_2^2} v_2$

$v_1= \dfrac{1.4^2}{2.8^2} v_2$

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Applying Bernoulli's equation

$\Delta P = \dfrac{1}{2}\rho (v_2^2-v_1^2)$

$\Delta P = \dfrac{1}{2}\rho (v_2^2-(0.25 v_2)^2)$

$\Delta P = \dfrac{1}{2}\rho v_2^2 (1 - 0.0625)$

$v_2=\sqrt{\dfrac{2\Delta P}{\rho(1 - 0.0625)}}$

$v_2=\sqrt{\dfrac{2\times 1200}{660 \times(1 - 0.0625)}}$

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A simple hydraulic lift is made by fitting a piston attached to a handle into a 3.0-cm diameter cylinder. The cylinder is connec

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Approximately $3.1 \times 10^4 \; \rm N$ (assuming that the acceleration due to gravity is $g = 9.81\; \rm kg \cdot N^{-1}$ .)

Explanation:

Let $A_1$ denote the first piston's contact area with the fluid. Let $A_2$ denote the second piston's contact area with the fluid.

Similarly, let $F_1$ and $F_2$ denote the size of the force on the two pistons. Since the person is placing all her weight on the first piston:

$F_1 = W = m \cdot g = 50\; \rm kg \times 9.81 \; \rm kg \cdot N^{-1} =495\; \rm N$ .

Since both pistons fit into cylinders, the two contact surfaces must be circles. Keep in mind that the area of a square is equal to $\pi$ times its radius, squared:

$\displaystyle A_1 = \pi \times \left(\frac{1}{2} \times 3.0\right)^2 = 2.25\, \pi\;\rm cm^{2}$ .
$\displaystyle A_2 = \pi \times \left(\frac{1}{2} \times 24\right)^2 = 144\, \pi\;\rm cm^{2}$ .

By Pascal's Law, the pressure on the two pistons should be the same. Pressure is the size of normal force per unit area:

$\displaystyle P = \frac{F}{A}$ .

For the pressures on the two pistons to match:

$\displaystyle \frac{F_1}{A_1} = \frac{F_2}{A_2}$ .

$F_1$ , $A_1$ , and $A_2$ have all been found. The question is asking for $F_2$ . Rearrange this equation to obtain:

$\displaystyle F_2 = \frac{F_1}{A_1} \cdot A_2 = F_1 \cdot \frac{A_2}{A_1}$ .

Evaluate this expression to obtain the value of $F_2$ , which represents the force on the piston with the larger diameter:

$\begin{aligned}F_2 &= F_1 \cdot \frac{A_2}{A_1} \\ &= 495\; \rm N \times \frac{2.25\, \pi\; \rm cm^2}{144\, \pi \; \rm cm^2} \approx 3.1 \times 10^4\; \rm N\end{aligned}$ .

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