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2 years ago

Quations (3.4) and (3.5) are equivalent expressions for Lagrange's equations.

1 answer:

3 0

Answer:Using the Nielsen form, determine the equation of motion for a mass m connected to a spring of constant k. Exercise 3.2 Using the

Explanation:

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When a severe weather watch has been issued by the National Weather Service, what should you do?

Harman [31]

Answer:

Begin to prepare

Explanation:

Just took the test!

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3 years ago

Read 2 more answers

A soccer ball kicked with a force of 12.5 N accelerates at 6.2 m/s’ to the right. What is the mass of the ball? Answer in units

stepladder [879]

Answer:

m = 2.01[kg]

Explanation:

This problem can be solved using Newton's second law which tells us that the force applied on a body is equal to the product of mass by acceleration.

$F =m*a$

where:

F = force = 12.5 [N]

m = mass [kg]

a = acceleration = 6.2 [m/s²]

$12.5=m*6.2\\m = 2.01[kg]$

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3 years ago

If 5 mm of rain falls in a 100 m2 field, what volume of rain, in m3, fell in the field?

Nina [5.8K]

The volume of rain that fells in the field is simply given by the area of the field, which is

$A=100 mm^2$

multiplied by the height of rain that fell, which is

$h=5.0 mm$

Therefore, the volume is

$V=hA=(5 mm)(100 mm^2)=500 mm^3$

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4 years ago

When checking for noncondensables inside a recovery cylinder why should the technician allow the temperature of the cylinder to

77julia77 [94]

Before taking a pressure reading, it is necessary for the technician to first allow the temperature of the cylinder to stabilize to room temperature because a comparison with a temperature-pressure chart is only valid and true when both temperature and pressure of the refrigerant are stable.

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3 years ago

Water is flowing in a pipe with a varying cross-sectional area, and at all points the water completely fills the pipe. At point

Salsk061 [2.6K]

Answer:

The fluids speed at a) $0.105\,m^{2}$ and b) $0.047\,m^{2}$ are $2.33\,\frac{m}{s^{2}}$ and $5.21\,\frac{m}{s^{2}}$ respectively

c) Th volume of water the pipe discharges is: $882\,m^{3}$

Explanation:

To solve a) and b) we should use flow continuity for ideal fluids:

$\Delta Q=0$ (1)

With Q the flux of water, but Q is $Av$ using this on (1) we have:

$A_{2}v_{2}-A_{1}v_{1}=0$ (2)

With A the cross sectional areas and v the velocities of the fluid.

a) Here, we use that point 2 has a cross-sectional area equal to $A_{2}=0.105\,m^{2}$ , so now we can solve (2) for $v_{2}$ :

$v_{2}=\frac{A_{1}v_{1}}{A_{2}}=\frac{(0.070)(3.5)}{0.105}\approx2.33\,\frac{m}{s^{2}}$

b) Here we use point 2 as $A_{2}=0.047\,m^{2}$ :

$v_{2}=\frac{A_{1}v_{1}}{A_{2}}=\frac{(0.070)(3.5)}{0.047}\approx5.21\,\frac{m}{s^{2}}$

c) Here we need to know that in this case the flow is the volume of water that passes a cross-sectional area per unit time, this is $Q=\frac{V}{t}$ , so we can write:

$A_{1}v_{1}=\frac{V}{t}$ , solving for V:

$V=A_{1}v_{1}t=(0.070m^{2})(3.5\frac{m}{s})(3600s)=882\,m^{3}$

3 0

4 years ago