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Anastasy [175]
3 years ago
8

The y component of velocity in a steady, incompressible flow field in the xy plane is v = -Bxy3, where B = 0.4 m-3 · s-1, and x

and y are measured in meters. (a) Find the simplest x component of velocity for this flow field. (b) Find the equation of the streamlines for this flow (use C as constant).

Engineering
1 answer:
love history [14]3 years ago
3 0

Answer:

attached below

Explanation:

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Gas is kept in a 0.1 m diameter cylinder under the weight of a 100 kg piston that is held down by a spring with a stiffness k =
Artyom0805 [142]

Answer:

The spring is compressed by 0.275 meters.

Explanation:

For equilibrium of the gas and the piston the pressure exerted by the gas on the piston should be equal to the sum of  weight of the piston and the force the spring exerts on the piston

Mathematically we can write

Force_{pressure}=Force_{spring}+Weight_{piston}

we know that

Force_{pressure}=Pressure\times Area=300\times 10^{3}\times \frac{\pi \times 0.1^2}{4}=750\pi Newtons

Weight_{piston}=mass\times g=100\times 9.81=981Newtons

Now the force exerted by an spring compressed by a distance 'x' is given by Force_{spring}=k\cdot x=5\times 10^{3}\times x

Using the above quatities in the above relation we get

5\times 10^{3}\times x+981=750\pi \\\\\therefore x=\frac{750\pi -981}{5\times 10^{3}}=0.275meters

5 0
3 years ago
Doubling the diameter of a solid, cylindrical wire doubles its strength in tension.
julsineya [31]

Answer:

True ❤️

-Solid by solid can make Cylindrical wire doubles Strengths in tension

4 0
3 years ago
By law, who is responsible for providing Safety Data Sheets?
nalin [4]

Answer:

OSHA Hazard Communication Standard is responsible.

Explanation:

5 0
3 years ago
Read 2 more answers
Find the power and the rms value of the following signal square: x(t) = 10 sin(10t) sin(15t)
ArbitrLikvidat [17]

Answer:

\mathbf{P_x =25 \ watts}

\mathbf{x_{rmx} = 5 \ unit}

Explanation:

Given that:

x(t) = 10 sin(10t) . sin (15t)

the objective is to find the power and the rms value of the following signal square.

Recall that:

sin (A + B) + sin(A - B) = 2 sin A.cos B

x(t) = 10 sin(15t) . cos (10t)

x(t) = 5(2 sin (15t). cos (10t))

x(t) = 5 × ( sin (15t + 10t) +  sin (15t-10t)

x(t) = 5sin(25 t) + 5 sin (5t)

From the knowledge of sinusoidial signal  Asin (ωt), Power can be expressed as:

P= \dfrac{A^2}{2}

For the number of sinosoidial signals;

Power can be expressed as:

P = \dfrac{A_1^2}{2}+ \dfrac{A_2^2}{2}+ \dfrac{A_3^2}{2}+ ...

As such,

For x(t), Power  P_x = \dfrac{5^2}{2}+ \dfrac{5^2}{2}

P_x = \dfrac{25}{2}+ \dfrac{25}{2}

P_x = \dfrac{50}{2}

\mathbf{P_x =25 \ watts}

For the number of sinosoidial signals;

RMS = \sqrt{(\dfrac{A_1}{\sqrt{2}})^2+(\dfrac{A_2}{\sqrt{2}})^2+(\dfrac{A_3}{\sqrt{2}})^2+...

For x(t), the RMS value is as follows:

x_{rmx} =\sqrt{(\dfrac{5}{\sqrt{2}} )^2 +(\dfrac{5}{\sqrt{2}} )^2 }

x_{rmx }=\sqrt{(\dfrac{25}{2} ) +(\dfrac{25}{2} ) }

x_{rmx }=\sqrt{(\dfrac{50}{2} )}

x_{rmx} =\sqrt{25}

\mathbf{x_{rmx} = 5 \ unit}

8 0
3 years ago
The net negative charge flowing through a device varies as q(t) = 2.2 t2 C. Find the current through the device at t = 0, 0.5, a
Leni [432]

Answer:

The current at t= 0 sec, is 0 A

The current at t= 0.5 sec, is 2.2 A

The current at t= 1 sec, is 4.4 A

Explanation:

Given that

q(t) = 2.2 t²

We know that:- the change in the charge w.r.t. time is known as current. So,

I=\dfrac{dq}{dt}

q(t) = 2.2 t²

\dfrac{dq}{dt}= 4.4 t

I= 4.4 t

1.

t = 0 s :

I = 4.4 x 0 = 0 A

<u>Therefore, the current at t= 0 sec, is 0 A</u>

2 .

t= 0.5 s :

I = 4.4 x 0.5 = 2.2 A

<u>Therefore, the current at t= 0.5 sec, is 2.2 A</u>

3.

t= 1 s

I = 4.4 x 1 =4.4 A

<u>Therefore, the current at t= 1 sec, is 4.4 A</u>

4 0
3 years ago
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