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3 years ago

A 30 kg buzzard is sitting on a cable. The cable is attached to two poles 25

Physics

1 answer:

sertanlavr [38]3 years ago

6 0

Answer:

T1 = 675 [N] ; T2 = 707.5 [N]

Explanation:

To solve this problem we must draw a free body diagram, where the bird is located, the posts and distances mentioned.

In the attached image we can see the free body diagram with different forces, angles and distances.

By conducting a distance analysis we can find the angles on both sides of the rope relative to the horizontal.

Then we perform a force analysis on the X-axis and the y-axis in order to find the two equations we need to find the stresses in the cables (T1 & T2)

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uppose that the terminal speed of a particular sky diver is 150 km/h in the spread-eagle position and 320 km/h in the nosedive p

AysviL [449]

Answer:

4.55

Explanation:

The terminal speed of a diver is given by:

$v_t=\sqrt{\frac{2mg}{C\rho A} } \\\\Where\ m=mass\ of \ driver,d=acceleration\ due\ to\ gravity,C=drag\ \\coefficient,A=cross\ sectional\ Area.\\\\Therefore:\\\\A=\frac{2mg}{C \rho v_t^2} \\\\For\ area\ with\ terminal\ speed\ in\ spread\ angle\ position(v_s):\\\\A_s=\frac{2mg}{C \rho v_s^2} \\\\For\ area\ with\ terminal\ speed\ in\ nose\ dive\ position(v_n):\\\\A_n=\frac{2mg}{C \rho v_n^2}\\\\Therefore\ since\ g,m,C,\rho\ are\ constant:\\\\$

$\frac{A_s}{A_n}= \frac{\frac{2mg}{C \rho v_s^2}}{\frac{2mg}{C \rho v_n^2}}\\\\\frac{A_s}{A_n}= \frac{v_n}{v_s} \\\\v_n=320\ km/h,v_s=150\ km/h\\\\\frac{A_s}{A_n}=\frac{320^2}{150^2} =4.55$

4 0

3 years ago

Suppose that a rectangular toroid has 1,500 windings and a self-inductance of 0.060 H. If the height of the rectangular toroid i

V125BC [204]

Answer:

Current in the toroid will be $I=17.32\times 10^{-3}A$

Explanation:

We have given number of winding in rectangular toroid N = 1500

Self inductance of toroid L = 0.06 H

Magnetic energy stored in toroid $E=9\times 10^{-16}J$

We have to find the current in the toroid

Magnetic energy stored is equal to $E=\frac{1}{2}Li^2$

$9\times 10^{-6}=\frac{1}{2}\times 0.06\times I^2$

$I=17.32\times 10^{-3}A$

So current in the toroid will be $I=17.32\times 10^{-3}A$

5 0

4 years ago

What is the efficiency of an engine that exhausts 440 J of heat to a cold reservoir and receives 570 J of heat from a hot reserv

katrin2010 [14]

Answer:

Efficiency = 77%

Explanation:

Input energy = 570 J

Output energy = 440 J

To find the efficiency;

$Efficiency = \frac {Out-put \; energy}{In-put \; energy} * 100$

Substituting into the equation, we have;

$Efficiency = \frac {440}{570} * 100$

$Efficiency = 0.7719 * 100$

Efficiency = 77.19 ≈ 77%

Therefore, the efficiency of the engine is 77 percent.

7 0

3 years ago

Which of the following statements is INCORRECT based on our Ray Model of Light?

viva [34]

Answer:

a and c

Explanation:

a)light moves in a straight line and distracts when it encounters an obstacle forming a shadow

c)the light rays from that object or also tgose from the image must enter your eye for you to see that image

8 0

3 years ago

A sample of an ideal gas is slowly compressed to one-half its original volume with no change in pressure. If the original root-m

Zigmanuir [339]

Answer:

$V_{rms_f}=\sqrt{\frac{3PV}{2nM}} = \sqrt{\frac{1}{2}} \sqrt{\frac{3PV}{M}} = \frac{1}{\sqrt{2}} \sqrt{\frac{3PV}{M}} = \frac{1}{\sqrt{2}} V_{rms_i}$

So then the final answer on this case would be:

$\frac{V_{rms_i}}{\sqrt{2}}$

Explanation:

From the kinetic theory model of gases we know that the velocity rms (speed of gas molecules) is given by:

$V_{rms}= \sqrt{\frac{3RT}{M}}$ (1)

Where V represent the velocity

R the constant for ideal gases

T the temperature

M the molecular weight of the gas

We also know from the ideal gas law that $PV= nRT$

If we solve for T we got: $T = \frac{PV}{nR}$

For the initial state we can replace T into the equation (1) and we got:

$V_{rms_i}= \sqrt{\frac{3R (\frac{PV}{nR})}{M}} = \sqrt{\frac{3PV}{M}}$

For the final state we know that : $V_f = \frac{V}{2}$ And the pressure not change , so then the final velocity would be:

$V_{rms_f}= \sqrt{\frac{3R (\frac{P(V/2)}{nR})}{M}} = \sqrt{\frac{3P(V/2)}{M}}$

$V_{rms_f}=\sqrt{\frac{3PV}{2nM}} = \sqrt{\frac{1}{2}} \sqrt{\frac{3PV}{M}} = \frac{1}{\sqrt{2}} \sqrt{\frac{3PV}{M}} = \frac{1}{\sqrt{2}} V_{rms_i}$

So then the final answer on this case would be:

$\frac{V_{rms_i}}{\sqrt{2}}$

6 0

3 years ago