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Hatshy [7]
3 years ago
10

Determine the tension developed in cables OD and OB and the strut OC, required to support the 500-lb crate. The spring OA has an

unstretched length of 0.2 ft and a stiffness of kOA

Physics
1 answer:
natali 33 [55]3 years ago
6 0

Answer:Fob= 86.2 lb, Foc= 175 lb, Fod= 506 lb

Explanation:firstly we need to get the spring stretch by subtracting the unstretched length from 1. The result will be (1-0.2=0.8ft). We then get the spring force by multiplying kOA by 0.8ft. therefore the spring force will be {350x0.8}= 280 lb. After that we can resolve each force as shown in the attachment.

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Work done in taking charge from one point of a conductor to is another point is called ​
Yuliya22 [10]

Answer:

⁸

Explanation:

electric potential

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6 0
3 years ago
A hockey player skates with an average speed of 8.25 m/s. The distance the player will travel in 5.60 s is __________ .
OverLord2011 [107]
<h2>Greetings!</h2>

To find this value, you need to remember the speed formula:

3 = 6 / 2

Speed = distance ÷ time

Rearrange to make distance the subject:

Distance = speed * time

Simply plug these values into this:

5.6 * 8.25 = 46.2

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7 0
3 years ago
If anyone here can help me with physics please comment. its about coulombs law
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I may be able to help you with This subject
3 0
3 years ago
Blood in a carotid artery carrying blood to the head is moving at 0.15 m/s when it reaches a section where plaque has narrowed t
sp2606 [1]

Answer:

26.9 Pa

Explanation:

We can answer this question by using the continuity equation, which states that the volume flow rate of a fluid in a pipe must be constant; mathematically:

A_1 v_1 = A_2 v_2 (1)

where

A_1 is the cross-sectional area of the 1st section of the pipe

A_2 is the cross-sectional area of the 2nd section of the pipe

v_1 is the velocity of the 1st section of the pipe

v_2 is the velocity of the 2nd section of the pipe

In this problem we have:

v_1=0.15 m/s is the velocity of blood in the 1st section

The diameter of the 2nd section is 74% of that of the 1st section, so

d_2=0.74d_1

The cross-sectional area is proportional to the square of the diameter, so:

A_2=(0.74)^2 A_1=0.548 A_1

And solving eq.(1) for v2, we find the final velocity:

v_2=\frac{A_1 v_1}{A_2}=\frac{A_1 (0.15)}{0.548 A_1}=0.274 m/s

Now we can use Bernoulli's equation to find the pressure drop:

p_1 + \frac{1}{2}\rho v_1^2 = p_2 + \frac{1}{2}\rho v_2^2

where

\rho=1025 kg/m^3 is the blood density

p_1,p_2 are the initial and final pressure

So the pressure drop is:

p_1 - p_2 = \frac{1}{2}\rho (v_2^2-v_1^2)=\frac{1}{2}(1025)(0.274^2-0.15^2)=26.9 Pa

8 0
3 years ago
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