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mrs_skeptik [129]
2 years ago
6

A vertebra is subjected to a shearing force of 535 N. Find the shear deformation, taking the vertebra to be a cylinder 3.00 cm h

igh and 4.00 cm in diameter. The shear modulus for vertebrae is 8.00 × 1010 N/m2.
Physics
1 answer:
Ganezh [65]2 years ago
5 0
4.00 is the answer i think im not sure either 4.00 or 8.00
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A car is traveling at 20 meters/second and is brought to rest by applying brakes over a period of 4 seconds. What is its average
frez [133]
 (u) = 20 m/s 
(v) = 0 m/s 
<span> (t) = 4 s 
</span>
<span>0 = 20 + a(4) 

</span><span>4 x a = -20 
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so, the answer is <span>-5 m/s^2. or -5 meter per second</span>
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3 years ago
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Differentiate between concave and convex lens. Write your own answer
Katyanochek1 [597]

Answer:

A <em>concave</em><em> </em><em>lens</em><em> </em><em>is</em><em> </em><em>thinner</em><em> </em><em>at</em><em> </em><em>the</em><em> </em><em>cen</em><em>ter</em><em> </em><em>and </em><em>thick</em><em>er</em><em> </em><em>at</em><em> </em><em>the</em><em> </em><em>edges</em><em> </em><em>while</em><em> </em><em>a</em><em> </em><em>convex </em><em>lens </em><em>is</em><em> </em><em>thicker</em><em> </em><em>at</em><em> </em><em>the</em><em> </em><em>centre</em><em> </em><em>and</em><em> </em><em>thinner</em><em> </em><em>at</em><em> </em><em>the</em><em> edges</em><em>.</em>

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2 years ago
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1. A cyclist accelerates from 0 m/s to 9 m/s in 3 seconds. What is his<br>acceleration?​
Lapatulllka [165]

The cyclist accelerates from 0 m/s to 9 m/s in 3 seconds with an acceleration of 3 m/s².

Answer:

Explanation:

Acceleration exerted by an object is the measure of change in speed or velocity of that object with respect to time. So the initial and final velocities play a major role in determining the acceleration of the cyclist. As here the initial velocity of the cyclist is the speed at rest and that is given as 0 m/s. Then after 3 seconds, the velocity of the cyclist changes to 9 m/s.

Then acceleration = change in velocity/Time.

Acceleration = \frac{Change in velocity}{Time taken}

Acceleration = (9-0)/3=9/3=3 m/s².

So the cyclist accelerates from 0 m/s to 9 m/s in 3 seconds with an acceleration of 3 m/s².

3 0
3 years ago
1. A student lifts a box of books that weighs 185 N. The box is
aksik [14]

1)  148 J

When lifting an object, the work done on the object is equal to its change in gravitational potential energy. Mathematically:

W = \Delta U = (mg) \Delta h

where

mg is the weight of the object

\Delta h is the change in height

For the box in this problem,

mg = 185 N

\Delta h = 0.800 m

Substituting into the equation, we find:

W=(185)(0.800)=148 J

2) (a) 28875 J

The work done by a force applied parallel to the direction of motion of the object is given by

W=Fd

where

F is the magnitude of the force

d is the displacement

In this problem,

F = 825 N is the force applied by the two students together

d = 35 m is the displacement of the car

Substituting,

W=(825)(35)=28875 J

2) (b) 57750 J

As seen previously, the equation that gives the work done by the force is

W=Fd

We see that the work done is proportional to the magnitude of the force: therefore, if the force is doubled, then the work done is also doubled.

The work done previously was

W = 28875 J

Now the force is doubled, so the new work done will be

W' = 2(28875)=57750 J

3) 4.4 J

In this case, the force acting on the ball is the force of gravity, whose magnitude is:

F = mg

where

m = 0.180 kg is the mass of the ball

g=9.8 m/s^2 is the acceleration of gravity

Solving the equation,

F=(0.180)(9.8)=1.76 N

Now we find the work done by gravity using the same formula applied before:

W=Fd

where d = 2.5 m is the displacement of the ball. We can apply this version of the formula since the force is parallel to the displacement. Substituting,

W=(1.76)(2.5)=4.4 J

4) 595.2 kg

In this case, we have the work done on the box:

W = 7.0 kJ = 7000 J

And we also know the change in height of the box:

\Delta h = 1.2 m

As we stated in part a), the work done on the box is equal to its change in gravitational potential energy:

W=mg \Delta h

Solving for m, we find

m=\frac{W}{g \Delta h}

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m=\frac{7000}{(9.8)(1.2)}=595.2 kg

5) They do the same work

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W=mg \Delta h

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This means that the work done on the box depends only on its initial and final position, not on the path taken. The two men carry the box along different paths, however the reach at the end the same position, and they started from the same position: this means that the value of \Delta h is the same for both of them, so the work they have done is exactly the same.

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3 years ago
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vlada-n [284]

Wind speed and air temperature are used to calculate a windchill factor.

<u>Explanation:</u>

<u></u>

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3 years ago
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