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Umnica [9.8K]
2 years ago
8

The cantilever beam AB has a rectangular cross section of 150 × 200 mm. Knowing that the tension in the cable BD is 10.8 kN and

neglecting the weight of the beam, determine the normal and shearing stresses at the three points indicated. (Round the final answers to two decimal places.) The normal stress at point a is MPa. The shearing stress at point a is MPa. The normal stress at point b is MPa. The shearing stress at point b is MPa. The normal stress at point c is MPa. The shearing stress at point c is MPa

Physics
1 answer:
tatyana61 [14]2 years ago
6 0

Answer:

Explanation:

Find attached the solution

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The table below shows data of sprints of animals that traveled 75 meters. At each distance marker, the animals' times were recor
mr_godi [17]

Answer:

Explanation:

Animal 1 because it takes 3s to go 25 meters 3.5s to go 50 meters and 5s to go 75 meters while the others take longer.

5 0
2 years ago
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A 6.0 kg box slides down an inclined plane that makes an angle of 39° with the horizontal. If the coefficient of kinetic frictio
dlinn [17]

Answer:

a = 4.72 m/s²  

Explanation:

given,

mass of the box (m)= 6 Kg

angle of inclination (θ) = 39°

coefficient of kinetic friction (μ) = 0.19

magnitude of acceleration = ?

box is sliding downward so,

F - f = m a                        

f is the friction force

m g sinθ - μ N = ma                        

m g sinθ - μ m g cos θ = ma            

a = g sinθ - μ g cos θ                    

a = 9.8 x sin 39° - 0.19 x 9.8 x cos 39°

a = 4.72 m/s²                                      

the magnitude of acceleration of the box down the slope is a = 4.72 m/s²  

3 0
2 years ago
A sample of tendon 3.00 cm long and 4.00 mm in diameter is found to break under a minimum force of 128 N. If instead the sample
Hatshy [7]

Answer:128 N

Explanation:

Sample of 3 cm and 4 mm diameter found to break under a minimum force of 128 N .

If sample is 1.5 cm long with same cross-sectional area then minimum force required to break is also 128 N because the applied force is same for any length and diameter of tendon.        

3 0
2 years ago
The initial velocity of a 4.0-kg box is 11 m/s, due west. After the box slides 4.0 m horizontally, its speed is 1.5 m/s. Determi
ankoles [38]

Answer:

F = - 59.375 N

Explanation:

GIVEN DATA:

Initial velocity = 11 m/s

final velocity = 1.5 m/s

let force be F

work done =  mass* F = 4*F

we know that

Change in kinetic energy = work done

kinetic energy = = \frac{1}{2}*m*(v_{2}^{2}-v_{1}^{2})

kinetic energy = = \frac{1}{2}*4*(1.5^{2}-11^{2}) = -237.5 kg m/s2

-237.5 = 4*F

F = - 59.375 N

7 0
2 years ago
Step 8: Observe How Changes in the Speed of the Bottle Affect Beanbag Height
lina2011 [118]

Answer:

When the speed of the bottle is 2 m/s, the average maximum height of the beanbag is <u>0.10</u>  m.

When the speed of the bottle is 3 m/s, the average maximum height of the beanbag is<u> 0.43</u>  m.

When the speed of the bottle is 4 m/s, the average maximum height of the beanbag is  <u>0.87</u> m.

When the speed of the bottle is 5 m/s, the average maximum height of the beanbag is  <u>1.25</u> m.

When the speed of the bottle is 6 m/s, the average maximum height of the beanbag is  <u>1.86</u> m.

Sorry for not answering early on! If anyone in the future needs help, I got these answers from 2020 egenuity, though I can't post the picture for proof. Stay Safe!

6 0
2 years ago
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