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Ira Lisetskai [31]
3 years ago
13

g A uniform ladder whose length is 4.4 m and whose weight is 330 N leans against a frictionless vertical wall. The coefficient o

f static friction between the level ground and the foot of the ladder is 0.46. What is the greatest distance the foot of the ladder can be placed from the base of the wall without the ladder immediately slipping

Physics
1 answer:
Damm [24]3 years ago
3 0

Answer:

Distance = 4.4 [m]

Explanation:

This problem can be easily solved using a static analysis of forces acting on the ladder, taking into account the respective distances. For easy understanding, a free body diagram should be made.

We perform a sum of force on the X-axis equal to zero, to find that the force exerted by the wall is equal to the friction force on the floor.

Then we perform a summation of forces on the Y axis, to determine that the normal force exerted by the floor is equal to the weight of the ladder.

We know that the friction force is equal to the product of normal force by the coefficient of friction.

In this way, by relating the friction force to the equations deduced above we can find the force exerted by the wall.

Then we make a summation of moments around the base point of the ladder, the equation realized can be seen in the attached image.

In the last analysis we can find the relationship between the horizontal and vertical distance of the ladder, with respect to the wall and the floor.

Then with the complementary analysis of the Pythagorean theorem we can find another additional equation.

The result of the greater distance is 4.4 [m]

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kondor19780726 [428]

Answer:

Answer is in the following attachment.

                                                                                                                                                                                                 

Explanation:

3 0
3 years ago
Determine the 3 standing waves for a 4 m length of rope.
strojnjashka [21]

Harmonics, Loop and Harmonic number

Hope this helps :)

7 0
4 years ago
A light ray in glass (n=1.5) hits the air-glass interface at an angle of 10 degrees from the normal. What angle from the normal
jekas [21]

Answer:

The angle from the normal is 15.1°.

Explanation:

We can find the angle by using Snell's law:

n_{1}sin(\theta_{1}) = n_{2}sin(\theta_{2})

Where:

n₁: is the first medium (glass) = 1.5

n₂: is the second medium (air) = 1.0

θ₁: is the first angle (in the glass) = 10°

θ₂: is the second angle (in the air) =?

\theta_{2} = arcsin(\frac{n_{1}sin(\theta_{1})}{n_{2}}) = arcsin(\frac{1.5*sin(10)}{1.0}) = 15.1 ^{\circ}

Therefore, the angle from the normal is 15.1°.

I hope it helps you!        

7 0
3 years ago
The device used to measure a masses of a body is kilogram . true or false​
olga nikolaevna [1]

Answer: false

Explanation:

While kilograms are the unit used to measure body mass, the device used is a scale.

Hope it helps :)

6 0
3 years ago
If a large housefly 2.4 m away from you makes a noise of 47.0 dB, what is the noise level (in dB) of 5400 flies at that distance
dem82 [27]

Answer:

the noise level of 5400 flies is equal to 84.32 dB

Explanation:

Noise made by house flies = 47 dB from the distance of 2.4 m

to calculate the noise of 5400 flies at a distance of 2.4 m.

the intensity of noise

  10 log_{10}(\dfrac{I}{I_0}) = 47 dB

now,  I' = 5400 I₀

=     10 log_{10}(\dfrac{5400 I}{I_0})

=    10 [log_{10}[5400] +log_{10}(\dfrac{I}{I_0}))      

= 10 ( 3.732 + 4.7 )                            

= 84.32 dB                                          

hence, the noise level of 5400 flies is equal to 84.32 dB

4 0
3 years ago
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