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lutik1710 [3]
3 years ago
8

A cup slides off a 1.1\,\text m1.1m1, point, 1, start text, m, end text high table with a speed of 1.3\,\dfrac{\text m}{\text s}

1.3 s m ​ 1, point, 3, start fraction, start text, m, end text, divided by, start text, s, end text, end fraction to the right. We can ignore air resistance. What was the cup's horizontal displacement during the fall?
Physics
1 answer:
Tju [1.3M]3 years ago
8 0

Answer:

<h3>0.145m</h3>

Explanation:

Using the equation of motion formula y = ut + 1/2gt² where;

y is the horizontal displacement

u is the initial velocity

g is the acceleration due to gravity

t is the time

To calculate the horizontal displacement, we need to first get the time t. Using the equation:

v = u+gt

1.3² = 0+(9.8)t

1.69 = 9.8t

t = 1.69/9.8

t = 0.172s

Substituting the time into the equation above to get y

y = 0+1/2(9.8)(0.172)²

y = 4.905(0.029584)

y = 0.145m

Hence the cup's horizontal displacement during the fall is 0.145m

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1) D

2) D.) Greater than \theta_c

Explanation:

1)

The phenomenon of total internal reflection occurs when a ray of light hitting the interface between two mediums is totally reflected back into the original medium, therefore no refraction into the second medium occurs.

This phenomenon occurs only if two conditions are satisfied:

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In picture 1, we have 4 different diagrams. In the diagrams:

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Total internal reflection occurs when there is no refraction, therefore when there is no green arrow: this occurs only in figure D, so this is the correct option. (in figure C, there is a refracted ray but it is parallel to the interface: this condition occurs when the angle of incidence is exactly equal to the critical angle, however in this problem, the angle of incidence is greater than the critical angle, so the correct option is D)

2)

As we stated in problem 1), total internal reflection occurs when the angle of incidence is equal or greater than the critical angle. Therefore in this case, the angle of incidence must be

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IRISSAK [1]

Answer:

43.83 N

Explanation:

Given that,

The mass of an object, m = 34.4 kg

The coefficients of static and kinetic frictions for plastic on wood are 0.53 and 0.40, respectively.

The force of static friction,

F_s=\mu_smg\\\\F_s=0.53\times 34.4\times 9.8\\\\F_s=178.67\ N

The force of kinetic friction,

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Net force acting on the object is :

F = 178.67-134.84

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