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zepelin [54]
2 years ago
5

The two blocks are connected by a light string that passes over a frictionless pulley with a negligible mass. The 2 kg block lie

s on a rough horizontal surface with a constant coefficient of kinetic friction 0.4. This block is connected to a spring with spring constant 4 N/m. The second block has a mass of 8 kg. The system is released from rest when the spring is unstretched, and the 8 kg block falls a distance h before it reaches the lowest point. The acceleration of gravity is 9.8 m/s2 . Note: When the 8 kg block is at the lowest point, its velocity is zero.
a. Calculate the falling distance h where the 8 kg blocks stops.
b. Consider the moment when the 8 kg block has descended by a distance of 21.168 m, where 21.168 m is less than h. At this moment calculate the sum of the kinetic energy for the two blocks. Answer in units of J.
Physics
1 answer:
frozen [14]2 years ago
5 0

Answer: YOU ARE A FAT NEGRO

Explanation:YOU ARE

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Allisa [31]

The statement that says "Psychological disorders can be very difficult to diagnose" is true about diagnosing psychological disorders.

<h2>What are psychological disorders?</h2>

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  • Mental disorders are psychiatric conditions that are expressed in a syndrome, verifiable from different diagnostic criteria.

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Therefore, we can conclude that a psychological disorder is an alteration in the mental balance of a person that requires specialized attention adapted to the characteristics of the dysfunction.

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3 0
2 years ago
A student in an undergraduate physics lab is studying Archimede's principle of bouyancy. The student is given a brass cylinder a
Readme [11.4K]

Answer:

V = 0.3724 m³

T = 27.836 N

Explanation:

Given :

m = 3.21 kg  , W= 3.21 * 9.81 m / s² = 31.4901 N

ρ = 8.62 g / cm ³  = 8620 kg / m³

V = m / ρ =  3.21 kg  /  8620 kg / m³

V = 0.3724 m³

when submerged the weight of brass cylinder is equal to the tension in string:

F =  (0.3724m³) * (1000 kg / m³) * (9.81 m/s²²) = 3.653 ≈ 3.65 N

T = 31.4901 N - 3.65 N  

T = 27.836 N

3 0
3 years ago
• A cable is suspended over a pulley. A 5.0 kg mass is attached to
Galina-37 [17]

Answer:

Tension= \frac{20g}{3}  (g=acceleration of gravity)

Explanation:

Given that,

A 5Kg and 10Kg are attached by a cable suspended over a pulley.

As 10Kg > 5Kg , the 10 kg mass accelerates down and the 5kg mass accelerates up, let it be a. Let the tension in the cable be T.

So, the equations of motion are

10g-T = 10a

T-5g=5a

Now adding them we get,

5g=15a

a=\frac{g}{3}

Substituting them back in the equation we get,

10g-10(\frac{g}{3} ) = T

T= \frac{20g}{3}

3 0
3 years ago
Calculate the separation between the two lowest levels for an O2 molecule in a one-dimensional container of length 5.0 cm. At wh
MariettaO [177]

Answer:

The separation between the two lowest levels =  1.24 * 10^{-39}J

The values of n where the energy of molecule reaches 1/2 kT at 300K = 2.2 * 10^{9}

The separation at this level = 1.8 * 10^{-30}J

Explanation:

Knowing the formula

En = \frac{n^{2} h^{2}  }{8 mL^{2} }

Mass of oxygen molecule

m (O2) = 32 amu * \frac{1.6605 * 10^{-27 kg} }{1 amu}

So the energy diference between the two lowest levels:

E2 - E1 = \frac{3h^{2} }{8mL^{2} }

E2 - E1 =  \frac{3 * (6.626 * 10^{-34} Js)^{2} } {8 * 32 amu * (\frac{1.6605 * 10^{-27 kg} }{1 amu})* (5*10^{-2})^{2}   } = 1.24 * 10^{-39}J

Now we should find n where the energy of molecule reaches 1/2 kT

En = \frac{n^{2} h^{2}  }{8 mL^{2} } = \frac{1}{2}kT

\frac{h^{2}  }{8 mL^{2} } = 4.13 * 10^{-14}J

n^{2} *  (4.13 * 10^{-14}J) = \frac{1}{2} (1.38 * 10^{-23}JK^{-1}) * 300K

n = 2.2 * 10^{9}

by the end is necessary to calculate the separation of the level

En - En-1 = (n^{2} - (n - 1)^{2}) * \frac{h^{2}  }{8 mL^{2} }

              = 1.8 * 10^{-30}J

4 0
2 years ago
A kayaker paddles at 4.0 m/s in a direction 30° south of west. He then turns and paddles at 3.7 m/s in a direction 20° west of s
user100 [1]
7.2 m/s 

49 degrees south of west

7 0
2 years ago
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