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EastWind [94]
3 years ago
5

Jvhftfcticgvigybvhbjnjndjanvjisn

Physics
2 answers:
nirvana33 [79]3 years ago
4 0

Answer:

dnndmssnxnxdnskkaamzmzma nzz.

fenix001 [56]3 years ago
4 0
SJJSKESKIWKALWLSOWOWOWKSKWWKW
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Tâm ống dẫn đặt dưới đường phân giới giữa nước và thuỷ ngân h1 = 820mm, chênh lệch chiều cao cột thuỷ ngân h2 = 880mm (Hg = 13,
gayaneshka [121]

Answer:

P = 79993.43245

Explanation:

Được -

Vị trí tâm ống dẫn dưới đường thuỷ ngân h1 = 820 mm

Chênh lệch chiều cao của cột thủy ngân h2 = 880mm

Tỷ trọng của thủy ngân = 13,5951 g / cm3

P = mật độ * gia tốc do trọng lực a * h

Thay thế các giá trị đã cho, chúng ta nhận được -

P = 13,5951 g / cm3 * 980,665 cm / s2 * (88-82) cm

P = 79993.43245

8 0
3 years ago
The distance a wave moves from resting position
tekilochka [14]

Answer:

waving

Explanation:

it waves go to a beach and see water doesn't rest it waves

6 0
3 years ago
Read 2 more answers
A penny rides on top of a piston as it undergoes vertical simple harmonic motion with an amplitude of 4.0cm. If the frequency is
BlackZzzverrR [31]

Answer:

the penny loses contact at the piston's highest point.

f = 2.5 Hz

Explanation:

Concepts and Principles  

1- Newton's Second Law: The net force F on a body with mass m is related to the body's acceleration a by  

∑F = ma                                                          (1)  

2- The maximum transverse acceleration of a particle in simple harmonic motion is found in terms of the angular speed w and the amplitude A as follows:  

a_max = -w^2A                                                (2)  

3- The angular frequency w of a wave is related to the frequency f by:  

w = 2π f                                                              (3)  

Given Data  

- The amplitude of the piston is: A = (4.0 cm) ( 1/ 100 cm)=  0.04 m.  

- The frequency of oscillation of the piston is steadily increased.

Required Data

<em>In part (a), we are asked to determine the point at which the penny first loses contact with the piston.  </em>

<em>In part (b), we are asked to determine the maximum frequency for which the penny just barely remains in place for a full cycle.  </em>

Solution  

(a)  

The free-body diagram in Figure 1 shows the forces acting on the penny; mi is the gravitational force exerted by the Earth on the penny andrt is the normal contact force exerted by the piston on the penny.  

figure 1 is attached

Apply Newton's second law from Equation (1) in the vertical direction to the penny:  

∑F_y -mg= ma        

Solve for n=m(g+a) The penny loses contact with the surface of the oscillating piston when the normal force n exerted by the piston is zero. So  

0 = m(g + a)

a = —g  

Therefore, the penny loses contact with the piston when the piston starts accelerating downwards. The piston first acceleratesdownward at its highest point and hence the penny loses contact at the piston's highest point.

(b)  

The maximum acceleration of the penny at the highest point of the piston is found from Equation (2):  

a = —w^2A  

where a = —g at the highest point. So  

g = w^2A  

Solve for w:  

w =√g/A

Substitute for w from Equation (3):

2πf =  √g/A

Solve for f :  

f = 1/2π√g/A

Substitute numerical values:  

f = 1/2π√9.8 m/s^2/0.04

f = 2.5 Hz

6 0
3 years ago
Someone please help on this? I just need the diagram to be drawn.
gregori [183]

Well the diagram would look like the water cycle I think


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A century ago, swords were made from various metals, especially steel. The property that makes steel a good choice for sword mak
8_murik_8 [283]
Hardness and strength
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3 years ago
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