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Ostrovityanka [42]
3 years ago
8

As shown above an adhesive has been applied to contacting faces of two blocks so that the blocks interact with an adhesive force

that has a magnitude, Fadhesion A tension, T, is exerted by a wire attached to the upper blocks causing the blocks to remain at rest The two blocks have weights, Wbottom and Wtop. Which of the following must be true? (1 point) ♡ O Wtop=T OT = Wtop - Fadhesion O Wbottom Fadhesion O T = Wbottom + Wtop + Fadhesion​
Physics
1 answer:
rusak2 [61]3 years ago
8 0

The magnitude of the adhesive force allows the top block to remain

attached to the bottom when the blocks and the wire are balanced.

The option that must be true is; \mathbf{W_{bottom} = F_{adhesion}}

Reason:

The tension exerted by the wire attached to the top block = T

Magnitude of the adhesive = F_{adhesion}

Weight of the top block = W_{top}

Weight of the bottom block = W_{bottom}

Given that with the exertion of the tension, the two blocks remain at rest, we have;

  • T =  W_{top} + W_{bottom}

The adhesive causes the bottom block to remain attached to the top block, we have;

Therefore, the magnitude of the adhesive force adds the bottom weight, to the top, weight, which gives;

The magnitude of the adhesive force = The weight of the bottom block

Therefore;

  • W_{bottom} = F_{adhesion}

Learn more here:

brainly.com/question/18907970

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A rope is shaken and produces 2 waves each second. Calculate the time period of the rope waves.
motikmotik

Answer:

0.5 s

Explanation:

From the question given above, the following data were obtained:

Number of circle (n) = 2

Time (t) = 1 s

Period =?

Period of a wave is simply defined as the time taken to make one complete oscillation. Mathematically, it can be expressed as:

T = t / n

Whereb

T => is the period

t => is the space time

n => is the number of circle or oscillation.

With the above formula, we can obtain the period of the wave as follow:

Number of circle (n) = 2

Time (t) = 1 s

Period =?

T = t / n

T = 1 / 2

T = 0.5 s

Thus, the period of the wave is 0.5 s

7 0
2 years ago
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Answer: 100 units

Explanation:

4 0
2 years ago
Squids and octopuses propel themselves by expelling water. They do this by keeping water in a cavity and then suddenly contracti
Alex

Answer:

A) The speed of the water must be 8.30 m/s.

B) Total kinetic energy created by this maneuver is 70.12 Joules.

Explanation:

A) Mass of squid with water = 6.50 kg

Mass of water in squid cavuty = 1.55 kg

Mass of squid = m_1=6.50 kg- 1.55 kg=4.95 kg

Velocity achieved by squid = v_1=2.60 m/s

Momentum gained by squid = P=m_1v_1

Mass of water = m_2=1.55 kg

Velocity by which water was released by squid = v_2

Momentum gained by water but in opposite direction = P'=m_2v_2

P = P'

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v_2=\frac{m_1v_1}{m_2}=\frac{4.95 kg\times 2.60 m/s}{1.55 kg}=8.30 m/s

B) Kinetic energy does the squid create by this maneuver:

Kinetic energy of squid = K.E  =\frac{1}{2}m_1v_1^{2}

Kinetic energy of water = K.E' = \frac{1}{2}m_2v_2^{2}

Total kinetic energy created by this maneuver:

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=\frac{1}{2}\times 4.95 kg\times (2.60 m/s)^2+\frac{1}{2}\times 1.55 kg\times (8.30 m/s)^2=70.12 Joules

4 0
3 years ago
If the given wave has a frequency of 100 Hz, what frequency will the sixth harmonic have?
alukav5142 [94]

Answer:

600Hz

Explanation:

In electrical systems of alternating current, the harmonics are, as in acoustics, frequencies multiples of the fundamental working frequency of the system and whose amplitude decreases as the multiple increases. For example, if we have systems fed by the 50 Hz network, harmonics of 100, 150, 200, etc. may appear.

In our case having a fundamental wave of 100Hz, I can have harmonics of 200,300,400, ..., 600Hz

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