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grandymaker [24]
3 years ago
8

Which of the phases below are able to be compressed? A) Liquid only. B) Liquid and Gas. C) Solid only. D) Gas only

Physics
2 answers:
noname [10]3 years ago
6 0
Answer - B
Liquid and Gas can be compressed because of the freely moving particles in them. Solids cannot be compressed as their structure is fixed.
amid [387]3 years ago
3 0
The phrase below that is able to be compressed is D. gas only.
Although it is possible to compress liquids, it requires a great deal of pressure to accomplish a little compression, which is why liquids and solids are often referred to as incompressible. 
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A simple generator is used to generate a peak output voltage of 25.0 V . The square armature consists of windings that are 7.0 c
4vir4ik [10]

Answer:

<h3>The 28 loops wound on the square armature</h3>

Explanation:

Peak output voltage \epsilon _{peak}  = 25 V

Area of square armature A = (7 \times 10^{-2} )^{2}  = 49 \times 10^{-4}

Magnetic field B = 0.490 T

Angular frequency \omega = 2\pi f = 2 \pi \times 60 = 120\pi

According to the law of electromagnetic induction,

     \epsilon _{peak} = NBA \omega

Where N = number of loops of wire.

  N = \frac{25}{49 \times 10^{-4} \times 0.49 \times 120\pi  }

  N = 27.6 ≅ 28

Thus, 28 loops of wire should be wound on the square armature.

6 0
3 years ago
Gravitational force between two objects increases as mass increases and distance
Tems11 [23]
Distance decreases.
6 0
3 years ago
Runner A is initially 5.7 km west of a flagpole and is running with a constant velocity of 8.9 km/h due east. Runner B is initia
Oksanka [162]

Answer:

B meet A 0.01 km east of flagpole

Explanation:

given data

distance A = 5.7 km  west

velocity V1 = 8.9 km/h

distance B = 4.5 km  east

velocity V2 = 7 km/h

to find out

How far runners from the flagpole, when paths cross

solution

we know A and B are 5.7 + 4.5 = 10.2 km apart

and we consider here B will run distance x km for meet

so time will be for B is

time B = distance / velocity

time B = x / 7     ...................1

and

for A distance for meet = ( 10.2 - x ) km

so time A = distance / velocity

time A = ( 10.2 - x )  / 8.9      .............2

now equating equation 1 and 2

time A = time B

x / 7  =   ( 10.2 - x )  / 8.9

x = 4.490

so distance of B run for meet is 4.490 km

so distance from the flagpole when their paths cross is 4.5 - 4.490 = 0.01 km

so B meet A 0.01 km east of flagpole

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3 years ago
What kind of radiation is stopped by a piece of paper
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