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Alexxandr [17]
3 years ago
12

Technician A says that shop air pressure usually ranges from 100–150 psi. Technician B says shop air pressure is around 300 psi.

Who is correct?
Physics
1 answer:
mr Goodwill [35]3 years ago
5 0

Answer:

technician A.

Explanation:

Commonly,  air pressure rating of standard shop air compressor's is between the range of 100 psi and 150 psi. And over the years since, the plant pressure creep's up , the level of pressure increases from 100 psi to at max 150 psi but not more than that. Therefore, in the given question technician A is correct and technician B is incorrect.

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In a Young's double-slit experiment the separation distance y between the second-order bright fringe and the central bright frin
atroni [7]

Answer:

Y = 0.0254 m = 25.4 mm

Explanation:

The formula for the fringe spacing in Young's Double-slit experiment is given by the following formula:

Y = \frac{\lambda L}{d}

where,

Y = fringe spacing = 0.0176 m

λ = wavelength = 425 nm = 4.25 x 10⁻⁷ m

L = Distance between screen and slits

d = slit separation

Therefore,

\frac{0.0176\ m}{4.25\ x\ 10^{-7}\ m} = \frac{L}{d}\\\\\frac{L}{d} =  41411.76

Now, for:

λ = 614 nm = 6.14 x 10⁻⁷ m

Y = (6.14\ x\ 10^{-7}\ m)(41411.76)\\

<u>Y = 0.0254 m = 25.4 mm</u>

8 0
3 years ago
Two long, straight parallel wires 8.2 cm apart carry currents of equal magnitude I. The parallel wires repel each other with a f
o-na [289]

Answer:

36.22 mA

Explanation:

i1 = I , i2 = I, d = 8.2 cm = 0.082 m

Force per unit length = 3.2 nN/m = 3.2 x 10^-9 N/m

μo = 4 π × 10^-7 Tm/A

The formula for the force per unit length between the two wires is given by

F = μo / 4π x (2 i1 x i2) / d

3.2 x 10^-9 = 10^-7 x 2 x I^2 / 0.082

I = 0.0362 A = 36.22 mA

4 0
3 years ago
The impulse experienced by a body is equivalent to the body’s change in?
vovangra [49]
<span>impulse =force*time=mass*acceleration*time=mass*... in momentum , I hope this helps you out!! Also have an amazing day and good luck on any further work !!!

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5 0
3 years ago
Consider the space between a point charge and the surface of a neutral spherical conducting shell. If the charge sits at the cen
Furkat [3]

Answer:

True

Explanation:

If a thin, spherical, conducting shell carries a negative charge, We expect the excess electrons to mutually repel one another, and, thereby, become uniformly distributed over the surface of the shell. The electric field-lines produced outside such a charge distribution point towards the surface of the conductor, and end on the excess electrons. Moreover, the field-lines are normal to the surface of the conductor. This must be the case, otherwise the electric field would have a component parallel to the conducting surface. Since the excess electrons are free to move through the conductor, any parallel component of the field would cause a redistribution of the charges on the shell. This process will only cease when the parallel component has been reduced to zero over the whole surface of the shell

According to Gauss law

∅ = EA =-Q/∈₀

Where ∅  is the electric flux through the gaussian surface and E is the electric field strength

If the gaussian surface encloses no charge, since all of the charge lies on the shell, so it follows from Gauss' law, and symmetry, that the electric field inside the shell is zero. In fact, the electric field inside any closed hollow conductor is zero

8 0
2 years ago
I'll mark brainliest
irga5000 [103]

Answer:

A) 35 ft

B) 5 ft

C) Net displacement = distance covered by the dog to retrieve the stick - distance covered before the dog starts chewing the stick

Explanation:

A) Total distance covered by the dog = 20 + 15

                                  = 35 ft

B) Since the other distance covered by the dog before chewing the stick, after the retrieval, was in an opposite direction to the initial direction, then;

total displacement of the dog = 20 - 15

                                  = 5 ft

C) Net displacement = distance covered by the dog to retrieve the stick + distance covered before the dog starts chewing the stick

But, displacement involves a specified direction. The distance covered before the dog starts chewing the stick was in an opposite direction to the initial direction.

Thus,

Net displacement = distance covered by the dog to retrieve the stick - distance covered before the dog starts chewing the stick

7 0
2 years ago
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