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1 year ago

7

If you will be rinsing your regulator after removing it from the cylinder, you must make sure that the is firmly i

n place. Select one: Mouthpiece plug Alternate-air-source retainer Dust cap None of the above

1 answer:

Rudik [331]1 year ago

5 0

If you will be rinsing your regulator after removing it from the cylinder, you must make sure that the dust cap is firmly in place.

<h3>What is Dust cap?</h3>

A dust cap is a gently curved dome mounted either in concave or convex orientation over the central hole of most loudspeaker diaphragms.

Thus, if you will be rinsing your regulator after removing it from the cylinder, you must make sure that the dust cap is firmly in place.

Learn more about dust cap here: brainly.com/question/10723016

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A 2.1 ✕ 103-kg car starts from rest at the top of a 5.9-m-long driveway that is inclined at 19° with the horizontal. If an avera

Lina20 [59]

Answer:

3.9 m/s

Explanation:

We are given that

Mass of car,m= $2.1\times 10^3 kg$

Initial velocity,u=0

Distance,s=5.9 m

$\theta=19^{\circ}$

Average friction force,f= $4.0\times 10^3 N$

We have to find the speed of the car at the bottom of the driveway.

Net force, $F_{net}=mgsin\theta-f=2.1\times 10^3\times 9.8sin19-4.0\times 10^3$

Where $g=9.8 m/s^2$

Acceleration, $a=\frac{F_{net}}{m}=\frac{2.1\times 10^3\times 9.8sin19-4.0\times 10^3}{2.1\times 10^3}$

$v=\sqrt{2as}$

$v=\sqrt{2\times \frac{2.1\times 10^3\times 9.8sin19-4.0\times 10^3}{2.1\times 10^3}\times 5.9}$

v=3.9 m/s

7 0

3 years ago

At a distance D from a very long (essentially infinite) uniform line of charge, the electric field strength is 1000 N/C. At what

Alla [95]

Answer:

The correct option is (a).

Explanation:

We know that, the E is inversely proportional to the distance as follows :

$E=\dfrac{k}{d^2}$

We can write it as follows :

$\dfrac{E_1}{E_2}=(\dfrac{d_2}{d_1})^2$

Put all the values,

$\dfrac{1000}{2000}=(\dfrac{d_2}{d})^2\\\\\sqrt{\dfrac{1000}{2000}}=(\dfrac{d_2}{D})\\\\0.7071=\dfrac{d_2}{d}\\\\d_1=0.7071D\\\\d_1=\dfrac{D}{\sqrt2}$

So, the correct option is (a).

5 0

2 years ago

Using a density of air to be 1.21kg/m3, the diameter of the bottom part of the filter as 0.15m (assume circular cross-section),

salantis [7]

Answer:

The drag coefficient is $D_z = 1.30512$

Explanation:

From the question we are told that

The density of air is $\rho_a = 1.21 \ kg/m^3$

The diameter of bottom part is $d = 0.15 \ m$

The power trend-line equation is mathematically represented as

$F_{\alpha } = 0.9226 * v^{0.5737}$

let assume that the velocity is 20 m/s

Then

$F_{\alpha } = 0.9226 * 20^{0.5737}$

$F_{\alpha } = 5.1453 \ N$

The drag coefficient is mathematically represented as

$D_z = \frac{2 F_{\alpha } }{A \rho v^2 }$

Where

$F_{\alpha }$ is the drag force

$\rho$ is the density of the fluid

$v$ is the flow velocity

A is the area which mathematically evaluated as

$A = \pi r^2 = \pi \frac{d^2}{4}$

substituting values

$A = 3.142 * \frac{(0.15)^2}{4}$

$A = 0.0176 \ m^2$

Then

$D_z = \frac{2 * 5.1453 }{0.0176 * 1.12 * 20^2 }$

$D_z = 1.30512$

3 0

2 years ago

HELP PLEASE BOYLES LAW

kogti [31]

Answer:

3 L

Explanation:

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

Initial volume (V₁) = 2 L

Initial pressure (P₁) = 0.75 atm

Final pressure (P₂) = 0.5 atm

Final volume (V₂) =?

Using the Boyle's law equation, the new volume (i.e final volume) of the Ne gas can be obtained as:

Initial volume (V₁) = 2 L

Initial pressure (P₁) = 0.75 atm

Final pressure (P₂) = 0.5 atm

Final volume (V₂) =?

P₁V₁ = P₂V₂

0.75 × 2 = 0.5 × V₂

1.5 = 0.5 × V₂

Divide both side by 0.5

V₂ = 1.5 / 0.5

V₂ = 3 L

Thus, the new volume of the Ne gas is 3 L

7 0

3 years ago

Kevin jumps straight up in the air to a height of 1 meter.At the top of his jump, he has potential energy of 1,000 joules.Answer

Llana [10]

Gravitational potential energy can be given by the equation
PE = mgh
where m is the mass,
g is the gravitational constant 9.81 or 10 depending on rounding
and h is the height

well weight is a force equiavlent to
W= m*g

so comparing that to the potential energy equation, divide the potential energy by the height and you will get weight in Newtons

3 0

2 years ago