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3 years ago

How can we fill ink inside a pen?

Physics

2 answers:

Novay_Z [31]3 years ago

7 0

hi here is your answer hope it helps or then sry

Explanation:

To fill, take off the blind cap on the end of the pen. Push and hold the button down, compressing the sac. Dip the nib into the ink, then release the button to fill the pen.

Tanya [424]3 years ago

4 0

Answer:

Open the cap and fill the ink ...

As simple as me

Explanation:

mark me as brainliest

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Post WWII Germany was divided into four zones with the Americans occupying one zone and all of the following countries occupying

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Answer:

b. Italy

Explanation:

Italy had sided with germany and lost in world war 2

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2 years ago

At a rock concert, the sound intensity 1.0 m in front of the bank of loudspeakers is 0.10 W/m². A fan is 30 m from the loudspeak

Klio2033 [76]

To solve this problem we will apply the concepts related to the Area, the power and the proportionality relationships between intensity and distance.

The expression for sound power is,

$P = AI$

Here,

A = Area

I = Intensity

P = Power

At the same time the area can be written as,

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

Now the intensity is inversely proportional to the square of the distance from the source, then

$I \propto \frac{1}{r^2}$

The expression for the intensity at different distance is

$\frac{I_1}{I_2}= \frac{r^2_2}{r_1^2}$

Here,

$I_1$ = Intensity at distance 1

$I_2$ = Intensity at distance 2

$r_1$ = Distance 1 from light source

$r_2$ = Distance 2 from the light source

If we rearrange the expression to find the intensity at second position we have,

$I_2 = I_1 (\frac{r_1^2}{r_2^2})$

If we replace with our values at this equation we have,

$I_2 = (0.10W/m^2)(\frac{1.0m^2}{30.0m^2})$

$I_2 = 1.11*10^{-4} W/m^2$

Now using the equation to find the area we have that

$A = \frac{\pi (8.4*10^{-3}m)^2}{4}$

$A = 5.5*10^{-5}m^2$

Finally with the intensity and the area we can find the sound power, which is

$P = AI$

$P = (5.5*10^{-5}m^2)(1.11*10^{-4}W/m^2)$

$P = 6.1*10^{-9}J/s$

Power is defined as the quantity of Energy per second, then

$E = 6.1*10^{-9}J$

8 0

3 years ago

Problem 1: Three beads are placed along a thin rod. The first bead, of mass m1 = 24 g, is placed a distance d1 = 1.1 cm from the

Svet_ta [14]

Answer:

b) $x_{cm}$ = 4.88 cm , c) $x_{cm}$ ’= 1/M (m₁ d₁ + m₃ d₃) and d)

$x_{cm}$ ’= 1.88 cm

Explanation:

The definition of mass center is

$x_{cm}$ = 1/M ∑ xi mi

Where mi, xi are the mass and distance from an origin for each mass and M is the total mass of the object.

Part b

Apply this equation to our case.

Body 1

They give us the mass (m₁ = 24 g) and the distance (d₁ = 1.1 cm) from the origin at the far left

Body 2

They give us the mass (m₂ = 12.g) and the distance relative to the distance of the body 1, let's look for the distance from the left end (origin)

D₂ = d₁ + d₂

D₂ = 1.1 + 1.9

D₂ = 3.0 cm

Body 3

Give the mass (m₃ = 56 g) and the position relative to body 2, let's find the distance relative to the origin

D₃ = D₂ + d₂

D₃ = 3.0 + 3.9

D₃ = 6.9 cm

With this data we substitute and calculate the center of mass

M = m₁ + m₂ + m₃

M = 24 + 12 + 56

M = 92 g

$x_{cm}$ = 1/92 (1.1 24 + 3.0 12 + 6.9 56)

$x_{cm}$ = 1/92 (448.8)

$x_{cm}$ = 4,878 cm

$x_{cm}$ = 4.88 cm

This distance is from the left end of the bar

Par c)

In this case we are asked for the same calculation, but the reference system is in the center marble, we have to rewrite the distance with the reference system in this marble.

Body 1

It is at d1 = -1.9 cm

It is negative for being on the left and the value is the relative distance of 1 to 2

Body 2

d2 = 0 cm

The reference system for her

Body 3

d3 = 3.9 cm

Positive because that is to the left of the reference system and is the relative distance between 2 and 3

Let's write the new center of mass (xcm')

$x_{cm}$ ’= 1/M (m₁ d₁ + m₂ d₂ + m₃ d₃)

$x_{cm}$ ’= 1/M (m₁ d₁ + m₃ d₃)

Part d) Let's calculate the value of the center of mass

$x_{cm}$ ’= 1/92 ((24 (-1.9) +56 3.9)

$x_{cm}$ ’= 1/92 (172.8)

$x_{cm}$ ’= 1.88 cm

This distance is to the right of the central marble

3 0

3 years ago

How can we fill ink inside a pen?​

How can we fill ink inside a pen?