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tankabanditka [31]
3 years ago
8

In printmaking, a(n) ______ is a surface on which a design is prepared before being transferred through pressure to a receiving

surface such as paper.
Physics
1 answer:
MaRussiya [10]3 years ago
6 0

Answer:Matrix

Explanation:

In print production, A matrix is like an object upon which a drawing has been created and which is then used to make a copy on a piece of paper, hence creating a print. The matrix is used with ink to hold the image that makes up the print

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What is the density of a 2 gallon milk jug that has a mass of 2.0 kg? Answer should be in g/ml.
kirill [66]

volume of milk is given as

V = 2 Gallon

we will convert it into mL unit

V = 2 Gal = 7570.82 ml

mass of the milk m = 2kg

m = 2000 g

now for the density we can use

\rho = \frac{m}{V}

\rho = \frac{2000}{7570.82}

\rho = 0.264 g/mL

<em>so the density is 0.264 g/mL for above sample</em>

8 0
4 years ago
If mr galans paces is equal to 1.8 m,how many paces would it take to get to the moon
Anettt [7]

Answer:

The number of paces it would take to get to the Moon is 213,555,556 paces

Explanation:

The given length of Mr Galan's paces = 1.8 m/pace

The distance from the Earth to the Moon is, 384,400 km = 384,400,000 m

Therefore, the number of paces, "n", it would take to get to the Moon from the Earth is given as follows;

n = (The distance from the Earth to the Moon)/(The length of each Mr Galan's paces)

∴ n = 384,400,000 m/(1.8 m/pace) = 213,555,556 paces

The number of paces it would take to get to the Moon = n = 213,555,556 paces

5 0
3 years ago
An electric pump has 2kW power . How much water will it lift every minute if the height is 10 m ?
valkas [14]

Answer:

Given that,

  • Power = 2000 W
  • time = 60 seconds
  • distance= 10m

Power = work done ÷ time

Here, since the movement is vertical, w = mgh

So,

Power = mgh÷t

2000 = (m × 9.8 ×10) ÷ 60

m = (2000 ×60) ÷98

m = 1224.5kg

3 0
3 years ago
One end of a metal rod is in contact with a thermal reservoir at 699. K, and the other end is in contact with a thermal reservoi
bulgar [2K]

Answer:

a)ΔS₁ = - 9.9 J/K

ΔS₂ = 69 J/K

b)The entropy change for the rod = 0 J/K

c)ΔS = 59.1 J/K

Explanation:

Given that

T₁ = 699 K

T₂= 101 K

Q= 6970 J

Change in entropy given as

\Delta S=\dfrac{Q}{T}

For 699 K:

\Delta S_1=\dfrac{Q}{T}

\Delta S_1=-\dfrac{6970}{699}

ΔS₁ = - 9.9 J/K  ( Negative because heat is leaving from the system)

For 101 K;

\Delta S_2=\dfrac{Q}{T}

\Delta S_2=\dfrac{6970}{101}

ΔS₂ = 69 J/K

The entropy change for the rod = 0 J/K

Entropy  change for the system

ΔS = ΔS₂  + ΔS₁

ΔS = 69 -9.9 J/K

ΔS = 59.1 J/K

8 0
3 years ago
Some material consisting of a collection of microscopic systems is kept at a high temperature, so that all excited states are po
Lelechka [254]

Answer:

The responses to this question can be defined as follows:

Explanation:

During energy exchange E=hv, electrodes spring through one orbit to another

Please find the image file in the attachment.

Its absorption layer comprises 0.3 eV, 0.5 eV., 0.8 eV, 2.0 eV, 2.5 eV again, as light passes via material at low temperature those lines absorbed in the strata called absorption stratum.

6 0
3 years ago
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