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LiRa [457]
2 years ago
14

THIS IS MY EXAM HURRY PLS

Physics
2 answers:
Galina-37 [17]2 years ago
8 0

Answer:

haha sucks for you, hope this helps

Explanation:

tangare [24]2 years ago
5 0

Answer:

elements in the same column have the same number of neutrons. elements with similar mass are placed in the same column.

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The SI unit for pressure is pascal (Pa). One hectopascal would equal how many pascals?
Marat540 [252]
1 hectopascal (hPa) is equivalent to 100 Pa
3 0
3 years ago
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PLEASE HELP ASAP!!!!!! <br> What should each experiment only have one of? <br> variable or constant
slega [8]

Answer:

the answer should be a constant

hope this helps!!

3 0
3 years ago
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Which statement describes a digital signal used to store information?
velikii [3]

Answer:

D. It is easy to copy.

Explanation:

It is not B, or A, nor C.

8 0
2 years ago
If a sample emits 2000 counts per second when the detector is 1 meter from the sample, how many counts per second would be obser
Alona [7]

Answer:

<h2><em>6000 counts per second</em></h2>

Explanation:

If a sample emits 2000 counts per second when the detector is 1 meter from the sample, then;

2000 counts per second = 1 meter ... 1

In order to know the number of counts per second that would be observed when the detector is 3 meters from the sample, we will have;

x count per second = 3 meter ... 2

Solving the two expressions simultaneously for x we will have;

2000 counts per second = 1 meter

x counts per second = 3 meter

Cross multiply to get x

2000 * 3 = 1* x

6000 = x

<em></em>

<em>This shows that 6000 counts per second would be observed when the detector is 3 meters from the sample</em>

5 0
3 years ago
Air contained in a rigid, insulated tank fitted with a paddle wheel, initially at 300 K, 2 bar, and a volume of 3 m3, is stirred
madam [21]

Answer:

a)P₂ =4 bar

b)W= - 1482.48 KJ

It means that work done on the system.

c)S₂ - S₁ = 3.42 KJ/K

Explanation:

Given that

T₁ = 300 K   ,V₁ = 3 m³  ,P₁=2 bar

T₂ = 600 K ,V₂=V₁ 3 m³

Given that tank is rigid and insulated.It means that volume of the gas will remain constant.

Lets take the final pressure = P₂

For ideal gas  P V = m R T

\dfrac{P_2}{P_1}=\dfrac{T_2}{T_1}

P_2=\dfrac{T_2}{T_1}\times P_1

P_2=\dfrac{600}{300}\times 2

P₂ =4 bar

Internal energy

ΔU = m Cv ΔT

Cv=0.71 KJ/kg.k for air

m=\dfrac{PV}{RT}

m=\dfrac{200\times 3}{0.287\times 300}\ kg

m= 6.96 kg

ΔU= 6.96 x 0.71 x (600 - 300)

ΔU=1482.48 KJ

From first law

Q= ΔU + W

Q= 0  Insulated

W = - ΔU

W= - 1482.48 KJ

It means that work done on the system.

Change in the entropy

S_2-S_1=mC_v\ \ln\dfrac{T_2}{T_1}

S_2-S_1=6.96\times 0.71\ \ln\dfrac{600}{300}

S₂ - S₁ = 3.42 KJ/K

 

5 0
3 years ago
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