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kakasveta [241]
2 years ago
11

Match the measuring instrument with the appropriate statement.

Physics
1 answer:
yaroslaw [1]2 years ago
7 0

<u>Answer:</u>

1A, 2E, 3C, 4F, 5B, 6D

<u>Explanation:</u>

  Tilt meter is a an equipment used to measure very small changes from vertical level. So it is a correct match for blank space in statement A.

  Richter Scale is a scale of measurement of earthquake strength, So it is a correct match for blank space in statement E.

 Mercalli Intensity Scale measures effect of an earthquake, like damage caused.So it is a correct match for blank space in statement C.

  Seismograph is earthquake wave measuring instrument. So it is a correct match for blank space in statement F.

  Correlation Spectrometer (C.O.S.P.E.C.) is used to measure sulfur dioxide content in smoke. So it is a correct match for blank space in statement B.

  Moment Magnitude Scale is an earthquake measuring scale for great earthquakes. So it is a correct match for blank space in statement D.


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A red ball, initially at rest, is simultaneously hit by a blue ball traveling from west to east at 5 m/s and a green ball travel
andrezito [222]

Answer:

C. Between North and West

Explanation:

Since all have equal masses and the red ball and green ball are moving in south and east direction, the blue ball would most likely be moving between the north and West direction.

8 0
2 years ago
At what condition does a body become weightless at equator
salantis [7]
Hope this is helpful <span>Weightlessness

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3 0
2 years ago
A rigid tank contains nitrogen gas at 227 °C and 100 kPa gage. The gas is heated until the gage pressure reads 250 kPa. If the a
aleksley [76]

Answer:

 T₂ =602  °C

Explanation:

Given that

T₁ = 227°C =227+273 K

T₁ =500 k

Gauge pressure at condition 1 given = 100 KPa

The absolute pressure at condition 1 will be

P₁ = 100 + 100 KPa

P₁ =200 KPa

Gauge pressure at condition 2 given = 250 KPa

The absolute pressure at condition 2 will be

P₂ = 250 + 100 KPa

P₂ =350 KPa

The temperature at condition 2 = T₂

We know that

\dfrac{T_2}{T_1}=\dfrac{P_2}{P_1}\\T_2=T_1\times \dfrac{P_2}{P_1}\\T_2=500\times \dfrac{350}{200}\ K\\

T₂ = 875 K

T₂ =875- 273 °C

T₂ =602  °C

5 0
2 years ago
Sarah is 14 years old and skips school twice a week without any written explanation. What can she be charged with?
harina [27]

Answer:

Too many people are unaware or indifferent to that.” Fines can cost up to $500 per truancy, due within 30 days unless a judge gives an extension. For many students and families, it's another debt they can't pay. And if fines aren't paid, they can convert into an arrest warrant when a student turns 17.

Explanation:

5 0
2 years ago
A slender uniform rod 100.00 cm long is used as a meter stick. Two parallel axes that are perpendicular to the rod are considere
Nataliya [291]

Answer:

The correct answer is D    I_{30} /I_{50} =   1.5

Explanation:

In this exercise the moment of inertia equation should be used

    I = ∫ r² dm

In addition to the parallel axis theorem

    I = I_{cm} + M D²

Where  I_{cm} is the moment of the center of mass, M is the total mass of the body and D the distance from this point to the axis of interest

Let's apply these relationships to our problem, the center of mass of a uniform rod coincides with its geometric center, in this case the rod is 1 m long, so the center of mass is in

    L = 100.00 cm (1m / 100 cm) = 1.0000 m

     x_{cm} = 50 cm = 0.50 m

Let's calculate the moment of inertia for this point, suppose the rod is on the x-axis and use the concept of linear density

    λ = M / L = dm / dx

    dm =  λ dx

Let's replace in the moment of inertia equation

    I = ∫ x² ( λ dx)

We integrate

    I =  λ x³ / 3

We evaluate between the lower limits x = -L/2 to the upper limit x = L/2

    I =  λ/3 [(L/2)³ - (-L/2)³] = lam/3  [L³/8 - (-L³ / 8)]

    I =  λ/3  L³/4

    I = 1/12  λ L³

Let's replace the linear density with its value

    I = 1/12  (M/L)  L³

    I = 1/12  M L²

Let's calculate with the given values

   I = 1/12  M 1²

   I = 1/12 M

This point is the center of mass of the rod

    Icm = I = 1/12 M  = 8.333 10-2 M

Now let's use the parallel axis theorem to calculate the moment of resection of the new axis, which is 0.30 m from one end, in this case the distance is

    D = x_{cm} - x

    D = 0.50 - 0.30

    D = 0.20  m

Let's calculate

   I_{30} =I_{cm} + M D²

   I_{30} = 1/12 M + M 0.202

   I_{30} = M (1/12 + 0.04)

   I_{30} = M 0.123

To find the relationship between the two moments of inertia, divide the quantities

  I_{30} / I_{50} = M 0.123 / (M 8.3 10-2)

   I_{30} /I_{50} = 1.48

The correct answer is d 1.5

6 0
2 years ago
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