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exis [7]
3 years ago
5

How are earthquakes measured quantitatively

Physics
1 answer:
AlexFokin [52]3 years ago
4 0

Answer:Intensity: The severity of earthquake shaking is assessed using a descriptive scale – the Modified Mercalli Intensity Scale. Magnitude: Earthquake size is a quantitative measure of the size of the earthquake at its source. The Richter Magnitude Scale measures the amount of seismic energy released by an earthquake.

Explanation:

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​A piston–cylinder assembly contains 5.0 kg of air, initially at 2.0 bar, 30 oC. The air undergoes a process to a state where th
vlada-n [284]

Answer:

Explanation:

The process is isothermic,  as P V = constant .

work done = 2.303 n RT log P₁ / P₂

= 2.303 x 5 / 29 x 8.3 x 303  log 2 / 1 kJ

= 300.5k J

This energy in work done by the gas will come fro heat supplied as internal energy is constant due to constant temperature.

heat supplied  = 300.5k J

specific volume is volume per unit mass

v / m

pv = n RT

pv  = m / M  RT

v / m = RT / p M

specific volume = RT / p M

option B is correct.

5 0
3 years ago
A U-tube open at both ends is partially filled withwater. Oil
ArbitrLikvidat [17]

Answer:

a)  h_w = 0.02139 m , b)      v₁ = 9.74 m / s

Explanation:

For this exercise we use the pascal principle that states that the pressure at one point is the same regardless of body shape.

At the initial moment (before emptying the oil), we fix the point on the surface of the liquid, in this case the left and right sides are in balance.

      P₁ = P₂ = P₀

Now we add the 5 cm (h₂ = 0.05 m) of oil, in this case the weight of the oil creates an extra pressure that pushes the water, let's look for how much the water moved (h_w). The weight of oil added is equal to the weight of displaced water

      W_w = W_oil

      m_w g= m_oil g

Density is defined

      ρ = m / V

we replace

        ρ_w V_w =  ρ_oil ​​W_oil

       V = A h

       ρ_w A h_w =  ρ_oil ​​A h_oil

      h_w = h_oil  ρ_oil ​​/  ρ_w

Now let's analyze the pressure at the initial reference height for both sides two had in U

Right side

We have the atmospheric pressure, with its decrease due to the lower height, plus the oil pressure above the reference level

       h’= 0.05 cm - h_w

      P = (P₀ -  ρ_air g (0.05-h_w)) +  ρ_oil ​​g (0.05-h_w)

Left side

We have at the same point, the atmospheric pressure with its reduction due to the height change plus the water pressure

        P = (P₀ -  ρ_air h h_w) +  ρ_w g h_w

As we have the same point we can equalize the pressure

(P₀ -ρ_air g (0.05-h_w)) +ρ_oil ​​g (0.05-h_w) = (Po -ρ_air h h_w) +ρ_w g h_w

        ρ_air g (h_w - (0.05-h_w)) =  ρ_w g h_w - ρ_oil ​​g (0.05-h_w)

       - ρ_air g 0.05 = h_w g ( ρ_w +  ρ _oil) - rho_oil ​​g 0.05

       h_w g ( ρ_w +  ρ_oil) = g 0.05 ( ρ_air -  ρ_oil)

calculate

       h_w = 0.05 (ρ_ oil- ρ_air)  / ( ρ_w +  ρ_oil)

       h_w = 0.05 (750 - 1.29) / (1000-750)

      h_w = 0.02139 m

The amount that decreases the height on one side is equal to the amount that increases the other

b) cover the right side and blow the air on the left side, let's use Bernoulli's equation, where index 1 will be for the left side and index 2 for the right side

      P₁ + 1/2 ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂

Since they indicate that both sides are at the same height y₁ = y₂, the right side is protected from the wind speed therefore v₂ = 0, let's write the equation

      P₁ + ½ ρ_air v₁² = P₂

    v₁² = (P₂-P₁) 2 / ρ_air

Let's analyze the pressure on each side of the tube, since the new equilibrium height is the height that was added of oil distributed between the two tubes, bone

       h’= 2.5 cm = 0.025 m

     

       P₁ = P₀ - ρ_w g h’

      P₂ = P₀ - ρ_oil ​​g h ’

      P₂-P₁ = g h’ (rho_w ​​- Rho_oil)

We replace

     v₁² = 2g h’ (ρ_w ​​–ρ_oil) / ρ_air

calculate

    v₁² = 2 9.8 0.025 (1000 - 750) /1.29

    v₁ = √ 94.96

    v₁ = 9.74 m / s

3 0
4 years ago
Atoms of elements at the top of a group on the periodic table are smaller than the atoms of elements at the bottom of the group.
serg [7]
Well formation of metallic bond depends on free electrons.smaal sized atoms hold their electrons more firmly as compared to large size atoms ,this z due to distance of outer shell electrons by nucleus..in this way no of free electrons affect strength of metallic bond..smaal sized atoms release less free electrons..
8 0
4 years ago
Read 2 more answers
a horse began running due east and covered 25km in 4.0 hours what is the average velocity of the horse
Komok [63]

25km / 4hr = <em>6.25 km/hr east</em>

6 0
3 years ago
Read 2 more answers
(A) What is the maximum tension possible in a 1.00-millimeter-diameter nylon tennis racket string?
attashe74 [19]

Complete Question

(A) What is the maximum tension possible in a 1.00- millimeter-diameter nylon tennis racket string?

(B) If you want tighter strings, what do you do to prevent breakage: use thinner or thicker strings? Why? What causes strings to break when they are hit by the ball?

The  tensile  strength of the nylon string is  600*10^{6} \  N/m^2

Answer:

A

   T = 471.3 \  N

B

 To prevent breakage the thickness of the string is increased

  String breakage when the racket hit the ball is as a result of the string not being thick enough to withstand the increase in tension

Explanation:

From the question we are told that

     The  diameter is  d =  1.00 \ mm  =  0.001 \  m  

       The  tensile strength of the nylon string is \sigma =  600 *10^{6} \  N/m^2

  Generally the radius is mathematically evaluated as

     r=  \frac{d}{2}

=>    r =  \frac{0.001}{2}

=>     r =  0.0005 \  m

The cross sectional area is mathematically represented as

     A = \pi  r^2

=>   A =  3.142  *  (0.005)^2

=>    A =  7.855*10^{-7}\  m^2

Generally the tensile strength of nylon is mathematically represented as

      \sigma  = \frac{T}{ A }

Where T is the tension on the maximum tension on the string

 So  

           T =  \sigma  *  A

=>          T =  600*10^{6} *  7.855*10^{-7}

=>         T = 471.3 \  N

Form the equation above  we  see that

        T  \  \alpha \  A

So if the tension is  increased to prevent breakage the thickness of the string is increased(i. e the cross-sectional  area )

String breakage when the racket hit the ball is as a result of the string not being thick enough to withstand the increase in tension

8 0
4 years ago
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