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lesya [120]
3 years ago
10

What is the ratio of the intensities of an earthquake P wave passing through the Earth and detected at two points 14 km and 49 k

m from the source?
Physics
1 answer:
Molodets [167]3 years ago
6 0

Answer:

\dfrac{I_1}{I_2}=12.25

Explanation:

r_1 = 14 km

r_2 = 49 km

Intensity of a wave is inversely proportional to distance

I\propto \dfrac{1}{r^2}

So,

\dfrac{I_1}{I_2}=\dfrac{r_2^2}{r_1^2}\\\Rightarrow \dfrac{I_1}{I_2}=\dfrac{49^2}{14^2}\\\Rightarrow \dfrac{I_1}{I_2}=12.25

The ratio of the intensities is \dfrac{I_1}{I_2}=12.25

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Two cars collide and come to a complete stop. where did all of their energy go?
MA_775_DIABLO [31]
Most of the energy will be absorbed by the materials that make up the cars, causing them to deform. The energy will also be converted into sound energy, causing a loud bang upon collision. Also, some energy will be converted to thermal energy, which will cause the cars to heat up slightly.
3 0
3 years ago
Please help asap asap<br> wdtwygcbhyeegdlu tcgbihjlnkdm;fnbkhjg hv
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Answer:

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6 0
3 years ago
A 0.73-m aluminum bar is held with its length parallel to the east-west direction and dropped from a bridge. Just before the bar
Tasya [4]

Answer:

A)  B = 5.4 10⁻⁵ T, B) the positive side of the bar is to the West

Explanation:

A) For this exercise we must use the expression of Faraday's law for a moving body

            fem = -  \frac{d \phi }{dt}

            fem = - \frac{d (B l y}{dt}= - B l v- d (B l y) / dt = - B lv

            B = - \frac{fem}{l \ v}

we calculate

             B = - 7.9 10⁻⁴ /(0.73 20)

             B = 5.4 10⁻⁵ T

B) to determine which side of the bar is positive, we must use the right hand rule

the thumb points in the direction of the rod movement to the south, the magnetic field points in the horizontal direction and the rod is in the east-west direction.

Therefore the force points in the direction perpendicular to the velocity and the magnetic field is in the east direction; therefore the positive side of the bar is to the West

4 0
3 years ago
A 7.5-cmcm-diameter horizontal pipe gradually narrows to 4.5 cmcm . When water flows through this pipe at a certain rate, the ga
tino4ka555 [31]

Answer

given,

diameter,d₁ = 7.5 cm

               d₂ = 4.5 cm

P₁ = 32 kPa

P₂ = 25 kPa

Assuming, we have calculation of flow in the pipe

using continuity equation

 A₁ v₁ = A₂ v₂

 π r₁² v₁ = π r₂² v₂

 v_1= \dfrac{r_2^2}{r_1^2} v_2

 v_1= \dfrac{2.25^2}{3.75^2} v_2

 v_1= 0.36 v_2

Applying Bernoulli's equation

 \Delta P = \dfrac{1}{2}\rho (v_2^2-v_1^2)

 P_1-P_2 = \dfrac{1}{2}\rho (v_2^2-(0.36 v_2)^2)

 32-25 = \dfrac{1}{2}1000\times v_2^2 (1 - 0.1269)

 v_2=\sqrt{\dfrac{2\times 7\times 10^3}{1000\times (0.8704)}}

 v_2=\sqrt{16.084}

       v₂ = 4.01 m/s

fluid flow rate

Q = A₂ V₂

Q = π (0.0225)²  x 4.01

Q = 6.38 x 10⁻³ m³/s

flow in the pipe is equal to 6.38 x 10⁻³ m³/s

4 0
3 years ago
A circular loop of wire of cross-sectional area 0.12 m2 consists of 200 turns, each carrying 0.50 A. It is placed in a magnetic
defon

Answer:

0.52 Nm

Explanation:

A = 0.12 m^2, N = 200, i = 0.5 A, B = 0.050 T

Angle between the plane of loop and magnetic field = 30 Degree

Angle between the normal of loop and the magnetic field = 90 - 30 = 60 degree

θ = 60°

Torque = N i A B Sinθ

Torque = 200 x 0.5 x 0.12 x 0.050 x Sin 60

Torque = 0.52 Nm

4 0
3 years ago
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