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3 years ago

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The velocity of the transverse waves produced by an earthquake is 5.05 km/s, while that of the longitudinal waves is 8.585 km/s.

A seismograph records the arrival of the transverse waves 56.4 s after that of the longitudinal waves. How far away was the earthquake? Answer in units of km.

1 answer:

sattari [20]3 years ago

8 0

Answer:

$d=691.71km$

Explanation:

The time lag between the arrival of transverse waves and the arrival of the longitudinal waves is defined as:

$t=\frac{d}{v_t}-\frac{d}{v_l}$

Here d is the distance at which the earthquake take place and $v_t, v_l$ is the velocity of the transverse waves and longitudinal waves respectively. Solving for d:

$t=d(\frac{1}{v_t}-\frac{1}{v_l})\\d=\frac{t}{\frac{1}{v_t}-\frac{1}{v_l}}\\d=\frac{56.4s}{\frac{1}{5.05\frac{km}{s}}-\frac{1}{8.585\frac{km}{s}}}\\d=691.71km$

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Answer:

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Explanation:

Given that,

Mass of block = 1.60 kg

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Distance = 6.55 m

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Coefficient of kinetic friction = 0.50

We need to calculate the amount of potential energy

Using formula of conservation of energy between point A and B

$U_{A}+k_{A}+w_{A}=U_{B}+k_{B}$

$U_{A}+0-fd=mgy+\dfrac{1}{2}mv^2$

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Put the value into the formula

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$U_{A}=88.8\ J$

Hence, The amount of potential energy that was initially stored in the spring is 88.8 J.

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Answer:

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Explanation:

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$\dot Q = \frac{2\cdot k\cdot L}{\ln \left(\frac{D_{o}}{D_{i}} \right)}\cdot (T_{i}-T_{o})$

Where:

$k$ - Thermal conductivity, measured in watts per meter-Celsius.

$L$ - Length of the cylinder, measured in meters.

$D_{i}$ - Inner diameter, measured in meters.

$D_{o}$ - Outer diameter, measured in meters.

$T_{i}$ - Temperature at inner surface, measured in Celsius.

$T_{o}$ - Temperature at outer surface, measured in Celsius.

Now we clear the thermal conductivity in the equation:

$k = \frac{\dot Q}{2\cdot L\cdot (T_{i}-T_{o})}\cdot \ln\left(\frac{D_{o}}{D_{i}} \right)$

If we know that $\dot Q = 40.8\,W$ , $L = 0.6\,m$ , $T_{i} = 50\,^{\circ}C$ , $T_{o} = 20\,^{\circ}C$ , $D_{i} = 0.01\,m$ and $D_{o} = 0.04\,m$ , the thermal conductivity of the biomaterial is:

$k = \left[\frac{40.8\,W}{2\cdot (0.6\,m)\cdot (50\,^{\circ}C-20\,^{\circ}C)}\right]\cdot \ln \left(\frac{0.04\,m}{0.01\,m} \right)$

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The thermal conductivity of the biomaterial is approximately 1.571 watts per meter-Celsius.

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castortr0y [4]

Answer:

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Explanation:

You apparently want the sum of forces ...

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Your angle reference is a bit unconventional, so we'll compute the components of the forces as ...

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This way, the 2nd quadrant angle that has a negative angle measure will have a positive y component.

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