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3 years ago

If the wave is detected 12.5 minutes after the earthquake, estimate the distance from the detector to the site of the quake

Physics

1 answer:

umka2103 [35]3 years ago

5 0

Answer:

Remember the relation:

Speed*Time = Distance.

We can estimate that the speed at which an earthquake "moves", in the surface, is:

S = 6km/s (this is a low estimation actually)

Then if the wave is detected 12.5 minutes after the earthquake, we know that it traveled for 12.5 minutes before reaching the detector.

So we know the speed of the wave and the time it took to reach the detector, then we can use the equation:

Speed*Time = Distance.

to find the distance.

First, we should write the time in seconds

we know that:

1 min = 60 s

then:

12.5 min = 12.5*(60 s) = 750 s

Then, the wave traveled with a speed of 6 km/s for 750 seconds until it reached the detector, then the distance that it traveled is:

(6km/s)*750s = 4500 km

The distance between the detector and the site of the quake is around 4500 km.

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A 0.85 N force exists between a 7.1 * 10 ^ - 6 * C charge 5.4 m away. What is the magnitude of the second charge ? Please show w

katovenus [111]

Answer:

Explanation:

Force between charge is given by the following expression

F = k Q₁ Q₂ / R² , k = 9 x 10⁹ , Q₁ and Q₂ are charges , R is distance between charges .

Putting the given values ,

.85 = 9 x 10⁹ x 7.1 x 10⁻⁶ x Q₂ / 5.4²

Q₂ = .85 x 5.4² / (9 x 10⁹ x 7.1 x 10⁻⁶ )

= .38788 x 10⁻³ C .

= 387.88 x 10⁻⁶ C .

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3 years ago

A parallel plate capacitor creates a uniform electric field of and its plates are separated by . A proton is placed at rest next

zalisa [80]

Complete Question

A parallel plate capacitor creates a uniform electric field of 5 x 10^4 N/C and its plates are separated by 2 x 10^{-3}'m. A proton is placed at rest next to the positive plate and then released and moves toward the negative plate. When the proton arrives at the negative plate, what is its speed?

Answer:

$V=1.4*10^5m/s$

Explanation:

From the question we are told that:

Electric field $B=1.5*10N/C$

Distance $d=2 x 10^{-3}$

At negative plate

Generally the equation for Velocity is mathematically given by

$V^2=2as$

Therefore

$V^2=\frac{2*e_0E*d}{m}$

$V^2=\frac{2*1.6*10^{-19}(5*10^4)*2 * 10^{-3}}{1.67*10^{-28}}$

$V=\sqrt{19.2*10^9}$

$V=1.4*10^5m/s$

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3 years ago

What is the energy per photon absorbed during the transition from n = 2 to n = 3 in the hydrogen atom?

adelina 88 [10]

Answer : The energy of one photon of hydrogen atom is, $3.03\times 10^{-19}J$

Explanation :

First we have to calculate the wavelength of hydrogen atom.

Using Rydberg's Equation:

$\frac{1}{\lambda}=R_H\left(\frac{1}{n_i^2}-\frac{1}{n_f^2} \right )$

Where,

$\lambda$ = Wavelength of radiation

$R_H$ = Rydberg's Constant = 10973731.6 m⁻¹

$n_f$ = Higher energy level = 3

$n_i$ = Lower energy level = 2

Putting the values, in above equation, we get:

$\frac{1}{\lambda}=(10973731.6)\left(\frac{1}{2^2}-\frac{1}{3^2} \right )$

$\lambda=6.56\times 10^{-7}m$

Now we have to calculate the energy.

$E=\frac{hc}{\lambda}$

where,

h = Planck's constant = $6.626\times 10^{-34}Js$

c = speed of light = $3\times 10^8m/s$

$\lambda$ = wavelength = $6.56\times 10^{-7}m$

Putting the values, in this formula, we get:

$E=\frac{(6.626\times 10^{-34}Js)\times (3\times 10^8m/s)}{6.56\times 10^{-7}m}$

$E=3.03\times 10^{-19}J$

Therefore, the energy of one photon of hydrogen atom is, $3.03\times 10^{-19}J$

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3 years ago

Two objects are electrically charged. The net charge on one object is doubled. Therefore, the electric force _____.

creativ13 [48]

If net charge on one object is doubled, then electric force will also get double. It is because they are directly proportional to each other.

Hope this helps!

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3 years ago

Read 2 more answers

Two 10 kg pucks head straight towards each other with velocities of 10 m/s and -20 m/s. They collide and stick together. Calcula

RUDIKE [14]

The final velocity of the two pucks is -5 m/s

Explanation:

We can solve the problem by using the law of conservation of momentum.

In fact, in absence of external force, the total momentum of the two pucks before and after the collision must be conserved - so we can write:

$m_1 u_1 + m_2 u_2 = (m_1 +m_2)v$