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MArishka [77]
3 years ago
8

The vibration produced days or even years before an earthquake

Physics
1 answer:
Margaret [11]3 years ago
7 0
I don't know I guess its the plate tectonics 
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Find the velocity v=ds/dt at t=3​
IRINA_888 [86]

Answer:

you need at least two out of the three to get any aenser

7 0
3 years ago
A spring with a force constant of 5.3 n/m has a relaxed length of 2.60 m. when a mass is attached to the end of the spring and a
olganol [36]
This is the equation for elastic potential energy, where U is potential energy, x is the displacement of the end of the spring, and k is the spring constant. 
<span> U = (1/2)kx^2 
</span><span> U = (1/2)(5.3)(3.62-2.60)^2 
</span> U = <span> <span>2.75706 </span></span>J
6 0
3 years ago
The temperature of a black body is 500 and its radiation is of wavelength 600 . If the number of oscillators with energy is 100
stiks02 [169]

Answer: An equation is missing in your question below is the missing equation

a) ≈ 8396

b) 150 nm/k

Explanation:

<u>A) Determine the number of Oscillators in the black body</u>

number of oscillators = 8395

attached below is the detailed solution

<u>b) determine the peak wavelength of the black body </u>

Black body temperature = 20,000 K

applying Wien's law / formula

λmax = b / T  ------ ( 1 )

T = 20,000 K

b = 3 * 10^6 nm

∴  λmax = 150 nm/k

4 0
2 years ago
Please help on this one?
butalik [34]

B. +Q, + W are the correct sign

7 0
3 years ago
Read 2 more answers
A new planet is discovered beyond Pluto at a mean distance to the sun of 4004 million miles. Using Kepler's third law, determine
AVprozaik [17]

Answer:

103239.89 days

Explanation:

Kepler's third law states that the square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of its orbit.

                               a³ / T² = 7.496 × 10⁻⁶  (a.u.³/days²)

where,

a is the distance of the semi-major axis in a.u

T is the orbit time in days

Converting the mean distance of the new planet to astronomical unit (a.u.)

                       1 a.u = 9.296 × 10⁷ miles        

                                      \frac{4004 * 10^{6}}{9.296 * 10^{7}}  =  43.07\ a.u.

Substituting the values into Kepler's third law equation;

                                    \frac{(43.07)^{3}}{T^{2}}  =  7.496 * 10^{-6}  

                                    T^{2} = \frac{(43.07)^{3}}{7.496 * 10^{-6}} (days)²

                                    T^{2} = \sqrt{\frac{(43.07)^{3}}{7.496 * 10^{-6}}}

                                    T = 103239.89 days

An estimate time T for the new planet to travel around the sun in an orbit is 103239.89 days

7 0
3 years ago
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