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PIT_PIT [208]
3 years ago
9

Where is the epicenter of the hypothetical earthquake as shown in the illustration below?

Physics
1 answer:
Genrish500 [490]3 years ago
5 0

Answer:

Point D

Explanation:

The epicenter of a hypothetical earthquake is located at the point where the earthquake begins.

(See the attached image).

Hope it helps!

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A dog barking at the sound of the postal carrier delivering mail is reacting to what type of cue?
Gwar [14]

Answer:

A. External

Explanation:

External stimulus includes touch/pain, vision, smell, taste, and sound.

6 0
2 years ago
How does what happens to the particles in a substance during melting differ from what happens during freezing?
sasho [114]
When in the melting process particles start to move more freely when in the freezing process particles tend to slow and vibrate in place
8 0
3 years ago
A uniform string of length 0.50 m is fixed at both ends. Find the
kozerog [31]

Answer:

configuration of string:

Node - Antinode - Node    or N-A-N

This is 1/2 wavelength since a full wavelength is N-A-N-A-N

f (fundamental) = V / wavelength

F0 = 300 m/s / 1 m = 100 / sec

F1 = 300 m/s / .5 m = 600 / sec

Each increase is a multiple of the fundamental since the wavelength

increases by 1/2 wavelength to keep nodes at both ends of the string

4 0
3 years ago
k) A stream of warm water is produced in a steady-flow mixing process by combining 1.0 kg s-1of cool water at 25°C with 0.8 kg s
Irina18 [472]

Answer:

T_ww = 43,23°C

Explanation:

To solve this question, we use energy balance and we state that the energy that enters the systems equals the energy that leaves the system plus losses. Mathematically, we will have that:

E_in=E_out+E_loss

The energy associated to a current of fluid can be defined as:

E=m*C_p*T_f

So, applying the energy balance to the system described:

m_CW*C_p*T_CW+m_HW*C_p*T_HW=m_WW*C_p*T_WW+E_loss

Replacing the values given on the statement, we have:

1.0 kg/s*4,18 kJ/(kg°C)*25°C+0.8  kg/s*4,18 kJ/(kg°C)*75°C=1.8 kg/s*4,18 kJ/(kg°C)*T_WW+30  kJ/s

Solving for the temperature Tww, we have:

(1.0 kg/s*4,18 kJ/(kg°C)*25°C+0.8 kg/s*4,18 kJ/(kg°C)*75°C-30 kJ/s)/(1.8 kg/s*4,18 kJ/(kg°C))=T_WW

T_WW=43,23 °C

Have a nice day! :D

6 0
3 years ago
The space shuttle approaches earth at a speed of 2.89*10^4 km/h when it sends a radio signal to earth 1.00*10^5 km away. The rad
zysi [14]

Answer:

3 secods later

Explanation:

6 0
3 years ago
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