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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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two 2.5 kg balls move away from each other one traveling 3 m/s to the right the other 4 m/s to the left what is the magnitude of
Juli2301 [7.4K]

Answer:

2.5 kg.m/s

Explanation:

Taking left side as positive while right side direction as negative then

Momentum, p= mv where m is the mass of the object and v is the velocity of travel

Momentum for ball moving towards right side=mv=2.5*-3=-7.5 kg.m/s

Momentum for the ball moving towards the left side=mv=2.5*4=10 kg.m/s

Total momentum=-7.5 kg.m/s+10 kg.m/s=2.5 kg.m/s

5 0
3 years ago
A 2500-lb vehicle has a drag coefficient of 0.35 and a frontal area of 20 ft2. What is the minimum tractive effort required for
emmasim [6.3K]

Answer:

i put this in the calculator and my answer is 600. hope this helps

Explanation:

4 0
3 years ago
Two transverse waves travel along the same taut string. Wave 1 is described by y1(x, t) = A sin(kx - ωt), while wave 2 is descri
Vadim26 [7]

Answer:

6) Wave 1 travels in the positive x-direction, while wave 2 travels in the negative x-direction.

Explanation:

What matters is the part kx \pm \omega t, the other parts of the equation don't affect time and space variations. We know that when the sign is - the wave propagates to the positive direction while when the sign is + the wave propagates to the negative direction, but <em>here is an explanation</em> of this:

For both cases, + and -, after a certain time \delta t (\delta t >0), the displacement <em>y</em> of the wave will be determined by the kx\pm\omega (t+\delta t) term. For simplicity, if we imagine we are looking at the origin (x=0), this will be simply \pm \omega (t+\delta t).

To know which side, right or left of the origin, would go through the origin after a time \delta t (and thus know the direction of propagation) we have to see how we can achieve that same displacement <em>y</em> not by a time variation but by a space variation \delta x (we would be looking where in space is what we would have in the future in time). The term would be then k(x+\delta x)\pm\omega t, which at the origin is k \delta x \pm \omega t. This would mean that, when the original equation has kx+\omega t, we must have that \delta x>0 for k\delta x+\omega t to be equal to kx+\omega\delta t, and when the original equation has kx-\omega t, we must have that \delta x for k\delta x-\omega t to be equal to kx-\omega \delta t

<em>Note that their values don't matter, although they are a very small variation (we have to be careful since all this is inside a sin function), what matters is if they are positive or negative and as such what is possible or not .</em>

<em />

In conclusion, when kx+\omega t, the part of the wave on the positive side (\delta x>0) is the one that will go through the origin, so the wave is going in the negative direction, and viceversa.

4 0
3 years ago
Convert 93.6 miles per hour. Convert this to kilometers per hour.
Readme [11.4K]

Answer:

150.6 km

Explanation:

One mile is about 1.61 km so multiply 93.6 by 1.6 which gives you above 150.6

3 0
3 years ago
: Does increasing temperature increase pressure?
Vesnalui [34]
Yes, an increase in temperature is accompanied by an increase in pressure. Temperature is the measurement of heat present and more heat means more energy. Molecules in hotter temperatures move faster and more often, eventually moving into the gaseous phase. The molecules would fill the container, and the hotter it got the more they would bounce off the walls, pushing outward, increasing the pressure.
I suppose you could measure this with some kind of loosely inflated balloon and subject it to different temperatures and then somehow measure the size/pressure of it.
5 0
3 years ago
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