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3 years ago

The three types of seismic waves produced by an earthquake are primary, secondary, and

Physics

2 answers:

Liono4ka [1.6K]3 years ago

7 0

The three types of seismic waves produced by an earthquake are primary, secondary, and (D.) surface.

Semmy [17]3 years ago

7 0

The three types of seismic waves produced by an earthquake are primary, secondary, and? The answer is, D. Surface.

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A brick and a feather fall to the earth at their respective terminal velocities. Which objectexperiences the greater force of ai

Karo-lina-s [1.5K]

Answer:

Under the reasonable assumption that the brick has more mass than the feather, the brick experiences a greater force of air friction.

Explanation:

Objects at terminal velocity, only under the influence of gravity, have maximized their speed and have an acceleration of zero. Thus, neither object is accelerating.

Recall Newton's second law: $\sum {\vec {F}}=m \vec {a}$

Since acceleration for each object is zero, the sum of the force acting on each of those objects must also be zero.

Since the only forces acting on the objects are gravity and the force of air friction, in order to zero out, the force of air friction must be equal in magnitude and opposite in direction to the force of gravity.

Recall that near the surface of the earth, $F_{gravity}=mg$ , so the Force of Gravity acting on an object is directly proportional to the object's mass. (A similar argument could be made even if this were not taking place on the surface of the earth, so long as the objects were the same distance from the object providing gravitational influence).

If the masses of the objects are different, the object with the greater mass will experience a larger force of gravity, and hence a larger force of air friction at terminal velocity.

Under the reasonable assumption that the brick has more mass than the feather, the brick experiences a greater force of air friction.

4 0

1 year ago

Two 3.0 μC charges lie on the x-axis, one at the origin and the other at What is the potential (relative to infinity) due to the

Airida [17]

Complete Question:

Two 3.0µC charges lie on the x-axis, one at the origin and the other at 2.0m. A third point is located at 6.0m. What is the potential at this third point relative to infinity? (The value of k is 9.0*10^9 N.m^2/C^2)

Answer:

The potential due to these charges is 11250 V

Explanation:

Potential V is given as;

$V =\frac{Kq}{r}$

where;

K is coulomb's constant = 9x10⁹ N.m²/C²

r is the distance of the charge

q is the magnitude of the charge

The first charge located at the origin, is 6.0 m from the third charge; the potential at this point is:

$V =\frac{9X10^9 X3X10^{-6}}{6} =4500 V$

The second charge located at 2.0 m, is 4.0 m from the third charge; the potential at this point is:

$V =\frac{9X10^9 X3X10^{-6}}{4} =6750 V$

Total potential due to this charges = 4500 V + 6750 V = 11250 V

6 0

3 years ago

If Earth were replaced by an object with the same mass but much smaller in size, would the moon continue to orbit the new object

Tanzania [10]

I am pretty sure but not 100% that it would still continue to orbit

3 0

3 years ago

Read 2 more answers

Si un planeta tuviese un periodo de traslación de 65 años terrestres a que distancia se encontraría del sol

Tju [1.3M]

Answer:

$R \approx 2.418\times 10^{9}\,km$

Explanation:

(The following exercise is written in Spanish and for that reason explanation will be held in Spanish)

Supóngase que el planeta tiene una órbita circular, el período de rotación del planeta es:

$T = \frac{2\pi}{\omega}$

Asimismo, la rapidez angular se describe como función de la aceleración centrípeta:

$\omega = \sqrt{\frac{a_{r}}{R} }$

Ahora se reemplaza en la ecuación de período:

$T = 2\pi \cdot \sqrt{\frac{R}{a_{r}} }$

La aceleración experimentada por el planeta es:

$a_{r} = G\cdot \frac{M_{sun}}{R^{2}}$

Se reemplaza en la ecuación de período:

$T = 2\pi \cdot \sqrt{\frac{R^{3}}{G\cdot M_{sun}} }$

La distancia del planeta con respecto al sol es finalmente despejada:

$R^{3} = G\cdot M_{sun}\cdot \left(\frac{T}{2\pi} \right)^{2}$

$R = \sqrt[3]{G\cdot M_{sun}\cdot \left(\frac{T}{2\pi} \right)^{2}}$

Finalmente, se sustituyen las variables y se determina la distancia:

$R = \sqrt[3]{\left(6.674\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}} \right)\cdot (1.989\times 10^{30}\,kg)\cdot \left[\frac{(65\,a)\cdot \left(365\,\frac{d}{a} \right)\cdot \left(86400\,\frac{s}{d} \right)}{2\pi} \right]^{2}}$

$R \approx 2.418\times 10^{12}\,m$

$R \approx 2.418\times 10^{9}\,km$

4 0

3 years ago

Give an example of mass making a difference in the amount of gravitational energy. Tell how you know the gravitational energy is

Andreas93 [3]

Answer:

The Gravitational potential energy at large distances is directly proportional to the masses and inversely proportional to the distance between them. The gravitational potential energy increases as r increases.

Examples of Gravitational Energy

A raised weight.

Water that is behind a dam.

A car that is parked at the top of a hill.

A yoyo before it is released.

5 0

2 years ago