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3 years ago

11

In Jacksonville, heavy amounts of rainfall can take place during hurricane season. In addition, a storm surge floods the beach a

reas before storms make landfall. Which of the following represents a possible interaction between the geosphere and the hydrosphere during a hurricane? *

1 answer:

charle [14.2K]3 years ago

3 0

Answer: B

THE FLOOD WATER WILL WASH AWAY SAND AND SOIL FROM THE BEACH.

Explanation:

The hydrosphere - the flood water is the main agent responsible for eroding or breaking down the geosphere - sand and beach soil (or rocks). Differences in the amount of precipitation each area receives and the state of the water (liquid or solid) where it contacts rocks result in different kinds of erosion and land forms.

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A ray of light traveling through air strikes a piece of diamond at an angle of incidence equal to 56 degrees. Calculate the angu

Montano1993 [528]

Answer:

The angle of separation is $\Delta \theta = 0.93 ^o$

Explanation:

From the question we are told that

The angle of incidence is $\theta _ i = 56^o$

The refractive index of violet light in diamond is $n_v = 2.46$

The refractive index of red light in diamond is $n_r = 2.41$

The wavelength of violet light is $\lambda _v = 400nm = 400*10^{-9}m$

The wavelength of red light is $\lambda _r = 700nm = 700*10^{-9}m$

Snell's Law can be represented mathematically as

$\frac{sin \theta_i}{sin \theta_r} = n$

Where $\theta_r$ is the angle of refraction

=> $sin \theta_r = \frac{sin \theta_i}{n}$

Now considering violet light

$sin \theta_r__{v}} = \frac{sin \theta_i}{n_v}$

substituting values

$sin \theta_r__{v}} = \frac{sin (56)}{2.46}$

$sin \theta_r__{v}} = 0.337$

$\theta_r__{v}} = sin ^{-1} (0.337)$

$\theta_r__{v}} = 19.69^o$

Now considering red light

$sin \theta_r__{R}} = \frac{sin \theta_i}{n_r}$

substituting values

$sin \theta_r__{R}} = \frac{sin (56)}{2.41}$

$sin \theta_r__{R}} = 0.344$

$\theta_r__{R}} = sin ^{-1} (0.344)$

$\theta_r__{R}} = 20.12^o$

The angle of separation between the red light and the violet light is mathematically evaluated as

$\Delta \theta = \theta_r__{R}} - \theta_r__{V}}$

substituting values

$\Delta \theta =20.12 - 19.69$

$\Delta \theta = 0.93 ^o$

6 0

4 years ago

A boat leaves the dock at t = 0.00 s and, starting from rest, maintains a constant acceleration of (0.461 m/s2)i relative to the

liberstina [14]

Answer:

At t=4.82 s, the boat is moving at 3.464 m/s.

At t=4.82 s, the boat is 13.112 m from the dock.

Explanation:

The speed of the boat in j'th direction remains constant for all times (vj=2.16 m/s), however, the speed in i'th direction is changing due to the constant acceleration (0.461 m/s^2)i.

In order to find the velocity of the boat a t=4.82 s, first we need to compute the velocity of the boat relative to the water in the i direction (vi_b) at t=4.82 s:

vi_b = a*t = (0.461 m/s^2)*(4.82 s) = 2.222 m/s

Now, we add this velocity to the velocity of the water in the i direction:

vi = vi_b + vi_w = 2.222 m/s + 0.486 m/s = 2.708 m/s

Therefore, the speed of the boat at t = 4.82 s is: v = (vi, vj) = (2.708, 2.16) m/s. Finally, to find its speed, we just calculate the magnitude of v and obtain that the speed is: 3.464 m/s.

For the second question, first we will find the distance that the boat moved in the i'th direction and then in the j'th direction.

The speed in the i'th direction, for all times, is given by:

(0.485 + 0.461*t) and in order to find the distance advanced in the i'th direction (di) during 4.82 s, we need to integrate this velocity:

di = 0.485*t + (0.461*t^2)/2 (evaluated from t=0 to t =4.82) = 0.485*(4.82) + (0.461*(4.82)^2)/2 = 2.337 + 5.634 = 7.971 m

The speed in j'th direction, for all times, is given by:

2.16 and in order to find the distance advanced in the j'th direction (dj) during 4.82 s, we need to integrate this velocity:

dj = 2.16*t (evaluated from t=0 to t =4.82) = (2.16)*(4.82) = 10.411 m

Using Pythagoras' Theorem, we find that the the boat is at 13.112 m from the dock at t = 4.82 s.

4 0

3 years ago

Neutron stars are extremely dense objects formed from the remnants of supernova explosions. Many rotate very rapidly. Suppose th

mihalych1998 [28]

Answer:

10989.55932 rad/s

Explanation:

m = Mass of object

M = Mass of neutron star = $2\times 1.989\times 10^{30}\ kg$

R = Radius of neutron star = 13000 m

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

$\omega$ = Angular speed

Here, the gravitational force will balance the centripetal force

$\dfrac{GmM}{R^2}=mR\omega^2\\\Rightarrow \omega=\sqrt{\dfrac{GM}{R^3}}\\\Rightarrow \omega=\sqrt{\dfrac{6.67\times 10^{-11}\times 2\times 1.989\times 10^{30}}{13000^3}}\\\Rightarrow \omega=10989.55932\ rad/s$

The greatest possible angular speed an object can have is 10989.55932 rad/s

8 0

4 years ago

Which of the following statements are true concerning the creation of magnetic fields?

Soloha48 [4]

Answer:

A moving electric charge creates a magnetic field at all points in the surrounding region.

An electric current in a conductor creates a magnetic field at all points in the surrounding region.

A permanent magnet creates a magnetic field at all points in the surrounding region.

Explanation:

Magnetic field can be produced by:

- moving charges (i.e. a moving electron, or a current in a conductor)

- A magnet

The strength of the magnetic field produced by a current-carrying wire is

$B=\frac{\mu_0 I}{2\pi r}$

where

I is the current

r is the distance from the wire

As we see from the formula, the magnetic field is produced at all points in the surrounding region, because B becomes zero only when r becomes infinite. The same is true for the magnetic field created by a single moving charge or by a magnet.

The following choices instead are not correct:

- A single stationary electric charge creates a magnetic field at all points in the surrounding region.

- A distribution of electric charges at rest creates a magnetic field at all points in the surrounding region.

Because they involve the presence of stationary charges, and stationary charges do not produce magnetic fields.

3 0

3 years ago

When large amplitude sound vibrates the ear significantly,

tensa zangetsu [6.8K]

Loud sounds ... d) willl be answer

5 0

4 years ago

Read 2 more answers