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3 years ago

Which option identifies the two natural wind patterns that move air masses in the United States?

Physics

2 answers:

grandymaker [24]3 years ago

5 1

Answer: option C is correct: Trade winds and prevailing westerlies.

Explanation:

COMPLETE QUESTION: Which option identifies the two natural wind patterns that move air masses in the United States?

(a). Prevailing westerlies and jet streams,

(b). Polar easterlies and jet streams,

(c). Trade winds and prevailing westerlies, (d). Trade winds and polar.

SOLUTION:

Wind:simply, wind is the movement of air in a definite direction.

Air mass: air mass can be reffered to as large air with uniform humidity and temperature.

So,.diving into the solutio ,the correct option here is option C, that is, trade winds and Prevailing westerlies.

Trade wind is also known as Easterlies. Easterlies are east-to-west prevailing winds. They blow continuosly and steady. Easterlies are air movement towards the equator. The are found at the equatorial region of latitude 30° North - 30° South.

The westerlies are also known as the Prevailing westerlies. They are found to occur at the latitudes of 30°-60°.

Both wind patterns, especially westerlies are responsible for weather movement in the United States.

nonya

2 years ago

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Darya [45]3 years ago

4 0

Answer: prevailing westerlies and jet streams

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A moving object has a kinetic energy of 150 j and a momentum with a magnitude of 30.0 kg•m/s. determine the mass and speed of th

SOVA2 [1]

To determine the answer to this item, we use two (2) equations.

Equation for kinetic energy:
KE = 0.5 mv²

Equation for momentum:
P = mv

From the second equation, we can deduced that,
m = P/v
Substituting the known values from the given above,

m = 30/v

Using this expression in the first equation,

KE = 0.5 mv²; 150 = 0.5(30/v)(v²)

The value of v from the equation is 10 m/s.

The mass is therefore calculated as such,
m = 30/v = 30/10 = 3 kg

Hence, the answers are,

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3 years ago

A segment A of wire stretched tightly between two posts a distance L apart vibrates in its fundamental mode with frequency f. A

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Answer:

option (c)

Explanation:

Fundamental frequency of segment A = f

Second harmonic frequency of B = fundamental frequency of A .

Tension in both the wires is same and the mass density is also same as the wires are identical.

fundamental frequency of wire A is given by

$f=\frac{1}{2L_{A}}{\sqrt{\frac{T}{m}}}$ .... (1)

Second harmonic of B is given by

$f=\frac{2}{2L_{B}}{\sqrt{\frac{T}{m}}}$ .... (2)

Equation (1) is equal to equation (2), we get

$\frac{1}{2L_{A}}=\frac{2}{2L_{B}}$

$L_{B}=2L_{A}$

So, LB = 2 L

Thus, the length of wire segment B is 2 times the length of wire segment A.

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3 years ago

A robotic rover on Mars finds a spherical rock with a diameter of 10 centimeters [cm]. The rover picks up the rock and lifts it

Makovka662 [10]

Answer: 5166.347

Explanation:

The specific gravity of a solid $SG$ (also called relative density) is the ratio of the density of that solid $\rho_{rock}$ to the density of water $\rho_{water}=1 kg/m^{3}$ (normally at $4\°C$ ):

$SG=\frac{\rho_{rock}}{\rho_{water}}$ (1)

On the other hand, the density of the rock is calculated by:

$\rho_{rock}=\frac{m_{rock}}{V_{rock}}$ (2)

Where:

$m_{rock}$ is the mass of the rock

$V_{rock}=\frac{4}{3} \pi r^{3}$ is the volume of the rock, since is spherical

Well, we already know the value of $\rho_{water}$ , but we need to find $\rho_{rock}$ in order to find the rock's specific gravity; and in order to do this, we firsly have to find $m_{rock}$ and then calculate $V_{rock}$ :

In the case of the mass of the rock, we can calclate it by the following equation:

$W_{rock}=m_{rock}g_{mars}$ (3)

Where:

$W_{rock}$ is the weight if the rock in mars

$g_{mars}=3.7 m/s^{2}$ is the acceleration due gravity in Mars

Isolating $m_{rock}$ :

$m_{rock}=\frac{W_{rock}}{g_{mars}}$ (4)

$m_{rock}=\frac{W_{rock}}{3.7 m/s^{2}}$ (5)

To find $W_{rock}$ we can use the following equation of the potential gravitational energy $U$ :

$U=W_{rock}H$ (6)

Where:

$U=2 J=2 Nm$ is the potential energy

$H=20 cm \frac{1m}{100 cm}=0.2 m$ is the height at which the rock has the mentioned potential energy

Isolating $W_{rock}$ :

$W_{rock}=\frac{U}{H}$ (7)

$W_{rock}=\frac{2 Nm}{0.2 m}$ (8)

$W_{rock}=10 N$ (9)

Substituting (9) in (5):

$m_{rock}=\frac{10 N}{3.7 m/s^{2}}$ (10)

$m_{rock}=2.702 kg$ (11)

Substituting (11) in (2):

$\rho_{rock}=\frac{2.702 kg}{V_{rock}}$ (12) At this point we only need to find the volume of the rock, knowing its diameter is $d=10 cm$ , hence its radius is $r=\frac{d}{2}=5 cm$

$V_{rock}=\frac{4}{3} \pi (5 cm)^{3}$ (13)

$V_{rock}=523.59 cm^{3} \frac{1 m^{3}}{(100 cm)^{3}}=0.000523 m^{3}$ (14)

Substituting (14) in (12):

$\rho_{rock}=\frac{2.702 kg}{0.000523 m^{3}}$ (15)

$\rho_{rock}=5166.34 kg/m^{3}$ (16)

Substituting (16) in (1):

$SG=\frac{5166.34 kg/m^{3}}{1 kg/m^{3}}$ (17)

Finally we obtain the specific gravity of the rock:

$SG=5166.347$

7 0

3 years ago

What percentage of the intensity gets through both polarizers?

sp2606 [1]

Answer:

if the two polarizers have the same direction the transmitted light is 50% of the incident and if the two polarizers are at 90º the transmitted light is zero

Explanation:

The incident light is generally random, that is, it does not have a polarization plane, when the first polarized stops by half, this already polarized light arrives at the second polarizer and the causticity passes

I = I₀ cos² θ

therefore if the two polarizers have the same direction the transmitted light is 50% of the incident and if the two polarizers are at 90º the transmitted light is zero

8 0

3 years ago