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3 years ago

Surface winds on Earth are primarily caused by differences in

2 answers:

6 0

Surface winds on Earth are primarily caused by differences in Air density due to unequal heating of Earth's surface. The answer is letter A. Surface currents are driven by winds that blow in certain patterns. They are caused by the Earth’s spin and the Coriolis effect. Evaporation adds water to the atmosphere. It is the change from liquid to gas wherein liquid molecules tends to go the gas phase and increases the number of molecules.

Explanation:

Causes of Different Pressure on the Earth's Exterior. Differences in pressures across the cover of the Earth are fundamentally produced by irregular heating of the surface by the Sun.

mylen [45]3 years ago

4 0

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Two parallel conducting plates are separated by 10.0 cm, and one of them is taken to be at zero volts. (a) what is the electric

Alex_Xolod [135]

The expression for the magnitude of the electric field between two uniform conducting plates is

$E=\frac{V}{d}$

Here, V is potential difference between plates and d is separation between plates.

As the potential 6.00 cm from the zero volt plate (and 4.00 cm from the other) is 420 V.

Therefore,

$E = \frac{420 \ V}{6 \times 10^{-2} m } = 70 \times 10^2 \ V/m$

Thus, the electric field strength between the plates is 7000 V/ m

8 0

3 years ago

A horizontal force of 12N is applied to 1.5kg of block which rests on a horizontal surface. If the coefficient friction is 0.3,

Rama09 [41]

Answer:

7 m/s²

Explanation:

From the question given above, the following data were:

Force applied (Fₐ) = 15 N

Mass (m) of block = 1.5 Kg

Acceleration due to gravity (g) = 10 m/s²

Coefficient of friction (μ) = 0.3

Acceleration (a) of block =?

Next, we shall determine the frictional force. This can be obtained as follow:

Mass (m) of block = 1.5 Kg

Acceleration due to gravity (g) = 10 m/s²

Coefficient of friction (μ) = 0.3

Normal reaction (R) = mg = 1.5 × 10 = 15 N

Frictional force (Fբ) =?

Fբ = μR

Fբ = 0.3 × 15

Fբ = 4.5 N

Next, we shall determine the net force acting on the block. This can be obtained as follow:

Force applied (Fₐ) = 15 N

Frictional force (Fբ) = 4.5 N

Net force (Fₙ) =.?

Fₙ = Fₐ – Fբ

Fₙ = 15 – 4.5

Fₙ = 10.5 N

Finally, we shall determine the acceleration produced in the block. This can be obtained as follow:

Mass (m) of block = 1.5 Kg

Net force (Fₙ) = 10.5 N

Acceleration (a) of block =?

Fₙ = ma

10.5 = 1.5 × a

Divide both side by a

a = 10.5 / 1.5

a = 7 m/s²

Therefore, the acceleration produced in the block is 7 m/s²

8 0

3 years ago

A meteoroid is traveling east through the atmosphere at 18. 3 km/s while descending at a rate of 11.5 km/s. What is its speed, i

Annette [7]

Answer:

The speed of meteoroid is 21.61 km/s in south-east.

Explanation:

Given that,

A meteoroid is traveling through the atmosphere at 18.3 km/s. while descending at a rate of 11.5 km/s it means 11.5 km/s in south.

We need to draw a diagram

Using Pythagorean theorem

$AC^2=AB^2+BC^2$

$AC^2=(18.3)^3+(11.5)^2$

$AC=\sqrt{(18.3)^2+(11.5)^2}$

$AC=21.61\ km/s$

Hence, The speed of meteoroid is 21.61 km/s in south-east.

6 0

3 years ago

Light is a form of what

mart [117]

Light is a form of energy.
(:

8 0

3 years ago

Assume the radius of an atom, which can be represented as a hard sphere, is r = 1.95 Å. The atom is placed in a (a) simple cubic

Nuetrik [128]

Answer:

(a) A = $3.90 \AA$

(b) $A = 4.50 \AA$

(d) $A = 9.02 \AA$

Solution:

As per the question:

Radius of atom, r = 1.95 $\AA = 1.95\times 10^{- 10} m$

Now,

(a) For a simple cubic lattice, lattice constant A:

A = 2r

A = $2\times 1.95 = 3.90 \AA$

(b) For body centered cubic lattice:

$A = \frac{4}{\sqrt{3}}r$

$A = \frac{4}{\sqrt{3}}\times 1.95 = 4.50 \AA$

$A = 2{\sqrt{2}}r$

$A = 2{\sqrt{2}}\times 1.95 = 5.51 \AA$

(d) For diamond lattice:

$A = 2\times \frac{4}{\sqrt{3}}r$

$A = 2\times \frac{4}{\sqrt{3}}\times 1.95 = 9.02 \AA$

6 0

3 years ago