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3 years ago

Wind blows from what pressure to what pressure

1 answer:

5 0

This is called the coriolis effect. So in the northern hemisphere, winds blow clockwise around an area of high pressure and counter-clockwise around low pressure. Not only do differences in air pressure help determine wind speed and direction, they help forecast precipitation and clear weather.

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A 2.4 mm -diameter copper wire carries a 37 A current (uniform across its cross section). Determine the magnetic field at the su

cluponka [151]

Answer:

Explanation:

We shall apply Ampere's circuital law to find out magnetic field . It is given as follows.

∫B.dl = μ₀ I , B is magnetic field , I is current , μ₀ is permeability .

Radius of the wire r = 1.2 x 10⁻³ m

magnetic field B will be circular in shape around the wire. If B is uniform

∫B.dl = B x 2πr

B x 2πr = μ₀ I

B = μ₀ I / 2πr

= 4π x 10⁻⁷ x 37 /2πx1.2 x 10⁻³

= 10⁻⁷ x 2x37 / 1.2 x 10⁻³

= 61.67 x 10⁻⁴ T

= 62 x 10⁻⁴ T

7 0

2 years ago

Which type of environmental science career involves collecting information about how human events impact the environment?

SVEN [57.7K]

C. environmental engineer

4 0

3 years ago

Read 2 more answers

When you step into a very full bath or hot tub, sometimes the water overflows. But when you step into a lake and swim, you don't

max2010maxim [7]

Answer:

This all simply due to the size of the two water bodies question and the your body size in a bath tub the water is small enough to allow the upthrust displace water out if the tub whereas in a lake the water is bigger and your body size is smaller to allow any noticeable upthrust that would cause an overflow

8 0

3 years ago

A 6.5 kg rock thrown down from a 120m high cliff with initial velocity 18 m/s down. Calculate

Olegator [25]

Answer:

See the answers below.

Explanation:

In order to solve this problem we must use the principle of energy conservation. Which tells us that the energy of a body will always be the same regardless of where it is located. For this case we have two points, point A and point B. Point A is located at the top at 120 [m] and point B is in the middle of the cliff at 60 [m].

$E_{A}=E_{B}$

The important thing about this problem is to identify the types of energy at each point. Let's take the reference level of potential energy at a height of zero meters. That is, at this point the potential energy is zero.

So at point A we have potential energy and since a velocity of 18 [m/s] is printed, we additionally have kinetic energy.

$E_{A}=E_{pot}+E_{kin}\\E_{A}=m*g*h+\frac{1}{2}*m*v^{2}$

At Point B the rock is still moving downward, therefore we have kinetic energy and since it is 60 [m] with respect to the reference level we have potential energy.

$E_{B}=m*g*h+\frac{1}{2}*m*v^{2}$

Therefore we will have the following equation:

$(6.5*9.81*120)+(0.5*6.5*18^{2} )=(6.5*9.81*60)+(0.5*6.5*v_{B}^{2} )\\3.25*v_{B}^{2} =4878.9\\v_{B}=\sqrt{1501.2}\\v_{B}=38.75[m/s]$

The kinetic energy can be easily calculated by means of the kinetic energy equation.

$KE_{B}=\frac{1}{2} *m*v_{B}^{2}\\KE_{B}=0.5*6.5*(38.75)^{2}\\KE_{B}=4878.9[J]$

In order to calculate the velocity at the bottom of the cliff where the reference level of potential energy (potential energy equal to zero) is located, we must pose the same equation, with the exception that at the new point there is only kinetic energy.

$E_{A}=E_{C}\\6.5*9.81*120+(0.5*9.81*18^{2} )=0.5*6.5*v_{C}^{2} \\v_{c}^{2} =\sqrt{2843.39}\\v_{c}=53.32[m/s]$