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1 year ago

What is mesothermal climates

Physics

1 answer:

Snowcat [4.5K]1 year ago

3 0

Answer:

Mesothermal regions have moderate climates. They are not cold enough to sustain a layer of winter snow but are also not remain warm enough to support flowering plants (and, thus, evapotranspiration) all year.

Explanation:

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Through which one of the following mediums is the velocity of a sound wave the greatest?

Colt1911 [192]

Answer: C. Steel

Explanation: When a sound wave travels through a solid body consisting

of an elastic material, the velocity of the wave is relatively

high. For instance, the velocity of a sound wave traveling

through steel (which is almost perfectly elastic) is about

5,060 meters per second. On the other hand, the velocity

of a sound wave traveling through an inelastic solid is

relatively low. So, for example, the velocity of a sound wave

traveling through lead (which is inelastic) is approximately

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2 years ago

A square loop of wire, with sides of length a, lies in the first quadrant of the xy plane, with one corner at the origin. In thi

Ainat [17]

Answer:

$emf=-\dfrac{1}{2}kta^5$

Explanation:

Given that

B(y, t) = k y ³t²

To find the total flux over the loop we have to integrate over the loop

$\phi =\int B.dS$

Given that loop is square,so

$\phi =\int B.dS$

B(y, t) = k y ³t²

$\phi =kt^2\int_{0}^{a}dx\int_{0}^{a}y^3dy$

$\phi =\dfrac{1}{4}kt^2a^5$

We know that emf given as

$emf=-\dfrac{d\phi }{dt}$

$\phi =\dfrac{1}{4}kt^2a^5$

$emf=-\dfrac{1}{2}kta^5$

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2 years ago

If a football quarterback can throw the ball at 20.0 m/s, what is the furthest distance the quarterback could throw the ball neg

kvasek [131]

The furthest distance the quarterback could throw the ball neglecting air resistance is 81.63 meters.

The situation given above is an example of projectile motion.

The maximum distance up to which a projectile can be thrown is known as its range, which is given by the formula,

Range(R) = 2u^2 sin(2θ)/g

Here u is the maximum velocity of the projectile= 20 m/s

θ is the angle at which the projectile is launched = 45°

g is the acceleration due to gravity = 9.8 m/s^2

Putting the above values in the given equation,

The range comes out to be 81.63 meters.

Hence, the furthest distance the quarterback could throw the ball neglecting air resistance is 81.63 meters.

To know more about "projectile motion", refer to the following link:

brainly.com/question/11422992?referrer=searchResults

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9 months ago

What would happen to temperature on the Earth If the Sun’s heat were not distributed throughout the atmosphere?

anzhelika [568]

It would be hotter near the Earth's equator and colder near Earth's poles.

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3 years ago

A 10.0-cm-long wire is pulled along a U-shaped conducting rail in a perpendicular magnetic field. The total resistance of the wi

Veronika [31]

A) 1.05 N

The power dissipated in the circuit can be written as the product between the pulling force and the speed of the wire:

$P=Fv$

where

P = 4.20 W is the power

F is the magnitude of the pulling force

v = 4.0 m/s is the speed of the wire

Solving the equation for F, we find

$F=\frac{P}{v}=\frac{4.20 W}{4.0 m/s}=1.05 N$

B) 3.03 T

The electromotive force induced in the circuit is:

$\epsilon=BvL$ (1)

where

B is the strength of the magnetic field

v = 4.0 m/s is the speed of the wire

L = 10.0 cm = 0.10 m is the length of the wire

We also know that the power dissipated is

$P=\frac{\epsilon^2}{R}$ (2)

where

$R=0.350 \Omega$ is the resistance of the wire

Subsituting (1) into (2), we get

$P=\frac{B^2 v^2 L^2}{R}$

And solving it for B, we find the strength of the magnetic field:

$B=\frac{\sqrt{PR}}{vL}=\frac{\sqrt{(4.20 W)(0.350 \Omega)}}{(4.0 m/s)(0.10 m)}=3.03 T$

5 0

3 years ago