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3 years ago

Describe two ways in which heat is transported in the biosphere

Physics

1 answer:

Trava [24]3 years ago

7 0

Answer and Explanation:

Warm air tends to rise, and cold air tends to fall, consequently, the hot air of the equator rises and at the same time the cold air of the poles descends to the ground. This generates wind currents that displace heat throughout the atmosphere.

The movement of water creates ocean currents that transport energy heat up the biosphere. Surface ocean currents heat or cool the air above

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A ball is dropped from a tower that is 206 m high, How long does it take to reach the ground?

Slav-nsk [51]

It will take 6.42 s for the ball that is dropped from a height of 206 m to reach the ground.

From the question given above, the following data were obtained:

Height (H) = 206 m

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

The time taken for the ball to get to the ground can be obtained as follow:

H = ½gt²

206 = ½ × 10 × t²

206 = 5 × t²

Divide both side by 5

$t^{2} = \frac{206}{5}\\\\ t^{2} = 41.2$

Take the square root of both side

$t = \sqrt{41.2}$

Therefore, it will take 6.42 s for the ball to get to the ground.

Learn more: brainly.com/question/24903556

7 0

2 years ago

What is your favorite thing to eat? is it healthy or unhealthy ?

denpristay [2]

Answer:

sometimes I like to eat unhealthy and healthy my healthy are fruits granola a grapes, strawberries, blueberries or make smoothie out fruits my unhealthy is pizza, streak fries, with everything on the side. so ways eaanation:

4 0

3 years ago

A block of mass 0.1 kg is attached to a spring of spring constant 21 N/m on a frictionless track. The block moves in simple harm

bogdanovich [222]

Answer:

A) 2.75 m/s B) 0.1911 m C) 0.109 s

Explanation:

mass of block = M =0.1 kg

spring constant = k = 21 N/m

amplitude = A = 0.19 m

mass of bullet = m = 1.45 g = 0.00145 kg

velocity of bullet = vᵇ = 68 m/s

as we know:

Angular frequency of S.H.M = ω₀ = $\sqrt\frac{k}{M}$

= $\sqrt\frac{21}{0.1}$

= 14.49 rad/sec

<h3>A) Speed of the block immediately before the collision:</h3>

displacement of Simple Harmonic Motion is given as:

$x = A sin (\omega t + \phi)\\$

Differentiating this to find speed of the block immediately before the collision:

$v=\frac{dx}{dt}= A\omega_{o} cos (\omega_{o}t =\phi}\\$

As bullet strikes at equilibrium position so,

φ = 0

t= 2nπ

⇒ cos (ω₀t + φ) = 1

⇒ $v= A\omega_{o}$

$v=(.19)(14.49)\\v= 2.75 ms^{-1}$

<h3>B) If the simple harmonic motion after the collision is described by x = B sin(ωt + φ), new amplitude B:</h3>

S.H.M after collision is given as :

$x= Bsin(\omega t + \phi)$

To find B, consider law of conservation of energy

$K.E = P.E\\K.E= \frac{1}{2}(m+M)v^{2} \\P.E = \frac{1}{2} kB^{2}$

$\frac{m+M}{k} v^{2} = B^{2} \\B =\sqrt\frac{m+M}{k} v\\B = \sqrt\frac{.00145+0.1}{21} (2.75)\\B = .1911m$

<h3>C) Time taken by the block to reach maximum amplitude after the collision:</h3>

Time period S.H.M is given as:

$T=2\pi \sqrt\frac{m}{k}\\ for given case\\m= m=M\\then\\T=2\pi \sqrt\frac{m+M}{k}$

Collision occurred at equilibrium position so time taken by block to reach maximum amplitude is equal to one fourth of total time period

$T=\frac{\pi }{2}\sqrt\frac{m+M}{k} \\T=0.109 sec$

5 0

3 years ago

If a roller coaster car had 40,000 J of gravitational potential energy when at rest on the top of a hill how much kinetic energy

Inessa [10]

Answer:

K.E = 30,000 J

Explanation:

Given,

The potential energy of the roller coaster car, P.E = 40000 J

The kinetic energy at height h/4, K.E = ?

According to the law of conservation of energy, the total energy of the system is conserved.

At height 'h', the total energy is,

P.E = mgh

K.E = 0

At height 'h/4', the total energy is

P.E + K.E = mgh

P.E = mgh/4

K.E = 1/2 mv²

Therefore,

mgh/4 + 1/2 mv² = mgh

gh/4 + v²/2 = gh

Hence,

v² = 3gh/2

Substituting in the K.E equation

K.E = 1/2 mv²

= 1/2 m (3gh/2)

= 3/4 mgh

= 3/4 x 40000

= 30000 J

Hence, the K.E of the roller coaster car is, K.E = 30000 J

6 0

3 years ago

Jacob is working on his physics portfolio and stands on a 0.45 m high step. If Jacob’s mass is 103 kg, what is his

Aleks [24]

$e_ p = m \times g \times h \\ \\ = 103kg \times 9.81 \frac{m}{s { }^{2} } \times 0.45m$

4 0

3 years ago