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2 years ago

When heat moves from the ocean to the surrounding air, which is this an example of?

1 answer:

6 0

B) The transfer of energy from the hydrosphere to the atmosphere

This is because oceans are part of the hydrosphere. As the air warms it flows up into the atmosphere.

Hope this helps!

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Bohr found experimental evidence for his atomic model by studying _____.

nevsk [136]

For the answer to te question above, it is a. The Electron Cloud.
For additional information, The Electron Cloud is an informal term in physics. It is used to describe where electrons are when they go around the nucleus of an atom.
I hope my answer helped you.

6 0

2 years ago

Read 2 more answers

Does there appear to be a simple mathematical relationship between the acceleration of an object (with fixed mass and negligible

KengaRu [80]

Answer:

the net force applied to an object is directly proportional to the acceleration undergone by that object

Explanation:

This verbal statement can be expressed in equation form as follows:

a = Fnet / m

7 0

2 years ago

A dragster travels 1/4 mi in 6.7 s. Assuming that acceleration is constant and the dragster is initially at rest, what is its ve

nlexa [21]

V0=0m/s (initially at rest)
t=6,7s
s=1/4mi=402,336m
s=(a*t^2)/2 -> a=2*s/t^2 -> a=2*402,336m/(6,7s)^2
a=804,672/44,89=17,93 m/s^2
<span>v=v0+at
</span>v=0+17,93 m/s^2 * 6,7s = 120,131 m/s = 432,4716 km/h

3 0

2 years ago

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A 57 kg boy and a 41 kg girl use an elastic rope while engaged in a tug-of-war on a frictionless icy surface. If the acceleratio

Sergeeva-Olga [200]

Answer:

Acceleration of the boy a₁:

$a_{1} = 1.87 \frac{m}{s^{2} }$

Explanation:

Conceptual analysis

We apply Newton's second law to the boy and the girl:

F = m*a (Formula 1)

F : Force in Newtons (N)

m : mass in kilograms (kg)

a : acceleration in meters over second square (m/s²)

Nomenclature

m₁ : boy mass

m₂ : girl mass

a₁ : boy acceleration

a₂ : girl acceleration

F₁ : boy acceleration

F₂ : girl acceleration

Known data

m₁ = 57 kg

m₂ = 41 kg

a₂ = 2.6 m/s²

Problem development

We apply to Newton's third law of action and reaction, then:

F₁ = F₂ , We apply the formula (1):

m₁*a₁ = m₂*a₂

$a_{1} = \frac{m_{2}* a_{2} }{m_{1} }$

$a_{1} = \frac{41* 2.6 }{ 57 }$

$a_{1} = 1.87 \frac{m}{s^{2} }$

5 0

2 years ago

A system where friction is considered and calculated

katrin [286]

Answer:

the answer is B :)

Explanation:

7 0

2 years ago