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Alona [7]
3 years ago
11

Under which condition would time periods of daylight and darkness be equal everywhere on Earth all year? A. if Earth revolved ar

ound the sun in a more circular orbit B. if Earth rotated on an axis that was not tilted with respect to Earth's orbit C. if Earth had two moons D. if Earth revolved in an orbit closer to the sun
Physics
2 answers:
Marizza181 [45]3 years ago
7 0
The correct answer is
<span>B. if Earth rotated on an axis that was not tilted with respect to Earth's orbit 

In fact, the fact that Earth's axis is tilted is the reason why the durations of day and night are different in every part of the Earth. If the axis was not tilted, we would have exactly 12 hours of day and night in every point of the Earth, for the whole year.</span>
user100 [1]3 years ago
5 0

Answer: The correct Answer is B.

Explanation: I just took the test! Have a good day

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Un cubo de madera de densidad 0.780 g/cm³ mide 11.2 cm en un lado. Cuando se coloca en agua, ¿qué altura del bloque flotará sobr
Stolb23 [73]

Answer:

2.464 cm above the water surface

Explanation:

Recall that for the cube to float, means that the volume of water displaced weights the same as the weight of the block.

We calculate the weight of the block multiplying its density (0.78 gr/cm^3) times its volume (11.2^3  cm^3):

weight of the block = 0.78 * 11.2^3  gr

Now the displaced water will have a volume equal to the base of the cube (11.2 cm^2) times the part of the cube (x) that is under water. Recall as well that the density of water is 1 gr/cm^3.

So the weight of the volume of water displaced is:

weight of water = 1 * 11.2^2 * x

we make both weight expressions equal each other for the floating requirement:

0.78 * 11.2^3 = 11.2^2 * x

then x = 0.78 * 11.2 cm = 8.736 cm

This "x" is the portion of the cube under water. Then to estimate what is left of the cube above water, we subtract it from the cube's height (11.2 cm) as follows:

11.2 cm - 8.736 cm = 2.464 cm

6 0
3 years ago
If the bonds in the reactants of Figure 7-3 contained 432 kJ of chemical energy and the bonds in the
Yuri [45]

Answer:

  • <u>The energy change would be 46kJ</u>
  • <u>The energy would be absorbed</u>

Explanation:

The <em>energy change </em>during a chemical reation, i.e. the reaction energy, is equal to the chemical energy stored in the<em> bonds of the products </em>less the chemical energy stored in the <em>bonds of the reactants</em>.

Hence:

  • <em>Energy change</em> = 478 kJ - 432kJ = 46kJ

The change is positive, this is, the chemical energy of the products is greater than the chemical energy of the reactants.

That corresponds to the second graph, where the level of the energy of the products in the graph is higher than the level of the energy of the reactants. Therefore, the conclusion is that the reaction <em>absorbed energy</em> and it is endothermic.

4 0
3 years ago
How does what happens to the particles in a substance during melting differ from what happens during freezing?
sasho [114]
When in the melting process particles start to move more freely when in the freezing process particles tend to slow and vibrate in place
8 0
3 years ago
A wheel has a radius of r = 2.0 m and it rolls down a smooth incline. The height of the incline is h = 8.0 m . What is the angul
zepelin [54]

The angular velocity of the wheel at the bottom of the incline is 4.429 rad/sec

The angular velocity (ω) of an object is the rate at which the object's angle position is changing in relation to time.

For a wheel attached to an incline angle, the angular velocity can be computed by considering the conservation of energy theorem.

As such the total kinetic energy (K.E) and rotational kinetic energy (R.K.E) at a point is equal to the total potential energy (P.E) at the other point.

i.e.

P.E = K.E + R.K.E

\mathbf{mgh = \dfrac{1}{2}m(r \times \omega)^2 + \dfrac{1}{2}\times I \times \omega^2}

\mathbf{gh = \dfrac{1}{2}(r \times \omega)^2 + \dfrac{1}{2}\times r^2 \times \omega^2}

\mathbf{2 \times \dfrac{gh}{r^2} =\omega^2 +  \omega^2}

\mathbf{2 \omega^2=2 \times \dfrac{9.81 \times 8 m }{2.0 ^2}  }

\mathbf{\omega^2=\dfrac{39.24 }{2}}

\mathbf{\omega=\sqrt{19.62 } \ rad/sec}

\mathbf{\omega=4.429 \ rad/sec}

Therefore, we can conclude that the angular velocity of the wheel at the bottom of the incline is 4.429 rad/sec

Learn more about angular velocity here:

brainly.com/question/1452612

4 0
2 years ago
Read 2 more answers
Will give brainliest
Ivan

Answer:

i cant see the the pic its glitched post a new one i will help

Explanation:

4 0
3 years ago
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