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3 years ago

If Earth was with no tilt, would we still have seasons at all? If so, how would they be different?

Physics

1 answer:

murzikaleks [220]3 years ago

8 0

Answer:

If earth had no tilt, we would have no seasons.

Explanation:

As stated in the answer, if the earth had no tilt we wouldn't have seasons. The earth all around the globe would maintain the same temperature,

And due to the no tilt it would also change our orbit to a bit larger slant, in January when we are at our closest to the sun we WOULD have a mini summer. For the North and South Pole, they would remain cold.

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A skater rotates with her arms crossed at an angular speed of 8.0 rad/s and she has a moment of inertia of 100 kg/m2. She extend

solong [7]

Answer:

When her hands extends, her momen of inertia is $4.28\ kg-m^2$ .

Explanation:

Given that,

Initial angular speed, $\omega_i=8\ rad/s$

Initial moment of inertia, $I_1=100\ kg-m^2$

Final angular speed, $\omega_f=7\ rad/s$

Initially, a skater rotates with her arms crossed and finally she extends her arms. The momentum remains conserved. Using the conservation of momentum as :

$I_1\omega_1=I_2\omega_2$

$I_2$ is final moment of inertia

$I_2=\dfrac{I_1\omega_1}{\omega_2}\\\\I_2=\dfrac{100\times 8}{7}\\\\I_2=114.28\ kg-m^2$

So, when her hands extends, her momen of inertia is $4.28\ kg-m^2$ . Hence, this is the required solution.

7 0

3 years ago

The human ear canal is about 2.8 cm long. If it is regarded as a tube that is open at one end and closed at the eardrum, what is

Diano4ka-milaya [45]

Answer:

f = 3.1 kHz

Explanation:

given,

length of human canal =2.8 cm = 0.028 m

speed of sound = 343 m/s

fundamental frequency = ?

The fundamental frequency of a tube with one open end and one closed end is,

$f = \dfrac{v}{4L}$

$f = \dfrac{343}{4\times 0.028}$

$f = \dfrac{343}{0.112}$

f = 3062.5 Hz

f = 3.1 kHz

hence, the fundamental frequency is equal to f = 3.1 kHz

8 0

3 years ago

xConsider the following reduction potentials: Cu2+ + 2e– Cu E° = 0.339 V Pb2+ + 2e– Pb E° = –0.130 V For a galvanic cell employi

slega [8]

Answer:

Approximately $\rm 90\; kJ$ .

Explanation:

Cathode is where reduction takes place and anode is where oxidation takes place. The potential of a electrochemical reaction ( $E^{\circ}(\text{cell})$ ) is equal to

$E^{\circ}(\text{cell}) = E^{\circ}(\text{cathode}) - E^{\circ}(\text{anode})$ .

There are two half-reactions in this question. $\rm Cu^{2+} + 2\,e^{-} \rightleftharpoons Cu$ and $\rm Pb^{2+} + 2\,e^{-} \rightleftharpoons Pb$ . Either could be the cathode (while the other acts as the anode.) However, for the reaction to be spontaneous, the value of $E^{\circ}(\text{cell})$ should be positive.

In this case, $E^{\circ}(\text{cell})$ is positive only if $\rm Cu^{2+} + 2\,e^{-} \rightleftharpoons Cu$ is the reaction takes place at the cathode. The net reaction would be

$\rm Cu^{2+} + Pb \to Cu + Pb^{2+}$ .

Its cell potential would be equal to $0.339 - (-0.130) = \rm 0.469\; V$ .

The maximum amount of electrical energy possible (under standard conditions) is equal to the free energy of this reaction:

$\Delta G^{\circ} = n \cdot F \cdot E^{\circ} (\text{cell})$ ,

where

$n$ is the number moles of electrons transferred for each mole of the reaction. In this case the value of $n$ is $2$ as in the half-reactions.
$F$ is Faraday's Constant (approximately $96485.33212\; \rm C \cdot mol^{-1}$ .)

$\begin{aligned}\Delta G^{\circ} &= n \cdot F \cdot E^{\circ} (\text{cell})\cr &= 2\times 96485.33212 \times (0.339 - (-0.130)) \cr &\approx 9.0 \times 10^{4} \; \rm J \cr &= 90\; \rm kJ\end{aligned}$ .

5 0

3 years ago

Blank parasites can cause serious Illnesses like West Nile virus and Lyme diseases

Greeley [361]

Endo I think but look it up jus in case

5 0

3 years ago

Three cannons are located at the top of a cliff above a level plain. Cannon A is aimed at an angle of 25° above the horizontal a

vodka [1.7K]

Answer:

The canon B hits the ground fast.

Explanation:

Given that,

Speed of cannon A = 85 m/s

Speed of cannon B= 100 m/s

Speed of cannon C = 75 m/s

We need to calculate the cannonballs will hit the ground with the greatest speed

Using conservation of energy

The final kinetic energy of canon depends on initial kinetic energy and potential energy.

The final velocity depends upon initial velocity and initial height.

So, the initial velocity of canon B is high.

Hence, The canon B hits the ground fast.

3 0

3 years ago