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3 years ago

Why doesnt the moon crash towards earths surface?

Physics

2 answers:

VARVARA [1.3K]3 years ago

6 0

It is because Centrifugal force of earth on moon has developed of equal magnitude as Centripetal.....

Sonja [21]3 years ago

3 0

According to Newton's 3rd law the moon also attracts the earth with equal force which prevents it from falling towards the earth, while the gravitational pull of earth results in The revolution of the moon around the earth the force of the moon is not noticed on earth as the mass of the earth is much greater than that of the moon.

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The Old Faithful geyser found in Yellowstone Park is an example of _____.

katen-ka-za [31]

Answer:

cone geyser

Explanation:

The Old Faithful geyser is the oldest discovered geyser in the Yellowstone national park. The eruptions of the geyser are particularly predictable. It is a cone type geyser.

Cone geysers generally have a spout through which the water ejects out. When super heated water in the tube then the water starts to boil and form bubbles of steam, after this process the eruption takes place.

5 0

3 years ago

A ski lift carries people along a 220-meter cable up the side of a mountain. Riders are lifted a total of 110 meters in elevatio

jeka94

The ideal mechanical advantage (IMA) can be determined by the following equation:

IMA= Input distance/Output distance

The Input distance and Output distance are:

Input distance=220 meters

Output distance=110 meters

When you substitute in the equation of the ideal mechanical advantage (IMA), you obtain:

IMA= Input distance/Output distance

IMA= 220 meters/110 meters

IMA=2

3 0

3 years ago

Which of the following are examples of energy being conserved? Choose all that apply A light bulb constantly changes electrical

Degger [83]

A plant collects sunlight to form glucose, and your friend proposes an idea for a fan. Conserved = saving

4 0

3 years ago

Read 2 more answers

A person, with a mass of 50.0 kg, stands on a weighing scale in a lift which is moving

svet-max [94.6K]

Answer:

Explanation:

Because I just had that answer

6 0

3 years ago

Find the quantity of heat needed

krok68 [10]

Answer:

Approximately $3.99\times 10^{4}\; \rm J$ (assuming that the melting point of ice is $0\; \rm ^\circ C$ .)

Explanation:

Convert the unit of mass to kilograms, so as to match the unit of the specific heat capacity of ice and of water.

$\begin{aligned}m&= 100\; \rm g \times \frac{1\; \rm kg}{1000\; \rm g} \\ &= 0.100\; \rm kg\end{aligned}$

The energy required comes in three parts:

Energy required to raise the temperature of that $0.100\; \rm kg$ of ice from $(-10\; \rm ^\circ C)$ to $0\; \rm ^\circ C$ (the melting point of ice.)
Energy required to turn $0.100\; \rm kg$ of ice into water while temperature stayed constant.
Energy required to raise the temperature of that newly-formed $0.100\; \rm kg$ of water from $0\; \rm ^\circ C$ to $10\;\ rm ^\circ C$ .

The following equation gives the amount of energy $Q$ required to raise the temperature of a sample of mass $m$ and specific heat capacity $c$ by $\Delta T$ :

$Q = c \cdot m \cdot \Delta T$ ,

where

$c$ is the specific heat capacity of the material,
$m$ is the mass of the sample, and
$\Delta T$ is the change in the temperature of this sample.

For the first part of energy input, $c(\text{ice}) = 2100\; \rm J \cdot kg \cdot K^{-1}$ whereas $m = 0.100\; \rm kg$ . Calculate the change in the temperature:

$\begin{aligned}\Delta T &= T(\text{final}) - T(\text{initial}) \\ &= (0\; \rm ^\circ C) - (-10\; \rm ^\circ C) \\ &= 10\; \rm K\end{aligned}$ .

Calculate the energy required to achieve that temperature change:

$\begin{aligned}Q_1 &= c(\text{ice}) \cdot m(\text{ice}) \cdot \Delta T\\ &= 2100\; \rm J \cdot kg \cdot K^{-1} \\ &\quad\quad \times 0.100\; \rm kg \times 10\; \rm K\\ &= 2.10\times 10^{3}\; \rm J\end{aligned}$ .

Similarly, for the third part of energy input, $c(\text{water}) = 4200\; \rm J \cdot kg \cdot K^{-1}$ whereas $m = 0.100\; \rm kg$ . Calculate the change in the temperature:

$\begin{aligned}\Delta T &= T(\text{final}) - T(\text{initial}) \\ &= (10\; \rm ^\circ C) - (0\; \rm ^\circ C) \\ &= 10\; \rm K\end{aligned}$ .

Calculate the energy required to achieve that temperature change:

$\begin{aligned}Q_3&= c(\text{water}) \cdot m(\text{water}) \cdot \Delta T\\ &= 4200\; \rm J \cdot kg \cdot K^{-1} \\ &\quad\quad \times 0.100\; \rm kg \times 10\; \rm K\\ &= 4.20\times 10^{3}\; \rm J\end{aligned}$ .

The second part of energy input requires a different equation. The energy $Q$ required to melt a sample of mass $m$ and latent heat of fusion $L_\text{f}$ is:

$Q = m \cdot L_\text{f}$ .

Apply this equation to find the size of the second part of energy input:

$\begin{aligned}Q_2&= m \cdot L_\text{f}\\&= 0.100\; \rm kg \times 3.36\times 10^{5}\; \rm J\cdot kg^{-1} \\ &= 3.36\times 10^{4}\; \rm J\end{aligned}$ .

Find the sum of these three parts of energy:

$\begin{aligned}Q &= Q_1 + Q_2 + Q_3 = 3.99\times 10^{4}\; \rm J\end{aligned}$ .

3 0

3 years ago