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3 years ago

What keeps earth from falling in to the sun ?

Physics

2 answers:

natta225 [31]3 years ago

4 0

Gravity in space and earth

77julia77 [94]3 years ago

4 0

The Earth is always being pulled towards the Sun by gravity. The Earth is not moving fast enough to "escape" the Sun's gravity and leave the solar system, but it is going too fast to be pulled into the Sun. Therefore, it keeps going around and around - orbiting the Sun

Your answer is pretty much "Gravity"

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Calculate the rate of heat generation in kW due to the burning of the fuel when you drive a car rated at 34 MPG (miles per gallo

alexira [117]

Answer:

Explanation:

Let us calculate gallon used in one hour .

It travels 70 miles in one hour

in 70 miles it uses 70 / 34 gallons of fuel

70 / 34 gallons = 70 / 34 x 3.7854 kg

= 7.8 kg

heat generated = 7.8 x 44 x 10⁶ J

= 343.2 x 10⁶ J

This is heat generated in one hour

heat generated in one second = 343.2 x 10⁶ / 60 x 60 J/s

= 95.33 x 10³ J /s

= 95.33 kW.

4 0

3 years ago

Suppose the Sun were suddenly to shrink in size but that its mass remained the same. According to the law of conservation of ang

labwork [276]

I think its angular velocity would increase

8 0

4 years ago

In 2017, the company SpaceX became the first private company to send supplies to the International Space Station with a reusable

pav-90 [236]

Answer:

Approximately $3.98\; \rm m \cdot s^{-2}$ .

Assumption: air resistance on the rocket is negligible. Take $g = \rm 9.81\; m \cdot s^{-2}$ .

Explanation:

By Newton's Second Law of Motion, the acceleration of the rocket is proportional to the net force on it.

$\displaystyle \text{Acceleration} = \frac{\text{Net Force}}{\text{Mass}}$ .

Note that in this case, the uppercase letter $\rm M$ in the units stands for "mega-", which is the same as $10^6$ times the unit that follows. For example, $\rm 1\; Mg = 10^6\; g$ , while $\rm 1\; MN = 10^6\; N$ .

Convert the mass of the rocket and the thrust of its engines to SI standard units:

The standard unit for mass is kilograms: $\displaystyle m = \rm 552\; Mg = 552 \times 10^6\; g \times \frac{1\; \rm kg}{10^3\; g} = 552 \times 10^3 \; kg$ .
The standard for forces (including thrust) is Newtons: $\text{Thrust} = \rm 7.61 \; MN = 7.61 \times 10^6\; N$ .

At launch, the velocity of the rocket would be pretty low. Hence, compared to thrust and weight, the air resistance on the rocket would be pretty negligible. The two main forces that contribute to the net force of the rocket would be:

Thrust (which is supposed to go upwards), and
Weight (downwards due to gravity.)

The thrust on the rocket is already known to be $\rm 7.61 \times 10^6\; N$ . Since the rocket is quite close to the ground, the gravitational acceleration on it should be approximately $9.81\; \rm m \cdot s^{-2} = 9.81 \; N \cdot kg^{-1}$ . Hence, the weight on the rocket would be approximately $9.81\; \rm N \cdot kg^{-1} \times 552 \times 10^3\; kg = 5.41412\times 10^6\; N$ .

The magnitude of the net force on the rocket would be

$\begin{aligned}&\text{Thrust} - \text{Weight} \\ &= 7.61 \times 10^6\; \rm N - 5.41412\times 10^6\; N \\ &\approx 2.19 \times 10^6\; \rm N\end{aligned}$ .

Apply the formula $\displaystyle \text{Acceleration} = \frac{\text{Net Force}}{\text{Mass}}$ to find the net force on the rocket. To make sure that the output (acceleration) is in SI units (meters-per-second,) make sure that the inputs (net force and mass) are also in SI units (Newtons for net force and kilograms for mass.)

$\begin{aligned}\displaystyle &\text{Acceleration} \\ &= \frac{\text{Net Force}}{\text{Mass}} \\ &= \frac{2.19 \times 10^6\; \rm N}{552 \times 10^3\; \rm kg} \\ &\approx \rm 3.98\; \rm m \cdot s^{-2}\end{aligned}$ .