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3 years ago

Is our universe the size of a grain of sand?

Physics

1 answer:

NeTakaya3 years ago

8 0

Answer:

Explanation:

Our universe is not the size of a grain of sand. A grain of sand is approximately 2 millimeters while the size of the universe is approximate 93 billion light years in diameter and recently estimated to be 250 times 93 billion light years which is at least 7 trillion light years.

Light-year is a unit of length used to show astronomical distances.

1 light-year is exactly 9460730472580800 meters.

Comparing the size of a grain of sand to that of the universe,

2 millimeters = 0.002 meters and

9460730472580800 meters = 9.460730472580800 e+18

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A train is traveling at a speed of 50km/hr. How.many hours will it take the train to travel 800 km?

bagirrra123 [75]

Answer:

It would take 16 hours

Explanation:

800 divided by 50 = 16

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3 years ago

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Gravity is greater when there is

Papessa [141]

-- more mass involved

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3 years ago

A sealed 26-m3 tank is filled with 6000 moles of oxygen gas (O2) at an initial temperature of 270 K. The gas is heated to a fina

TiliK225 [7]

Answer:

The final pressure of oxygen gas is 8.33 atm.

Explanation:

From the given data

V=26 m^3 or 26000 L

T1=270K

T2=440K

n1=6000 moles

R=0.0821 L.atm/molK

Now from the ideal gas equation

$P_2V_2=nRT_2\\P_2=\frac{nRT_2}{V_2}\\P_2=\frac{6000*0.0821*440}{26000}\\\\P_2=8.33 atm\\$

As the options given are not defined in which unit thus the final pressure of oxygen gas is 8.33 atm.

<em>*The options are provided for a different question where </em>

3 0

4 years ago

If a car is traveling along a highway at a speed of 30 meters/second, what distance will it cover in one minute? 0.5 meter 2 met

meriva

D=v×t
=30×60
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7 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.)

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

6 0

3 years ago