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Kamila [148]
3 years ago
8

What is the most common frame of reference used by humans

Physics
1 answer:
valkas [14]3 years ago
4 0

Answer:

All measurements of motion will be compared to a frame of reference. Therefore, the most commonly used frame of reference is Earth itself, even though it moves.

Explanation:

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Today's topic
babunello [35]

Answer:

While a body is said to be in motion if it changes its position with respect to immediate surroundings.

A body is said to be in uniform motion if it covers equal distances in equal interval of time.

A body is said to be in non-uniform motion if it covers unequal distances in equal interval of time or vice-versa

4 0
2 years ago
A planet has been observed orbiting a nearby star. This star has a mass of 3.5 solar masses, and the planet is 4.2 AU from the s
kondaur [170]

Answer:

4.6 years

Explanation:

This is solved using Kepler's third law which says:

T^2=\frac{4\pi ^2}{GM} a^2

Where

T = Orbital period of the planet (in seconds)

a = Distance from the star (in meters)

G = Gravitational constant

M = Mass of the parent star (in kg)

From the information given

M = 3.5M_{sun} = 6.96*10^{30} kg

a=4.2AU = 6.28*10^{11} meters

G = 6.67*10^{-11}m^3kg^{-1}s^{-2}

We put this into Kepler's law and get:

T=\sqrt{\frac{4\pi ^2}{6.67*10^{-11}*6.96*10^{30}} (6.28*10^{11})^3}=145,128,196 seconds.

This when converted to years is 4.6 years.

4 1
3 years ago
Read 4 more answers
When devising a model, scientists can only use the information available during their lifetime. This means that the current mode
11111nata11111 [884]

no, it not useless. we still learn Bohr's model in HS n dats almost 200 yr old! while there may be new models, previous one is good for explaining the basics. it is also useful to learn previous model n see how our understanding improves over time.


6 0
3 years ago
Read 2 more answers
"
Volgvan

Answer:

The height is : 60.025 m

Explanation:

The flowerpot falls off the balcony with zero launch angle

Given the time of fright as 3.5 s then ;

The formula to apply is ;

T=\sqrt{\frac{2H}{g} }\\\\3.5=\sqrt{\frac{2H}{9.8} }

3.5²= 2H/9.8

12.25 =2H/9.8

12.25 * 9.8 = 2H

120.05 = 2H

120.05/2 = H

60.025 =H

7 0
3 years ago
An offshore oil well is 2 kilometers off the coast. The refinery is 4 kilometers down the coast. Laying pipe in the ocean is twi
shusha [124]

Answer:

Rectangular path

Solution:

As per the question:

Length, a = 4 km

Height, h = 2 km

In order to minimize the cost let us denote the side of the square bottom be 'a'

Thus the area of the bottom of the square, A = a^{2}

Let the height of the bin be 'h'

Therefore the total area, A_{t} = 4ah

The cost is:

C = 2sh

Volume of the box, V = a^{2}h = 4^{2}\times 2 = 128            (1)

Total cost, C_{t} = 2a^{2} + 2ah            (2)

From eqn (1):

h = \frac{128}{a^{2}}

Using the above value in eqn (1):

C(a) = 2a^{2} + 2a\frac{128}{a^{2}} = 2a^{2} + \frac{256}{a}

C(a) = 2a^{2} + \frac{256}{a}

Differentiating the above eqn w.r.t 'a':

C'(a) = 4a - \frac{256}{a^{2}} = \frac{4a^{3} - 256}{a^{2}}

For the required solution equating the above eqn to zero:

\frac{4a^{3} - 256}{a^{2}} = 0

\frac{4a^{3} - 256}{a^{2}} = 0

a = 4

Also

h = \frac{128}{4^{2}} = 8

The path in order to minimize the cost must be a rectangle.

8 0
3 years ago
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