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3 years ago

Which Law is connected with inertia?

Physics

2 answers:

Ilya [14]3 years ago

6 0

Newton's first law of motion is connected with inertia :)

GuDViN [60]3 years ago

4 0

Hey friends
The answer to this question is newtons law of motion.
Hope i helped
~Katie

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Which equation describes the sum of the vectors plotted below?

worty [1.4K]

Answer:

It's the third one in the picture ...

r = 3x - 2y

Explanation:

5 0

2 years ago

For a satellite of mass mS in a circular orbit of radius rS around the Earth, determine its kinetic energy, K . Express your ans

agasfer [191]

Since it is asking you to find the kinetic energy in relation to the mass, radius, mechanical energy (total energy), and constants, you will need to setup an equation first to "find" the Mechanical Energy, so that we can then solve for the kinetic energy, as from my experience with high school physics, there is only the graviational potential energy equation and force in relation to celestial bodies.

Knowing the ME is the total energy, we add up the energies of the system. Since it is being influenced by the Earth, as per the problem stating the satellite has circular orbit around the Earth, we know there is gravitational potential. Since it is orbiting, we can assume some type of velocity. Nothing else that we need to worry about should be occuring at this level of physics, leaving you with

ME= Ug+K

from here we solve for K, as plugging in could get confusing and messy at the moment.

ME-Ug=K

now using the equations presumably given in class, if not then using this equation, we can find the Ug

Ug=(-(Gm*M)/r) note that M is the mass of the Earth and m is the satellite

this should give us

ME-(-(GmM)/r)=K

since there is a negative being subracted, we can change that to

ME+(GmM)/r=K

I believe this should be fine, as the Earth's mass is constant, but if not, then all you need to figure left is how to get rid of the M in the equation, as the rest of the terms and constants are for sure within the requirements.

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2 years ago

A hot-air balloon and its basket are accelerating upward at 0.265 m/s2, propelled by a net upward force of 688 N. A rope of negl

Sergeu [11.5K]

Question: Find, separately, them mass of the balloon and the basket (incidentally, most of the balloon's mass is air)

Answer:

The mass of the balloon is 2295 kg, and the mass of the basket is 301 kg.

Explanation:

Let us call the mass of the balloon $m_1$ and the mass of the basket $m_2$ , then according to newton's second law:

$(1). \:F = (m_1+m_2)a$ ,

where $a =0.265m/s^2$ is the upward acceleration, and $F = 688N$ is the net propelling force (counts the gravitational force).

Also, the tension $T$ in the rope is 79.8 N more than the basket's weight; therefore,

$(2). \:T = m_2g+79.8$

and this tension must equal

$T -m_2g =m_2a$

$(3). \:T = m_2g +m_2a$

Combining equations (2) and (3) we get:

$m_2a = 79.8$

since $a =0.265m/s^2$ , we have

$\boxed{m_2 = 301.13kg}$

Putting this into equation (1) and substituting the numerical values of $F$ and $a$ , we get:

$688N = (m_1+301.13kg)(0.265m/s^2)$

$\boxed{m_1 = 2295 kg}$

Thus, the mass of the balloon and the basket is 2295 kg and 301 kg respectively.

8 0

3 years ago

A model air rocket with a mass of 50.g is free to travel along a horizontal track. It begins from rest. After 2.0s, the rocket h

katen-ka-za [31]

Answer:

$v_r=5.89\ m.s^{-1}$

Explanation:

Given:

mass of rocket, $m_r=50\ g$

time of observation, $t=2\ s$

mass lost by the rocket by expulsion of air, $m_a=10\%\ of m_r=5\ g$

velocity of air, $v_a=53\ m.s^{-1}$

Now the momentum of air will be equal to the momentum of rocket in the opposite direction: (Using the theory of elastic collision)

$m_a.v_a=(m_r-m_a)\times v_r$

$5\times 53=(50-5)\times v_r$

$v_r=5.89\ m.s^{-1}$

7 0

3 years ago

Alex is standing still and throws a football with a speed of 10 m/s to his friend, who is also standing still. The two friends a

Phantasy [73]

The question is incomplete. It comes with a set of answer choices.

These are the answer choices:

Alex observes it as 10 m/s, and his friend observes it as less than 10 m/s.

Alex observes it as less than 10 m/s, and his friend observes it as 10 m/s.

Both Alex and his friend observe it as 10 m/s.

Both Alex and his friend observe it as less than 10 m/s.

Answer: Both Alex and his friend observe it as 10 m/s.

Justification:

1) The speed is relative to the frame of reference.

2) It is said that the both Alex and his friend are standing still.

3) Then, the speed they both see is the same, 10 m/s, respect the Earth (where they are standing still).

Of course, Alex is watching the ball moving away and his friend is seing it approaching, but it is not relevant for the question, as it deals with the speed which is only about magnitude, not direction.

7 0

3 years ago

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