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3 years ago

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A 730 kg Mini runs into a stationary 2500 kg SUV. If the Mini was moving at 10. M , how s fast will it be moving after making a

completely inelastic collision with the SUV?

1 answer:

spayn [35]3 years ago

5 0

Answer:

2.26m/s

Explanation:

Since it is an inelastic collision, the mini and the Suv move together after the collision. The law of conservation of momentum states that the total momentum before and after the collision is conserved:

Using law of conservation of momentum,

$m_{1} u_{1}+m_{2} u_{2}=\left(m_{1}+m_{2}\right) v$

Given that,

$m_{1}$ = 730 kg

$u_{1}$ = 10 m/s

$u_{2}$ = 0

$m_2 =$ 2500kg

v = ?

When substituting the given values in the above equation, we get

$730(10) + 2500(0) = (730+2500)v\\\\7300 + 0 = 3230v\\\\v = \frac{7300}{3200} \\\\= 2.26m/s$

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Please help.. urgent Which statement is equivalent to Newton's first law? a. 15,300 N b. 1.20*10^3 N c. 2,030 N d. 1,560 N

taurus [48]

According to Newton laws of motion,
F = m*a
Here, m = 1,560 Kg
a = 1.30 m/s²

Substitute their values,
F = 1,560 * 1.30
F = 2028 N ~ 2030 N [ Closest value ]

In short, Your Answer would be Option C

Hope this helps!

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3 years ago

How did J.J. Thomson's experiments demonstrate that Dalton's model of the atom was incorrect?

morpeh [17]

The correct option is C.
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Read 2 more answers

Astronomers suspect that a galaxy’s type can be affected both by the conditions in the protogalactic cloud from which it forms (

Katen [24]

Answer:

The items here are describing either a condition in a later interacton or a protogalactic cloud. The results matching with spiral and elliptical galaxy are:

For spiral galaxy are options 6,3,2 and 5.

and for elliptical galaxy are options 4 and 1.

Explanation:

Here it is given that astrnomers suspect that types of galaxy can be affected both by the conditions which occurs due to protogalactic cloud and then from it forms the initial conditions and then by the later interactions with the other galaxies.

so, both types of galaxies are matched with their respective items given:

A. Spiral galaxy:

2. A galaxy collision results tostripping of gas.

3. The protogalactic cloud rotates in a very slow motion.

5. The density of protogalactic cloud is very high.

6. when the protogalactic cloud shrinks cloud forms very rapidly.

B. Elliptical galaxy:

1. The protogalactic cloud has high angular momentum.

4. Most of the protogalactic gases settles down into a disk.

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3 years ago

How to find resultant force?

lys-0071 [83]

Assuming you're working in a 3D cartesian coordinate system, i.e. each point in space has an x, y, and z coordinate, you add up the forces' x/y/z components to find the resultant force.

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3 years ago

4) A satellite, mass m, is in circular orbit (radius r) around the earth, which has mass ME and radius Re. The value of r is lar

defon

<h2>Answers:</h2>

(a) The kinetic energy of a body is that energy it possesses due to its movement and is defined as:

$K=\frac{1}{2}m{V}{2}$ (1)

Where $m$ is the mass of the body and $V$ its velocity.

In this specific case of the satellite, its kinetic energy $K_m$ taking into account its mass $m$ is:

$K_{m}=\frac{1}{2}m{V}^{2}$ (2)

On the other hand, the velocity of a satellite describing a circular orbit is constant and defined by the following expression:

$V=\sqrt{G\frac{ME}{r}}$ (3)

Where $G$ is the gravity constant, $ME$ the mass of the Earth and $r$ the radius of the orbit <u>(measured from the center of the Earth to the satellite). </u>

Now, if we substitute the value of $V$ from equation (3) on equation (2), we will have the final expression of the kinetic energy of this satellite:

$K_{m}=\frac{1}{2}m{\sqrt{G\frac{ME}{r}}}^{2}$ (4)

Finally:

$K_{m}=\frac{1}{2}Gm\frac{ME}{r}$ (5) >>>>This is the kinetic energy of the satellite

(b) According to Kepler’s 2nd Law applied in the case of a circular orbit, its Period $T$ is defined as:

$T=2\pi\sqrt{\frac{r^{3}}{\mu}}$ (6)

Where $\mu$ is a constant and is equal to $GME$ . So, this equation in these terms is written as:

$T=2\pi\sqrt{\frac{r^{3}}{GME}}$ (7)

As we can see, <u>the Period of the orbit does not depend on the mass of the satellite </u> $m$ , it depends on the mass of the greater body (the Earth in this case) $ME$ , the radius of the orbit and the gravity constant.

(c) The gravitational force described by the law of gravity is a central force and therefore is <u>a conservative force</u>. This means:

1. The work performed by a gravitational force to move a body from a position A to a position B <u>only depends on these positions and not on the path followed to get from A to B. </u>

2. When the path that the body follows between A and B is a c<u>losed path or cycle</u> (as this case with a <u>circular orbit</u>), <u>the gravitational work is null or zero</u>.

<h2>This is because the gravity force that maintains an object in circular motion is a centripetal force, that is, <u>it always acts perpendicular to the movement</u>. </h2>

Then, in the case of the satellite orbiting the Earth in a circular orbit, its movement will always be perpendicular to the gravity force that attracts it to the planet, at each point of its path.

(d) The total Mechanical Energy $E$ of a body is the sum of its Kinetic Energy $K$ and its Potential Energy $P$ :

$E=K+P$ (8)

But in this specific case of the circular orbit, its kinetic energy will be expresses as calculated in the first answer (equation 5):

$K_{m}=\frac{1}{2}Gm\frac{ME}{r}$ (5)

And its potential energy due to the Earth gravitational field as:

$P_{m}=-G\frac{mME}{r}$ (9)

This energy is negative by definition.

So, the total mechanical energy of the orbit, also called the Orbital Energy is:

$E=\frac{1}{2}Gm\frac{ME}{r}+(- G\frac{mME}{r})$ (10)

Solving equation (10) we finally have the Orbital Energy:

$E=-\frac{1}{2}mME\frac{G}{r}$ (11)

At this point, it is necessary to clarify that a satellite (or any other celestial body) orbiting another massive body, can describe one of these types of orbits depending on its Orbital Total Mechanical Energy $E$ :

-When $E=0$ :

We are talking about an <u>open orbit</u> in which the satellite escapes from the attraction of the planet's gravitational field. The shape of its trajectory is a parabola, fulfilling the following condition:

$K_{m}=-P_{m}$

Such is the case of some comets in the solar system.

-When $E>0$ :

We are also talking about <u>open orbits</u>, which are hyperbolic, being $K_{m}>P_{m}$

<h2>-When $E$ : >>>><u>This case</u></h2>

We are talking about <u>closed orbits</u>, that is, the satellite will always be "linked" to the gravitational field of the planet and will describe an orbit that periodically repeats with a shape determined by the relationship between its kinetic and potential energy, as follows:

-Elliptical orbit: Although $E$ is constant, $K_m$ and $P_m$ are changing along the trajectory .

-Circular orbit: When at all times both the kinetic energy $K_m$ and the potential $P_m$ remain constant, resulting in a total mechanical energy $E$ as the one obtained in this exercise. This means that the speed is constant too and <u>is the explanation of why this Energy has a negative sign. </u>

3 0

4 years ago