Ask question

Ask question

All categories

3 years ago

6

Jocko the clown, whose mass is 60-Kg, stands on a skateboard. A 20-Kg ball is thrown at Jocko at 3m/s, and when he catches the b

all, he and the ball move on the skateboard. How fast do Jocko and the ball move after he catches the ball?

1 answer:

Mekhanik [1.2K]3 years ago

7 0

Answer:

The speed of the Jocko and the ball move after he catches the ball is 0.75 m/s.

Explanation:

Given that,

Mass if Jocko, m = 60 kg

Mass of the ball, m' = 20 kg

Speed of the ball, v = 3 m/s

Let V is the speed of Jocko and the ball move after he catches the ball. The momentum of the system remains conserved. Using the conservation of momentum as :

$m'v'=(m+m')V\\\\V=\dfrac{m'v'}{(m+m')}\\\\V=\dfrac{20\times 3}{(60+20)}\\\\V=0.75\ m/s$

So, the speed of the Jocko and the ball move after he catches the ball is 0.75 m/s.

You might be interested in

What does gravitational potential energy depend on?

jolli1 [7]

It depends on the mass of an object and acceleration because of the gravity and the height of an object

3 0

3 years ago

What is the resolution of an analog-to-digital converter with a word length of 12 bits and an analogue signal input range of 100

Elena L [17]

Answer:

The resolution of an analog-to-digital converter is 24.41 mV

Explanation:

Resolution of an analog-to-digital = (analogue signal input range)/2ⁿ

where;

n is the number or length of bit, and in this question it is given as 12

Also, the analogue signal input range is 100V

Resolution of an analog-to-digital = 100V/2¹²

2¹² = 4096

Resolution of an analog-to-digital = 100V/4096

Resolution of an analog-to-digital = 0.02441 V = 24.41 mV

Therefore, the resolution of an analog-to-digital converter is 24.41 mV

5 0

3 years ago

Which part of an atom makes up most of its volume ?

alexandr1967 [171]

Remember that like charges repel each other. That is, positive repels positive and negative repels negative. Similar to how the north poles of magnets repel each other and south poles repel. However, at the atomic scale, protons, which have positive charge, are more influenced by the "Strong Force," which binds them close together. If they were to be separated ever so slightly, then the electromagnetic force would take over and they would repel each other like you'd expect.

Neutrons are also held together via the Strong Force, but don't have a charge so when separated, don't have an electromagnetic force pushing them away from each other.

However, electrons act differently. There is no "Strong Force" just the electromagnetic force. So, they keep a great distance from each other.

So in an atom, protons and neutrons stay close to each other, taking up little volume, while electrons take up a lot of volume.

BTW, the reason why electrons and protons act differently when they are close together is because protons are made up of smaller particles the carry this Strong Force. For electrons, there is no smaller constituent. And therefore, all you have is the electromagnetic force to influence it. That's it.

Hope that helps.

3 0

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

3 years ago

How much force is required to accelerate a 22Kg mass at 6 m/s?

GuDViN [60]

Answer:

F = 132N

Explanation:

7 0

3 years ago

Read 2 more answers