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3 years ago

1. According to Newton's third law of motion, how are action

Physics

1 answer:

san4es73 [151]3 years ago

8 0

Answer:

you couldn't do this on your own or search it up on google

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A particle is moving with (SHM) of period 8.0s and amplitude5.0m

nadezda [96]

Answer:

$velocity(x)=15\,\frac{\pi}{4}\,cos(\frac{\pi}{4}x)$

Max speed = $\frac{15\, \pi}{4} \,\, \frac{m}{s}$

Max acceleration = $\frac{15\,\pi^2}{16} \,\,\frac{m}{s^2}$

Explanation:

Given the description of period and amplitude, the SHM could be described by:

$f(x)=5\,sin(\frac{\pi}{4}x)$

and its angular velocity can be calculated doing the derivative:

$f(x)=5\, \,sin(\frac{\pi}{4}x)\\f'(x)=5\,\frac{\pi}{4}\,cos(\frac{\pi}{4}x)$

And therefore, the tangential velocity is calculated by multiplying this expression times the radius of the movement (3 m):

$velocity(x)=15\,\frac{\pi}{4}\,cos(\frac{\pi}{4}x)$ and is given in m/s.

Then the maximum speed is obtained when the cosine function becomes "1", and that gives:

Max speed = $\frac{15\, \pi}{4} \,\, \frac{m}{s}$

The acceleration is found from the derivative of the velocity expression, and therefore given by:

$acceleraton(x)=-15\,\frac{\pi^2}{16}\,sin(\frac{\pi}{4}x)$

and the maximum of the function will be obtained when the sine expression becomes "-1", which will render:

Max acceleration = $\frac{15\,\pi^2}{16} \,\,\frac{m}{s^2}$

6 0

3 years ago

A hose directs a horizontal jet of water, moving with a velocity of 20m/s, on to a vertical wall. The

just olya [345]

Force is defined as the rate of change of momentum.
The initial amount of momentum is $mv$ because water stops when it hit the wall total change of momentum must be $\Delta p=mv$ .
Now let's calculate the force.
$F= \frac{dp}{dt}=\frac{d(mv)}{dt}=\frac{dm}{dt}v$
We need to find $\frac{dm}{dt}$ . This is the amount of water hiting the wall per second.
$\frac{dm}{dt}=\rho Av$
Our final formula would be:
$F=\rho Avv=\rho Av^2$
And now we can calculate the answer:
$F=1000\cdot5\cdot 10^{-4}\cdot(20)^2=200 N$

6 0

3 years ago

he membrane that surrounds a certain type of living cell has a surface area of 6.0 x 10-9 m2 and a thickness of 1.6 x 10-8 m. As

k0ka [10]

Answer:

$1.54481175\times 10^{-12}\ C$

Explanation:

$\epsilon_0$ = Permittivity of free space = $8.85\times 10^{-12}\ F/m$

A = Area = $6\times 10^{-9}\ m^2$

d = Thickness = $1.6\times 10^{-8}\ m$

k = Dielectric constant = 5.4

V = Voltage = 86.2 mV

Charge is given by

$Q=CV\\\Rightarrow Q=k\epsilon\dfrac{A}{d}V\\\Rightarrow Q=5.4\times 8.85\times 10^{-12}\times \dfrac{6\times 10^{-9}}{1.6\times 10^{-8}}\times 86.2\times 10^{-3}\\\Rightarrow Q=1.54481175\times 10^{-12}\ C$

The charge on the outer surface is $1.54481175\times 10^{-12}\ C$

4 0

3 years ago

Wich usage of water uses the least amount in a year in the united states

Lapatulllka [165]

Live Stock because in 2010 live stock used 10,000 millions gallons of water per day but everything else was higher and irrigation is the highest with 115,000 million gallons per day.

6 0

3 years ago

Read 2 more answers

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 $K_{m}>P_{m}$

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