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3 years ago

A car accelerate from 25m/s to 50m/s over a time of 10 second.what is acceleration of the car

Physics

1 answer:

Ann [662]3 years ago

5 0

A=(vf-vi)/t
a=(50-25)/10
a=2.5m/s^2

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How to determine the parametric equation using Newtons second law

GenaCL600 [577]

Answer:

Newton's second law of motion states that the acceleration of a system is directly proportional to and in the same direction as the net external force acting on the system, and inversely proportional to its mass. In equation form, Newton's second law of motion is a=Fnetm a = F net m .

Explanation:

6 0

2 years ago

Starburst galaxies:

alexgriva [62]

Answer:

are often associated with a galaxy that is colliding with another galaxy.

Explanation:

A starburst galaxy is a galaxy that undergoes very fast formation of stars. The rate at which stars are born is 100 times more than 3 solar masses per year of the Milky Way. The starburst is stage of the formation of a galaxy. After this stage is complete the stars will have used almost all the gas in it. As the star formation rate is very fast the difference between the age of the stars and the galaxy itself is very less. The star formation is triggered by mergers and tidal interactions between gas-rich galaxies.

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3 years ago

A cutting tool several forces acting on it. One force is F=-axy^2 j , a force in the negative y-direction whose magnitude depend

liq [111]

The force on the tool is entirely in the negative-y direction.
So no work is done during any moves in the x-direction.

The work will be completely defined by

   (Force) x (distance in the y-direction),

and it won't matter what route the tool follows to get anywhere.
Only the initial and final y-coordinates matter.

We know that F = - 2.85 y². (I have no idea what that ' j ' is doing there.)
Remember that 'F' is pointing down.

From y=0 to y=2.40 is a distance of 2.40 upward.

Sadly, since the force is not linear over the distance, I don't think
we can use the usual formula for Work = (force) x (distance).
I think instead we'll need to integrate the force over the distance,
and I can't wait to see whether I still know how to do that.

Work = integral of (F·dy) evaluated from 0 to 2.40

= integral of (-2.85 y² dy) evaluated from 0 to 2.40

   = (-2.85) · integral of (y² dy) evaluated from 0 to 2.40 .

Now, integral of (y² dy) = 1/3 y³ .

Evaluated from 0 to 2.40 , it's (1/3 · 2.40³) - (1/3 · 0³)

      = 1/3 · 13.824 = 4.608 .

And the work = (-2.85) · the integral

   = (-2.85) · (4.608)

   =    - 13.133 .

-- There are no units in the question (except for that mysterious ' j ' after the 'F',
which totally doesn't make any sense at all).
If the ' F ' is newtons and the 2.40 is meters, then the -13.133 is joules.

-- The work done by the force is negative, because the force points
DOWN but we lifted the tool UP to 2.40. Somebody had to provide
13.133 of positive work to lift the tool up against the force, and the force
itself did 13.133 of negative work to 'allow' the tool to move up.

-- It doesn't matter whether the tool goes there along the line x=y , or
by some other route. WHATEVER the route is, the work done by ' F '
is going to total up to be -13.133 joules at the end of the day.

As I hinted earlier, the last time I actually studied integration was in 1972,
and I haven't really used it too much since then. But that's my answer
and I'm stickin to it. If I'm wrong, then I'm wrong, and I hope somebody
will show me where I'm wrong.

3 0

3 years ago

12. A las 10 de la mañana Elena sale a 100 Km/h de una ciudad A con dirección a Madrid. A la misma hora sale Javier desde otra c

Klio2033 [76]

Answer:

a) t = 3.3 [h]

b ) Hora = 13:18 o 1:18 [pm]

c) x = 327.79 [km/h] (Elena)

x = 196.6 [km] (Javier)

Explanation:

Para poder solucionar este problema debemos hacer un planteamiento inicial de ubicacion de las ciudades, este planteamiento nos ayudara a entender el problema de una manera mas facil.

Tenemos las ciudades A & B y la ciudad de Madrid que esta a una distancia x con respecto de B, (ver esquema adjunto).

De manera logica debemos deducir que la Ciudad A debe estar mas lejos de Madrid que la ciudad B de la misma Madrid, ya que en caso contrario Javier nunca alcanzara a Elena, ya que Elena va mas rapido que Javier.

a) Ahora debemos de utilizar la siguiene ecuacion de la cinematica, cuando los cuerpos se mueven a velocidad constante.

$x=x_{o}+v*t$

Donde:

x -xo = Distancia entre el punto inicial y punto final.

v = velocidad [m/s]

t = tiempo [s]

Debemos convertir las velocidades de kilometros por hora a metros por segundo.

$100 [\frac{km}{h} ]*1[\frac{h}{3600s}]*1000[\frac{m}{1km} ] = 27.77[m/s]\\60[\frac{km}{h} ]*1[\frac{h}{3600s}]*1000[\frac{m}{1km} ] = 16.66[m/s]$

Seguidamente formulamos una ecuacion por cada movimiento, luego debemos igualar estas ecuaciones en funcion de la variable x que sera el punto donde se encuentren ambas personas.

Para Elena

$(132000+x) = 27.77*t\\x = 27.77*t - 132000$

Para Javier

$x - xo = 16.66*t\\xo = 0\\x = 16.66*t$

Igualamos las variables x de ambas ecuaciones.

$16.66*t = 27.77*t -132000\\27.77*t - 16.66*t = 132000\\11.11*t = 132000\\t = 11880.83 [s] = 3.3 [h]$

b) La hora facilmente se puede encontrar sumando el tiempo con las 10:00am

hora = 10 + 3 = 13 [hrs]

La parte decimal debe convertirse a tiempo.

$0.3 [hr]*60[\frac{min}{1hr} ]= 18 min$

Hora = 13:18 o 1:18 [pm]

c) Para encontrar estas distancias utilizamos el tiempo encontrado en el item a.

Para Elena

$x = v*t\\x = 27.77*11800.83 = 327791.6 [m] = 327.79 [km]\\$

Para Javier

$x = v*t\\x = 16.66*11800.83 = 196601.8 [m] = 196.6 [km]$

6 0

2 years ago

A pulsar is a rapidly rotating neutron star that emits a radio beam the way a lighthouse emits a light beam. We receive a radio

Angelina_Jolie [31]

a) -0.259 rad/s/y

b) 1732.8 years

c) 0.0069698 s

Explanation:

The angular acceleration of a rotating object is equal to the rate of change of angular velocity of the object.

Mathematically, it is given by

$\alpha=\frac{\Delta \omega}{\Delta t}$

where

$\Delta \omega$ is the change in angular velocity

$\Delta t$ is the time elapsed

The angular velocity can be written as

$\omega=\frac{2\pi}{T}$

where T is the period of rotation of the object.

Therefore, the change in angular velocity can be written as

$\Delta \omega = \frac{2\pi}{T'}-\frac{2\pi}{T}=2\pi (\frac{1}{T'}-\frac{1}{T})$

In this problem:

T = 0.0140 s is the initial period of the pulsar

The period increases at a rate of 8.09 x 10-6 s/y, so after 1 year, the new period is

$T'=T+8.09\cdot 10^{-6} =0.01400809 s$

Therefore, the change in angular velocity after 1 year is

$\Delta \omega =2\pi (\frac{1}{0.01400809}-\frac{1}{0.0140})=-0.259 rad/s$

So, the angular acceleration of the pulsar is

$\alpha = \frac{-0.259 rad/s}{1 y}=-0.259 rad/s/y$

To solve this part, we can use the following equation of motion:

$\omega'=\omega + \alpha t$

where

$\omega'$ is the final angular velocity

$\omega$ is the initial angular velocity

$\alpha$ is the angular acceleration

t is the time

For the pulsar in this problem:

$\omega=\frac{2\pi}{T}=\frac{2\pi}{0.0140}=448.8 rad/s$ is the initial angular velocity

$\omega'=0$ , since we want to find the time t after which the pulsar stops rotating

$\alpha = -0.259 rad/s/y$ is the angular acceleration

Therefore solving for t, we find the time after which the pulsar stops rotating:

$t'=-\frac{\omega}{\alpha}=-\frac{448.8}{-0.259}=1732.8 y$

As we said in the previous part of the problem, the rate of change of the period of the pulsar is

$\frac{\Delta T}{\Delta t}=8.09\cdot 10^{-6} s/y$

which means that the period of the pulsar increases by

$\Delta T=8.09\cdot 10^{-6} s$

For every year:

$\Delta t=1 y$

From part A), we also know that the current period of the pulsar is

T = 0.0140 s

The current period is related to the initial period of the supernova by

$T=T_0+\frac{\Delta T}{\Delta t}\Delta t$

where $T_0$ is the original period and

$\Delta t=869 y$

is the time that has passed; solving for T0,

$T_0=T-\frac{\Delta T}{\Delta t}\Delta t=0.0140 - (8.09\cdot 10^{-6})(869)=0.0069698 s$

6 0

3 years ago