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tamaranim1 [39]
3 years ago
11

An electric vehicle starts from rest and accelerates at a rate of 2.4 m/s2 in a straight line until it reaches a speed of 27 m/s

. The vehicle then slows at a constant rate of 1.3 m/s2 until it stops. (a) How much time elapses from start to stop? (b) How far does the vehicle move from start to stop?
Physics
1 answer:
DENIUS [597]3 years ago
6 0
A) use v=u+at for both

First section, v=27, u=0, a=2.4. You should get 11seconds.
Second section, v=0, u=27, a=-1.3. You should get 21seconds.
This means that the total time is 22seconds.

b) You can either use s=ut+0.5at^2 or v^2=u^2+2as. Personally, I would use the second one as you are not relying on your previous answer.

First section, v=27, u=0, a=2.4. You should get 152m.
Second section, v=0, u=27, a=-1.3. You should get 280m.
This makes your overall displacement 432m.
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A droplet of pure mercury has a density of 13.6 g/cm3. What is the density of a sample of pure mercury that is 10 times as large
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A droplet of pure mercury has a density of 13.6 g/cm3. What is the density of a sample of pure mercury that is 10 times as large as the droplet?

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3 years ago
A projectile is shot directly away from Earth's surface. Neglect the rotation of the Earth. What multiple of Earth's radius RE g
natali 33 [55]

(a) 5.65 times the Earth's radius

The escape velocity for a projectile on Earth is

v_e=\sqrt{\frac{2GM}{R}}

where

G is the gravitational constant

M is the Earth's mass

R is the Earth's radius

If the projectile has an initial speed of 0.421 escape speed,

v=0.421 v_e

So its initial kinetic energy will be

K=\frac{1}{2}m(0.421 v)^2=0.089 m(\sqrt{\frac{2GM}{R}})^2=0.177 \frac{GMm}{R}

where m is the mass of the projectile

At the point of maximum altitude, all this energy is converted into gravitational potential energy:

K=U\\0.177 \frac{GMm}{R}=\frac{GMm}{r}

where r is the distance from the Earth's centre reached by the projectile. We can write r as a multiple of R, the Earth's radius:0.177 \frac{GMm}{R}=\frac{GMm}{nR}

And solving the equation we find

n=\frac{1}{0.177}=5.65

So, the projectile reaches a radial distance of 5.65 times the Earth's radius.

b) 2.36 times the Earth's radius

The kinetic energy needed to escape is:

K=\frac{1}{2}mv_e^2 = \frac{1}{2}m(\sqrt{\frac{2GM}{R}})^2=\frac{GMm}{R}

This time, the projectile has 0.421 times this energy:

K=0.421 \frac{GMm}{R}

Again, at the point of maximum altitude, all this energy will be converted into potential energy:

0.421 \frac{GMm}{R}=\frac{GMm}{nR}

and by solving for n we find

n=\frac{1}{0.421}=2.36

So, the projectile reaches a radial distance of 2.36 times the Earth's radius.

c) E=U=\frac{GMm}{R}

The least initial mechanical energy needed for the projectile to escape Earth is equal to the gravitational potential energy of the projectile at the Earth's surface:

E=U=\frac{GMm}{R}

Indeed, the kinetic energy of the projectile must be equal to this value. In fact, if we use the formula of the escape velocity inside the formula of the kinetic energy, we find

K_e=\frac{1}{2}mv_e^2 = \frac{1}{2}m(\sqrt{\frac{2GM}{R}})^2=\frac{GMm}{R}

6 0
2 years ago
Una carga positiva de 4 x 10-5 C, se encuentra a 0.05 m de otra carga positiva de 2 x 10-5 C. Calcular la fuerza que se ejerce e
KATRIN_1 [288]

Answer:

La fuerza que se ejerce entre las dos cargas es 2880 N.

Explanation:

La ley de Coulomb indica que los cuerpos cargados sufren una fuerza atractiva o repulsiva al acercarse. La fuerza es atractiva si las cargas son del signo opuesto y repulsión si son del mismo signo. El valor de la fuerza es proporcional al producto del valor de sus cargas e inversamente proporcional al cuadrado de la distancia que los separa. Esto se expresa matemáticamente como:

F=k*\frac{Q*q}{r^{2} }

donde:

  • F es la fuerza eléctrica de atracción o repulsión. Se mide en Newtons (N).
  • Q y q son los valores de las dos cargas puntuales. Se miden en culombios (C).
  • r es el valor de la distancia que los separa. Se mide en metros (m).
  • k es una constante de proporcionalidad llamada constante de la ley de Coulomb.

En este caso:

  • F= ?
  • Q= 4*10⁻⁵ C
  • q= 2*10⁻⁵ C
  • r= 0.05 m
  • k= 9*10⁹ \frac{N*m^{2} }{C^{2} }

Reemplazando:

F=9*10^{9} \frac{N*m^{2} }{C^{2} }*\frac{4*10^{-5} C*2*10^{-5}C }{(0.05 m)^{2} }

F= 2880 N

<u><em>La fuerza que se ejerce entre las dos cargas es 2880 N.</em></u>

7 0
3 years ago
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