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2 years ago

10

Un cuerpo se mueve, partiendo del reposo, con una aceleración constante de 8 m/s². Calcular la velocidad que tiene al cabo de 5s

.

1 answer:

nataly862011 [7]2 years ago

8 0

Answer:

v = 40 m / s

Explanation:

Let's use the expressions for accelerated motion

v = v₀ + a t

where vo is the initial velocity, at the acceleration and t is the time.

as the body starts from rest its initial velocity is zero

v = 0 + at

let's calculate

v = 8 5

v = 40 m / s

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In a Hydrogen atom an electron rotates around a stationary proton in a circular orbit with an approximate radius of r =0.053nm.

leonid [27]

Answer:

(a): $F_e = 8.202\times 10^{-8}\ \rm N.$

(b): $F_g = 3.6125\times 10^{-47}\ \rm N.$

(c): $\dfrac{F_e}{F_g}=2.27\times 10^{39}.$

Explanation:

Given that an electron revolves around the hydrogen atom in a circular orbit of radius r = 0.053 nm = 0.053 $\times 10^{-9}$ m.

Part (a):

According to Coulomb's law, the magnitude of the electrostatic force of interaction between two charged particles of charges $q_1$ and $q_2$ respectively is given by

$F_e = \dfrac{k|q_1||q_2|}{r^2}$

where,

$k$ = Coulomb's constant = $9\times 10^9\ \rm Nm^2/C^2.$
$r$ = distance of separation between the charges.

For the given system,

The Hydrogen atom consists of a single proton, therefore, the charge on the Hydrogen atom, $q_1 = +1.6\times 10^{-19}\ C.$

The charge on the electron, $q_2 = -1.6\times 10^{-19}\ C.$

These two are separated by the distance, $r = 0.053\times 10^{-9}\ m.$

Thus, the magnitude of the electrostatic force of attraction between the electron and the proton is given by

$F_e = \dfrac{(9\times 10^9)\times |+1.6\times 10^{-19}|\times |-1.6\times 10^{-19}|}{(0.053\times 10^{-9})^2}=8.202\times 10^{-8}\ \rm N.$

Part (b):

The gravitational force of attraction between two objects of masses $m_1$ and $m_1$ respectively is given by

$F_g = \dfrac{Gm_1m_2}{r^2}.$

where,

$G$ = Universal Gravitational constant = $6.67\times 10^{-11}\ \rm Nm^2/kg^2.$
$r$ = distance of separation between the masses.

For the given system,

The mass of proton, $m_1 = 1.67\times 10^{-27}\ kg.$

The mass of the electron, $m_2 = 9.11\times 10^{-31}\ kg.$

Distance between the two, $r = 0.053\times 10^{-9}\ m.$

Thus, the magnitude of the gravitational force of attraction between the electron and the proton is given by

$F_g = \dfrac{(6.67\times 10^{-11})\times (1.67\times 10^{-27})\times (9.11\times 10^{-31})}{(0.053\times 10^{-9})^2}=3.6125\times 10^{-47}\ \rm N.$

The ratio $\dfrac{F_e}{F_g}$ :

$\dfrac{F_e}{F_g}=\dfrac{8.202\times 10^{-8}}{3.6125\times 10^{-47}}=2.27\times 10^{39}.$

6 0

2 years ago

You wad up a piece of paper and throw it into the wastebasket. How far will

vitfil [10]

The range of the piece of paper is C) 1.4 m

Explanation:

The motion of the piece of paper is the motion of a projectile, which consists of two separate motions:

- A uniform motion along the horizontal direction, with constant velocity

- A uniformly accelerated motion along the vertical direction, with constant acceleration (the acceleration of gravity, $g=9.8 m/s^2$ )

From the equation of motion, it is possible to find an expression for the range (the total horizontal distance covered) of a projectile, which is given by:

$d=\frac{u^2 sin 2\theta}{g}$

where

u is the initial velocity

$\theta$ is the angle of projection

g is the acceleration of gravity

For the piece of paper in this problem,

u = 4.3 m/s

$\theta=65^{\circ}$

Substituting,

$d=\frac{(4.3)^2 sin(2\cdot 65^{\circ})}{9.8}=1.45 m \sim 1.4 m$

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

6 0

2 years ago

Read 2 more answers

When an object slides across the floor, it is slowed down by friction and the floor surface gets warmer. what energy conversion

OverLord2011 [107]

The movement of the object is considered to be kinetic energy while the object getting warmer indicates that there is thermal (heat) energy formed.

Based on this, as the object slides across the floor, friction slows down this motion and the object becomes warmer as kinetic energy is converted into thermal energy.

6 0

2 years ago

A proton moves through a magnetic field at 20.3 % of the speed of light. At a location where the field has a magnitude of 0.0062

galina1969 [7]

Answer:

Force on the proton will be $.73\times 10^{-14}N$

Explanation:

We have given speed of proton is 20.3% of speed of light

Speed of light $c=3\times 10^8m/sec$

So speed of proton $v=\frac{3\times 10^8\times 23}{100}=6.9\times 10^7m/sec$

Magnetic field B = 0.00629 T

Charge on proton $q=1.6\times 10^{-16}C$

Angle between velocity and magnetic field $\Theta =137^{\circ}$

Force on the proton is equal to $F=qvBsin\Theta =1.6\times 10^{-19}\times 6.9\times 10^7\times 0.00629\times sin(137^{\circ})=4.73\times 10^{-14}N$

4 0

2 years ago

A 90kg cannon fires 100kg shell with muzzle velocity of 75m/s. Calculate the recoil velocity of the cannon relative to the groun

Gala2k [10]

Answer: 83.8 m/s

Explanation:

momentum of shell = momentum of cannon

M(s) × V(s) = M(c) × V(c)

M(s) = mass of shell, V(s) = velocity of shell

M(c) = mass of cannon, V(c) = velocity of cannon

100kg × 75m/s = 90kg × V(c)

7500 = 90 × V(c)

7500 ÷ 90 = V(c)

83.3 = V(c)

5 0

2 years ago