A person must pull a rope 20 meters with a force of 200 N what is the work input?

Consider an electron that is 10-10 m from an alpha particle (9 = 3.2 x 10-19 C). (Enter the magnitudes.) (a) What is the electri

Studentka2010 [4]

Answer:

a) $E=2.88*10^{11}N/C$

b) $E=1.44*10^{11}N/C$

c) $F=4.61*10^{-8}N$

Explanation:

We use the definition of a electric field produced by a point charge:

$E=k*q/r^2$

a)Electric Field due to the alpha particle:

$E=k*q_{alpha}/r^2=9*10^9*3.2*10^{-19}/(10^{-10})^2=2.88*10^{11}N/C$

b)Electric Field due to electron:

$E=k*q_{electron}/r^2=9*10^9*1.6*10^{-19}/(10^{-10})^2=1.44*10^{11}N/C$

c)Electric Force on the alpha particle, on the electron:

The alpha particle and electron feel the same force but with opposite direction:

$F=k*q_{electron}*q_{alpha}/r^2=9*10^9*1.6*10^{-19}*3.2*10^{-19}/(10^{-10})^2=4.61*10^{-8}N$

4 0

3 years ago

Two kids are playing on a newly installed slide, which is 3 m long. John, whose mass is 30 kg, slides down into William (20 kg),

yuradex [85]

Answer:

$v=3.564\ m.s^{-1}$

$\Delta v =2.16\ m.s^{-1}$

Explanation:

Given:

mass of John, $m_J=30\ kg$

mass of William, $m_W=30\ kg$

length of slide, $l=3\ m$

(A)

height between John and William, $h=1.8\ m$

Using the equation of motion:

$v_J^2=u_J^2+2 (g.sin\theta).l$

where:

v_J = final velocity of John at the end of the slide

u_J = initial velocity of John at the top of the slide = 0

Now putting respective :

$v_J^2=0^2+2\times (9.8\times \frac{1.8}{3})\times 3$

$v_J=5.94\ m.s^{-1}$

Now using the law of conservation of momentum at the bottom of the slide:

Sum of initial momentum of kids before & after collision must be equal.

$m_J.v_J+m_w.v_w=(m_J+m_w).v$

where: v = velocity with which they move together after collision

$30\times 5.94+0=(30+20)v$

$v=3.564\ m.s^{-1}$ is the velocity with which they leave the slide.

(B)

frictional force due to mud, $f=105\ N$

Now we find the force along the slide due to the body weight:

$F=m_J.g.sin\theta$

$F=30\times 9.8\times \frac{1.8}{3}$

$F=176.4\ N$

Hence the net force along the slide:

$F_R=71.4\ N$

Now the acceleration of John:

$a_j=\frac{F_R}{m_J}$

$a_j=\frac{71.4}{30}$

$a_j=2.38\ m.s^{-2}$

Now the new velocity:

$v_J_n^2=u_J^2+2.(a_j).l$

$v_J_n^2=0^2+2\times 2.38\times 3$

$v_J_n=3.78\ m.s^{-1}$

Hence the new velocity is slower by

$\Delta v =(v_J-v_J_n)$

$\Delta v =5.94-3.78= 2.16\ m.s^{-1}$

8 0

3 years ago

Imagine that an electron in an excited state in a nitrogen molecule decays to its ground state, emitting a photon with a frequen

mash [69]

Since energy cannot be created nor destroyed, the change in energy of the electron must be equal to the energy of the emitted photon.

The energy of the emitted photon is given by:
$E=hf$
where
h is the Planck constant
f is the photon frequency
Substituting $f=8.88 \cdot 10^{14}Hz$ , we find
$E=hf=(6.6 \cdot 10^{-34} Js)(8.88 \cdot 10^{14} Hz)=5.86 \cdot 10^{-19} J$

This is the energy given to the emitted photon; it means this is also equal to the energy lost by the electron in the transition, so the variation of energy of the electron will have a negative sign (because the electron is losing energy by decaying from an excited state, with higher energy, to the ground state, with lower energy)
$\Delta E= -5.86 \cdot 10^{-19} J$

6 0

3 years ago

Read 2 more answers

Which term describes the number of joules used per second by a circuit?

jasenka [17]

Answer:

B) Power

Explanation:

The power is defined by the following equation:

P = W / t

where:

W = work = Force * Distance = [Newton] * [meter]

t = time = seconds

The units for work are give en Newton per second, which is equal to Joules

And for power the unit used commonly is Watts, therefore:

Watts = (Joule/second)

5 0

3 years ago

Read 2 more answers

I need to figure out A but i’m not sure how

Damm [24]

Answer:

cc

Explanation:

4 0

3 years ago