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3 years ago

Mechanical energy is lost:

1 answer:

7 0

3. Due to the fact that friction is not converted to kinetic energy nor potential energy. The energy is converted into heat energy which is lost and can’t be put back

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An initially stationary electron is accelerated by a uniform 640 N/C Electric Field. a) Calculate the kinetic energy of the elec

bulgar [2K]

Answer:

(a) 1.298 * 10^(-4) J

(b) 5.82 * 10^6 m/s

Explanation:

Parameters given:

Electric field, E = 640 N/C

Distance traveled by electron, r = 15 cm = 0.15 m

Mass of electron, m = 9.11 * 10^(-31) kg

Electric charge of electron, q = 1.602 * 10^(-19) C

(a) The kinetic energy of the electron in terms of Electric field is given as:

K = (q² * E² * r²) / 2m

Therefore, Kinetic energy, K, is:

K = [(1.602 * 10^(-19))² * 640² * 0.15²] / [2 * 9.11 * 10^(-31)]

K = {23651.981 * 10^(-38)} / [18.22 * 10^(-31)]

K = 1298.13 * 10^(-7) J = 1.298 * 10^(-4) J

(b) To find the final velocity of the electron, we have to first find the acceleration of the electron. This can be gotten by using the equations of force.

Force is generally given as:

F = ma

Electric force is given as:

F = qE

Therefore, equating both, we have:

ma = qE

a = (qE) / m

a = (1.602 * 10^(-19) * 640) / (9.11 * 10^(-31))

a = 112.54 * 10^(12) m/s² = 1.13 * 10^(14) m/s²

Using one of the equations of motion, we have that:

v² = u² + 2as

Since the electron started from rest, u = 0 m/s

Therefore:

v² = 2 * 1.13 * 10^(14) * 0.15

v² = 3.39 * 10^(13)

v = 5.82 * 10^6 m/s

The velocity of the electron after moving a distance of 15 cm is 5.82 * 10^6 m/s.

8 0

3 years ago

Two vehicles are traveling when they enter an intersection and crash and stick together. Both have a mass of 1,650 kg and both a

seropon [69]

Answer:

10.61 m/s

hope this helps!

8 0

3 years ago

Why might farmer plant purebred crop seeds?

S_A_V [24]

Answer:

They can be relied upon to produce the same traits generation after generation

Explanation:

7 0

3 years ago

Read 2 more answers

A spring with a spring constant kk of 100 pounds per foot is loaded with 1-pound weight and brought to equilibrium. it is then s

tino4ka555 [31]

Given:

k = 100 lb/ft, m = 1 lb / (32.2 ft/s) = 0.03106 slugs

Solution:

F = -kx
mx" = -kx
x" + (k/m)x = 0

characteristic equation:
r^2 + k/m = 0
r = i*sqrt(k/m)

x = Asin(sqrt(k/m)t) + Bcos(sqrt(k/m)t)

ω = sqrt(k/m)
2π/T = sqrt(k/m)
T = 2π*sqrt(m/k)
T = 2π*sqrt(0.03106 slugs / 100 lb/ft)
T = 0.1107 s (period)

x(0) = 1/12 ft = 0.08333 ft
x'(0) = 0

1/12 = Asin(0) + Bcos(0)
B = 1/12 = 0.08333 ft

x' = Aω*cos(ωt) - Bω*sin(ωt)
0 = Aω*cos(0) - (1/12)ω*sin(0)
0 = Aω
A = 0

So B would be the amplitude. Therefore, the equation of motion would be x = 0.08333*cos[(2π/0.1107)t]

5 0

4 years ago

Bare free charges do not remain stationary when close together. To illustrate this, calculate the acceleration of two isolated p

Marta_Voda [28]

Answer:

Acceleration, $a=9.91\times 10^{15}\ m/s^2$

Explanation:

It is given that,

Separation between the protons, $r=3.73\ nm=3.73\times 10^{-9}\ m$

Charge on protons, $q=1.6\times 10^{-19}\ C$

Mass of protons, $m=1.67\times 10^{-27}\ kg$

We need to find the acceleration of two isolated protons. It can be calculated by equating electric force between protons and force due to motion as :

$ma=\dfrac{kq^2}{r^2}$

$a=\dfrac{kq^2}{mr^2}$

$a=\dfrac{9\times 10^9\times (1.6\times 10^{-19})^2}{1.67\times 10^{-27}\times (3.73\times 10^{-9})^2}$

$a=9.91\times 10^{15}\ m/s^2$

So, the acceleration of two isolated protons is $9.91\times 10^{15}\ m/s^2$ . Hence, this is the required solution.

3 0

3 years ago