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zavuch27 [327]
3 years ago
15

Two thin conducting plates, each 24.0 cm on a side, are situated parallel to one another and 3.4 mm apart. If 1012 electrons are

moved from one plate to the other, what is the electric field between the plates?
Physics
1 answer:
adoni [48]3 years ago
8 0

Answer:

E=3.307×10⁻⁴N/C

Explanation:

Given data

Length of the plate side L=24.0 cm =0.24 m

Distance between the plates d= 3.4 mm

Number of electron moves from one plate to others n=1012 electrons

Permittivity of free space ε₀=8.85×10⁻¹²c²/N.m²

Electron charge e=-1.6×10⁻¹⁹C

To find

Electric field between the plates

Solution

E=σ/ε₀

E=\frac{(Q/A)}{E}\\ E=\frac{Q}{EA}\\ E=\frac{ne}{EA}\\ E=\frac{1012*()1.6*10^{-19} }{8.85*10^{-12}(0.24)(0.24)}\\ E=3.307*10^{-4}N/C

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Arrow_forward
garri49 [273]

Explanation:

(a) Hooke's law:

F = kx

7.50 N = k (0.0300 m)

k = 250 N/m

(b) Angular frequency:

ω = √(k/m)

ω = √((250 N/m) / (0.500 kg))

ω = 22.4 rad/s

Frequency:

f = ω / (2π)

f = 3.56 cycles/s

Period:

T = 1/f

T = 0.281 s

(c) EE = ½ kx²

EE = ½ (250 N/m) (0.0500 m)²

EE = 0.313 J

(d) A = 0.0500 m

(e) vmax = Aω

vmax = (0.0500 m) (22.4 rad/s)

vmax = 1.12 m/s

amax = Aω²

amax = (0.0500 m) (22.4 rad/s)²

amax = 25.0 m/s²

(f) x = A cos(ωt)

x = (0.0500 m) cos(22.4 rad/s × 0.500 s)

x = 0.00919 m

(g) v = dx/dt = -Aω sin(ωt)

v = -(0.0500 m) (22.4 rad/s) sin(22.4 rad/s × 0.500 s)

v = -1.10 m/s

a = dv/dt = -Aω² cos(ωt)

a = -(0.0500 m) (22.4 rad/s)² cos(22.4 rad/s × 0.500 s)

a = -4.59 m/s²

3 0
3 years ago
A 200.0 g block rests on a frictionless, horizontal surface. It is pressed against a horizontal spring with spring constant 4500
kenny6666 [7]

Answer:

 6 m/s

Explanation:

Given that :

mass of the block   m =  200.0 g  = 200 × 10⁻³ kg

the horizontal spring constant   k  =  4500.0 N/m

position of the block (distance x) = 4.00 cm  = 0.04 m

To determine the speed the block will be traveling when it leaves the spring; we applying the  work done on the spring as it is stretched (or compressed) with the kinetic energy.

i.e \frac{1}{2} kx^2  = \frac{1}{2} mv^2

kx^2 = mv^2

4500* 0.04^2 = 200*10^{-3} *v^2

7.2 =200*10^{-3}*v^{2}

v^{2}   =\frac{7.2}{200*10^{-3}}

v   =\sqrt{\frac{7.2}{200*10^{-3}}}

v = 6 m/s

Hence,the speed the block will be traveling when it leaves the spring is  6 m/s

5 0
4 years ago
A speed boat travels from the dock to the first buoy a distance of 20 meters in 18 seconds it began the trip at a speed of 0 m/s
Lunna [17]

Answer:

1.11 m/s

Explanation:

The motion of the boat is an example of accelerated motion, since the velocity is not constant. However, we don't need to find the acceleration, because we are only interested in the average velocity of the boat, which is given by:

v=\frac{d}{t}

where d is the total distance covered and t the time taken. In this problem, the boat covered a distance of d = 20 m and it takes t = 18 s, therefore the average velocity is

v=\frac{20 m}{18 s}=1.11 m/s

6 0
3 years ago
before colliding, the momentum of block A is +15.0 kg m/s. after, block A has a momentum -12.0 kg*m/s. what is the momentum of b
Helen [10]

Answer:

The momentum of block B = 27 Kg m/s

Explanation:

Given,

The initial momentum of block A, MU = 15 Kg m/s

The final momentum of block A, MV = -12 Kg m/s

Consider the block B is initially at rest.

Therefore, the initial momentum of block B, mu = 0

According to the laws of conservation of linear momentum, the momentum of the body before impact is equal to the momentum of the body after impact.

                               <em> MU + mu = MV + mv</em>

                                15  +  (0) = (-12) + mv

                                         mv = 15 + 12

                                              =  27 Kg m/s

Hence, the momentum of the block B after impact is, mv = 27 Kg m/s

3 0
3 years ago
What will the kinetic energy of a pile driver ram be if it starts from rest and undergoes a 10 kJ decrease in potential energy?
taurus [48]

Answer:

KE_f = 10 kJ

Explanation:

As per mechanical energy conservation law we can say that sum of kinetic energy and potential energy will always remains constant

so here we can say

KE_i + PE_i = KE_f + PE_f

here we have

KE_i = 0

as initially it starts from rest

so now we have

KE_f = PE_i - PE_f

now we know that its final potential energy is decreased by 10 kJ

so we have

PE_f = PE_i - 10 kJ

PE_i - PE_f = 10 kJ

so the final kinetic energy will be

KE_f = 10 kJ

4 0
3 years ago
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