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olchik [2.2K]
3 years ago
9

Moving the probe 1 cm towards the non-grounded electrode changes the value the potential from about 0.90 V to about 1.2 V. Expla

in how you can get the magnitude of the average electric field between these two points on the paper, and give the value of this field in Newtons/Coulomb. Show your calculations.
Physics
1 answer:
svp [43]3 years ago
6 0

Answer:

-30 N/C

Explanation:

Since the potential changes from 0.90 V to 1.2 V when I move the probe 1 cm closer to the non-grounded electrode, the electric field is the gradient between the two points is given by E = -ΔV/Δx where ΔV = change in electric potential and Δx = distance of potential change = 1 cm = 0.01 m

Now ΔV = final potential - initial potential = 1.2 V - 0.90 V = 0.30 V

Since E = -ΔV/Δx

substituting the values of the variables into the equation, we have

E = -ΔV/Δx

E = -0.30 V/0.01 m

E = -30 V/m

Since 1 V/m = 1 N/C.

E = -30 N/C

So, the average electric field is -30 N/C

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The subject of the second part of messiah is:
Zielflug [23.3K]
The second part is about <span>his suffering and the spread of his doctrine.</span>
5 0
3 years ago
A stretched string has a mass per unit length of 5.00 g/cm and a tension of 10.0 N. A sinusoidal wave on this string has an ampl
8_murik_8 [283]

Answer:

0.12 mm ; 140.50 rad/m ; 628.32 rad/sec ; +

Explanation:

Given the wave equation of the form :

y(x, t) = ym sin(kx ± ωt)

Mas per unit length (u) = 5 g/cm = (5÷1000)kg / 0.01m) = 0.005kg/0.01m = 0.5kg/m

Tension, T = 10 N

Amplitude, A = 0.12 mm

Frequency, F = 100 Hz

Comparing with the general wave equation :

y = Asin(kx ± ωt)

A = amplitude = ym = 0.12 mm

2.) k = 2π / λ

Recall :

v = fλ

v = sqrt(T/u) = sqrt(10/0.5) = sqrt(20) = 4.472

λ = v/ f = 4.472 / 100 = 0.04472

Hence,

k = (2 * π) / 0.04472

k = 140.50 rad/m

3.) Angular frequency, ω

ω = 2πf = 2 * 3.14 * 100 = 628.32 rad/sec

4.) sign is +ve

Direction of wave propagation as given is in the negative x axis

4 0
2 years ago
A pulley with a radius of 3.0 cm and a rotational inertia of 4.5 x 10–3 kg∙m2 is suspended from the ceiling. A rope passes over
vekshin1

Answer:

22J

Explanation:

Given :

radius 'r'= 3cm

rotational inertia 'I'=4.5 x 10^{-3} kgm²

mass on one side of rope 'm_{1'= 2kg

mass on other side of rope'm_{2' =4kg

velocity'v' of mass m_{2' = 2m/s

Angular velocity of the pulley is given by

‎ω = v /r => 2/ 3x 10^{-2

‎ω = 66.67 rad/s

For the rotating body, we have

KE = \frac{1}{2} I ω²

KE_p = \frac{1}{2} (4.5 *10^{-3} )(66.67^{2} )

KE_p = 10J

Next is to calculate kinetic energy of the blocks :

KE_{b} = \frac{1}{2} (m_1 + m_2).v^2\\KE_b= \frac{1}{2} (2+4).2^2

KE_b=12J

Therefore, the total kinetic energy will be

KE = KE_p + KE_b =10 + 12

KE= 22J

6 0
3 years ago
Which of the following is same dimension quantity?
hram777 [196]

Answer:

Find the dimension of each and every quantity in all the options to check whether they are the same or not. We can use any one formula of each identity to find its dimension.

Complete step by step solution:

To find the dimension of a quantity, we can use any formula related to that quantity but we will use the easiest ones to save time.

Force-

from Newton’s law of motion,

F=maF=ma

Dimension of force =[M][LT−2]=[MLT−2]=[M][LT−2]=[MLT−2]

Work done-

W=F×sW=F×s

Dimension of work=[MLT−2][L]=[ML2T−2]=[MLT−2][L]=[ML2T−2]

Momentum-

p=mvp=mv

Dimension of momentum=[M][LT1]=[MLT−1]=[M][LT1]=[MLT−1]

Impulse-

I=F×tI=F×t

Dimension of impulse=[MLT−2][T]=[

3 0
3 years ago
Suppose you put an ice cube into a cup of hot tea. In what direction does energy in the form of heat flow? What happens to the i
Semenov [28]

Answer:

The energy flows between the ice and the tea equally. The table below shows the temperatures of several different objects made of the same material.

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