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Sonbull [250]
3 years ago
5

Question Part Points Submissions Used The difference in potential between the accelerating plates of a TV set is about 218 V. If

the distance between these plates is 1.29 cm, find the magnitude of the uniform electric field (N/C) in this region. Consider entering your answer using scientific notation.
Physics
1 answer:
ludmilkaskok [199]3 years ago
3 0

Answer:

1.69\cdot 10^4 N/C

Explanation:

The relationship between electric field strength and potential difference is:

E=\frac{V}{d}

where

E is the electric field strength

V is the potential difference

d is the distance between the plates

Here we have

V = 218 V

d = 1.29 cm = 0.0129 m

So, the electric field is

E=\frac{218 V}{0.0129 m}=16900 N/C = 1.69\cdot 10^4 N/C

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7 0
3 years ago
Friction always acts____ and in the ____ direction of the motion of an object ​
Fed [463]

Answer:

Friction force always acts tangent to the surface at points of contact. Friction force acts opposite to the direction of motion. There are 2 types of friction: Static friction: If the two surfaces in contact do not move relative to each other, one has static friction.

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2 years ago
An electric dipole is formed from ± 5.0 nC point charges spaced 3.0 mm apart. The dipole is centered at the origin, oriented alo
Ymorist [56]

Answer:

The electric field strength at point (x,y) = ( 20 mm ,0cm) is =<u>16321.0769 N/C</u>

The electric field strength at point (x,y) = (0cm, 20 mm) is =<u>35321.58999 N/C</u>

Explanation:

Question: What is the electric field strength at point (x,y) = ( 20 mm ,0cm)?

Answer:

The electric field at any given point of the dipole is given as:

E= (KP) ÷ (r^2 + a^2)^3/2

Where:

K = 9x10^9 Nm^2/c^2 (coloumb constant)

P = (0.003) (5x10^-9c) which is the movement of the dipole

(0.003) is arrived at when mm is converted to m. 3.0 mm space apart was converted to a meter.

r= the point, in the question above is 20mm = 0.02m

Now, the electric field, E can be calculated by putting the values in the formula above:

E = (KP) ÷ (r^2 + a^2)^3/2

= (9x10^9 Nm^2/c^2) (0.003 m) (5x10^-9c) ÷ [ (0.02m)^2 + (0.003)^2]^3/2

= 0.135 ÷ (8.271513x10^-6)

=<u>16321.0769 N/C</u>

 Question: What is the electric field strength at point (x,y) = (0cm, 20 mm )?

Answer:

Here, the electric field, E= 2krp ÷ (r^2 - a^2)^2

E= 2 (9x10^9 Nm^2/c^2) (0.02m) (0.003 m) (5x10^-9c) ÷ [(0.02m)^2 - (0.003)^2]^2

= 0.0054 ÷  0.000000152881

=<u>35321.58999 N/C</u>

8 0
2 years ago
Speed=100m/sec<br> Frequency=10 Hz<br> Wavelength=?
elena-14-01-66 [18.8K]

Answer:

Wavelength = 10 m

Explanation:

Given:

Speed = 100 ms^{-1}

Frequency = 10 Hz = 10 s^{-1}

To find : Wavelength = ?

We know that the relationship between wavelength λ, frequency f and speed v is given by the equation

    v = fλ

Therefore wavelength λ = v/f

                                        = 100 ms^{-1} / 10 ms^{-1}

                                        = 10 m

Hence wavelength = 10 m

4 0
3 years ago
If an object is propelled upward from a height of 128 feet at an initial velocity of 112 feet per​ second, then its height h aft
Marina CMI [18]

Explanation:

The equation of motion of an object is given by :

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Where

t is the time in seconds

We need to find the time when the object hits the ground. When the object hits the ground, h(t) = 0

So,

-16t^2+112t+128=0

-t^2+7t+8=0

On solving above equation using online calculator, t = 8 seconds. So, the object hit the ground after 8 seconds. Hence, this is the required solution.

8 0
3 years ago
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