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Lapatulllka [165]
4 years ago
11

determine the potential difference between two charged parallel plates that are .10 cm apart and have an electric field strength

of 12V/cm
Physics
1 answer:
mote1985 [20]4 years ago
6 0
The potential difference between two parallel plates is related to the electric field intensity by
\Delta V = Ed
where
\Delta V is the potential difference
E is the electric field intensity
d is the distance between the two plates.

In our problem,
d=0.10 cm
and 
E=12 V/cm
so we can calculate the potential difference as
\Delta V= Ed = (12 V/cm)(0.10 cm)=1.2 V
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A uniform metal tube of length 5m and mass 9kg is suspended by two vertical wires attached at 50cm and 150cm respectively from t
Elena-2011 [213]

Answer:

force (tension) of 29.4 N (upward)  in 100 cm

force (tension) of 58.4 N (upward)  in 200 cm

Explanation:

Given:

Length of tube = 5 m (500 cm)

Mass of tube = 9

Suspended vertically from 150 cm and 50 cm.

Computation:

Force = Mass × gravity acceleration.

Force = 9.8 x 9

Force = 88.2 N

So,

Upward forces = Downward forces

D1 = 150 - 50 = 100 cm

D2 = 150 + 50 = 200 cm

And F1 = F2

F1 x D1 = F2 x D2

F1 x 100 = F2 x 200

F = 2F

Total force = Upward forces + Downward forces

3F = 88.2

F = 29.4 and 2F = 58.8 N

force (tension) of 29.4 N (upward)  in 100 cm

force (tension) of 58.4 N (upward)  in 200 cm

4 0
4 years ago
(NEED HELP PLEASE) A physics student goes to the roof of the school, 24.15 m above the ground, and drops a pumpkin straight down
slavikrds [6]

Answer:

t = 2.2 s

Explanation:

Given that,

Height of the roof, h = 24.15 m

The initial velocity of the pumpkin, u = 0

We need to find the time taken for the pumpkin to hit the ground. Let the time be t. Using second equation of kinematics to find it as follows :

h=ut+\dfrac{1}{2}at^2

Here, u = 0 and a = g

h=\dfrac{1}{2}gt^2\\\\t=\sqrt{\dfrac{2h}{g}} \\\\t=\sqrt{\dfrac{2\times 24.15}{9.8}} \\\\t=2.22\ s

So, it will take 2.22 s for the pumpkin to hit the ground.

7 0
3 years ago
The edge of a flying disc with a radius of 0.13 m spins with a tangential speed of 3.3 m/s. The centripetal acceleration of the
Marat540 [252]

Answer:

Centripetal acceleration = 83.77m/s²

Explanation:

<u>Given the following data;</u>

Radius, r = 0.13m

Velocity, v = 3.3m/s

To find centripetal acceleration;

Centripetal acceleration is given by the formula;

Acceleration, a = \frac {v^{2}}{r}

Substituting into the equation, we have;

Centripetal \; acceleration, a = \frac {3.3^{2}}{0.13}

Centripetal \; acceleration, a = \frac {10.89}{0.13}

<em>Centripetal acceleration = 83.77m/s²</em>

<em>Therefore, the centripetal acceleration of the edge of the disc is 83.77 m/s². </em>

5 0
3 years ago
Read 2 more answers
A projector is placed on the ground 22 ft. away from a projector screen. A 5.2 ft. tall person is walking toward the screen at a
Stella [2.4K]

Answer:

y = 67.6 feet,   y = 114.4/ (22 - 3t)

Explanation:

For this exercise let's use that light travels in a straight line and some trigonometric relationships, the symbols are in the attached diagram

Large triangle Projector up to the screen

         tan θ = y / L

For the small triangle. Projector up to the person

         tan θ = y₀ / (L-d)

The angle is the same, so we equate the two equations

         y₀ / (L -d) = y / L

         y = y₀  L / (L-d)

The distance from the screen (d), we look for it with kinematics

         v = d / t

        d = v t

we replace

         y = y₀ L / (L - v t)

         y = 5.2 22 / (22 - 3 t)

         y = 114.4 (22 - 3t)⁻¹

This is the equation of the shadow height change as a function of time

For the suggested distance the shadow has a height of

           y = 114.4 / (22-13)

           y = 67.6 feet

7 0
3 years ago
What type of electromagnetic wave does a light bulb and a radio antenna release?
andrew-mc [135]
I believe that a light bulb releases visible light and a radio antenna releases a radio waves
4 0
3 years ago
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