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lyudmila [28]
2 years ago
6

Cathode ray tubes (CRTs) used in old-style televisions have been replaced by modern LCD and LED screens. Part of the CRT include

d a set of accelerating plates separated by a distance of about 1.52 cm. If the potential difference across the plates was 24.0 kV, find the magnitude of the electric field (in V/m) in the region between the plates.
________V/m
Physics
1 answer:
Kaylis [27]2 years ago
7 0

The electric filed in V/m is 1.58 * 10^6 V/m

<h3>What is the electric field?</h3>

We know that the electric field is obtained as the ratio of the voltage to the distance that separates the plates.

Thus;

E = V/d

E = electric field

V = voltage

d = distance of separation

E = 24 * 10^3 V/1.52 * 10^-2 m

E = 1.58 * 10^6 V/m

Learn more about electric field :brainly.com/question/8971780

#SPJ1

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3 years ago
You drag a crate across a floor with a rope. The force applied is 750 N and the angle of the rope is 25.0° above the horizontalH
frez [133]

a.

The work done by a constant force along a rectilinear motion when the force and the displacement vector are not colinear is given by:

W=F\cos\theta\cdot d

where F is the magnitude of the force, theta is the angle between them and d is the distance.

The problen gives the following data:

The magnitude of the force 750 N.

The angle between the force and the displacement which is 25°

The distance, 26 m.

Plugging this in the formula we have:

\begin{gathered} W=\left(750\right)\left(\cos25\right)\left(26\right) \\ W=17673 \end{gathered}

Therefore the work done is 17673 J.

b)

The power is given by:

P=\frac{W}{t}

the problem states that the time it takes is 6 s. Then:

\begin{gathered} P=\frac{17673}{6} \\ P=2945.5 \end{gathered}

Therefore the power is 2945.5 W

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1 year ago
Letícia leaves the grocery store and walks 150.0 m to the parking lot. Then, she turns 90° to the right and walks an additional
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Answer:

165.529454

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