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3 years ago

The direction of an electric field is the direction (5 points)

1 answer:

5 0

The direction of an electric field is determined from the behavior of a positive test charge that is set free in the electric field.This charge moves along a distinct vector showing the direction of the electric field Therefore the answer is b. a positive charge will move in the field.

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A tugboat tows a ship with a constant force of magnitude F1. The increase in the ship's speed during a 10 s interval is 3 km/h.

Amanda [17]

Answer:

$F_{1}=\frac{1}{5}F_{2}$ or $F_{2}=5F_{1}$

In other words, $F_{1}$ is one fifth of $F_{2}$ or $F_{2}$ is five times as big as $F_{1}$

Explanation:

In order to solve this problem we must start by sketching the situation (refer to the attached picture).

When the ship is pulled only by force 1, it will change its speed by 3km/hr in 10 seconds. So in order to use these values we need to either turn the km/hr in km/s or turn the seconds to hours. Let's turn the seconds to hours:

$10s*\frac{1hr}{3600s}=\frac{1}{360} hr$

so we can now use the acceleration formula to find the acceleration of the boat so we get:

$a=\frac{\Delta v}{\Delta t}$

which will give us an accceleration of:

$a=\frac{3km/hr}{\frac{1}{360}hr}=1080km/hr^{2}$

once we got the acceleration we can for sure say taht:

$F_{1}=ma=m*1080\frac{km}{hr^{2}}$

Now, if we take a look at the second drawing we can see that the resultant force applied to the boat is found by adding the two forces, force one and force two, so we get:

$F_{1}+F_{2}=ma$

in this case the acceleration changes because the change in velocity is of 18km/hr in the same 10 seconds, so we get that:

$a=\frac{\Delta v}{\Delta t}$

$a=\frac{18km/hr}{\frac{1}{360}hr}=6480km/hr^{2}$

so we can say that:

$F_{1}+F_{2}=m*6480km/hr^{2}$

we can substitute the first force into this equation so we get:

$m*1080km/hr^{2}+F_{2}=m*6480km/hr^{2}$

and solve for the second force, so we get:

$F_{2}=m*6480km/hr^{2}-m*1080km/hr^{2}$

which yields:

$F_{2}=m*5400km/hr^{2}$

Now we can compare theh two forces, force 1 and force 2 by dividing them:

$\frac{F_{1}}{F_{2}}=\frac{m*1080km/hr^{2}}{m*5400km/hr^{2}}$

which yields:

$\frac{F_{1}}{F_{2}}=\frac{1}{5}$

when solving for the first force we get:

$F_{1}=\frac{1}{5}F_{2}$

which tells us that the second force is one fifth of the first force.

and when solving for the second force we get that:

$F_{2}=5F_{1}$

which means that the second force is 5 times as big as the first force.

8 0

3 years ago

Can anyone help me? Physics

olga2289 [7]

I cannot read please show closer

5 0

3 years ago

Light waves are electromagnetic waves that travel at 3.00 Light waves are electromagnetic waves that travel 108 m/s. The eye is

svlad2 [7]

(a) $5.45 \cdot 10^{14} Hz$

The relationship between frequency and wavelength of an electromagnetic wave is given by

$c=f \lambda$

where

$c=3.00 \cdot 10^8 m/s$ is the speed of light

$f$ is the frequency

$\lambda$ is the wavelength

In this problem, we are considering light with wavelength of

$\lambda=5.50 \cdot 10^{-7} m$

Substituting into the equation and re-arranging it, we can find the corresponding frequency:

$f=\frac{c}{\lambda}=\frac{3.00 \cdot 10^8 m/s}{5.50 \cdot 10^{-7} m}=5.45 \cdot 10^{14} Hz$

(b) $1.83\cdot 10^{-15} s$

The period of a wave is equal to the reciprocal of the frequency:

$T=\frac{1}{f}$

And using $f=5.45 \cdot 10^{14} Hz$ as we found in the previous part, we can find the period of this wave:

$T=\frac{1}{5.45 \cdot 10^{14} Hz}=1.83\cdot 10^{-15} s$

5 0

3 years ago

If the distance between two charged particles is increased 2.69 times, the ratio of new to old electric force is _______ times t

Tasya [4]

Answer:

F = K Q1 Q2 / R^2 force between 2 charged partices

F2 / F1 = (R1 / R2)^2 = (1 / 2.69)^2 = .139

F2 = .139 F1

8 0

2 years ago

What is less dense water or oil steel or water helium or air or oil or water

Leni [432]

Answer:

air because their is nothing contained within the air other than all the solutions that you have listed

Explanation:

4 0

3 years ago

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