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nlexa [21]
2 years ago
8

Classify each of the following statements as a characteristic (a) of electric forces only, (b) of magnetic forces only, (c) of b

oth electric and magnetic forces, or (d) of neither electric nor magnetic forces. (viii) The magnitude of the force depends on the charged object's direction of motion.
Physics
1 answer:
kakasveta [241]2 years ago
5 0

(b) Magnetic forces only

When an object is placed in an electric field it experiences an electric force given by :

F_{E} = |q| × E

Similarly when it is placed in a magnetic field it experiences a magnetic force given by :

F_{B} = |q|×v×Bsinθ

When a particle is placed in electric field the particle gets accelerated.

The electric field has a direction, positive to negative. This is the direction that the electric field will cause a positive charge to accelerate. For example If a positive charge is moving in the same direction as the electric field vector the particle's velocity will increase. If it is moving in the opposite direction it will decelerate.  ( No change in direction of motion in this case)

When a particle is moved in magnetic field ,the magnetic force is perpendicular to velocity and magnetic field . Since force is perpendicular to velocity ,it only changes the direction of motion.

To know more about electric and magnetic forces kindly refer to the below link:

brainly.com/question/1594186

#SPJ4

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" <em>Energy is never created or destroyed.</em> "

All the rest is commentary.

7 0
3 years ago
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Sheila uses a 45N force on her bowling ball across a 15m lane. What work did she do on the bowling ball? Show your work.​
Maurinko [17]

Answer:

675J

Explanation:

Given parameters:

Force  = 45N

Distance  = 15m

Unknown:

Work done by Sheila  = ?

Solution:

Work done by a body is the amount of force applied to make a body move through a distance;

         Work done  = Force x distance

 Now;

          Work done  = 45 x 15  = 675J

6 0
2 years ago
Suppose you are standing at the exact center of a park surrounded by a circular road. An ambulance drives completely around this
vladimir2022 [97]

Answer: The pitch of the sound does not change.

Explanation: The pitch of sound wave is dependent on the frequency of the sound wave. The frequency of sound wave when their is a relative motion between an observer and a source is given by Doppler effect.

Doppler effect is a mathematical equation that gives the relationship between the observed frequency by an observer from any sound source and the relative motion between the observer and the source.

Mathematically,

f'= (v+v') /(v-vs) * f

Where f' = observed frequency of sound wave

v= speed of sound in air

v'= velocity of observer relative to sound source

vs= velocity of sound source relating to observer

f= frequency of sound produced by source

The observer is at the center, thus the distance between the observer and the source is constant (according to mensuration, the radius is constant for any given circle and since the car is moving along a circular path and the observer is at the center, thus the distance between them is constant), thus making the relative velocity between the observer and the source constant (vs=constant).

Also the frequency of sound wave produce by the source is a constant (f=constant)

The speed of sound in air is also a constant (v=336m/s)

The observer is standing at the center thus he is not moving, hence the relative motion between observer and source is also constant (v'=constant)

Since all parameters are constant, then the observed frequency will be constant too.



4 0
3 years ago
What phase difference between two identical traveling waves, moving in the same direction along a stretched string, results in t
Tom [10]

Answer:

Explanation:

Let the amplitude of individual wave be I and resultant amplitude be 1.703 I . Let the phase difference be Ф in terms of degree

From the formula of resultant vector

(1.703I)² = I² + I² +  2 I² cosФ

2.9 I² = 2I² + 2 I² cosФ

.9I² = 2 I² cosФ

cosФ = .9 / 2

= .45

Ф = 63.25 .

5 0
3 years ago
kristine speeds past a parked police car at 32 m/s. The police car starts from rest with a uniform acceleration of 2.5 m/s^2. Ho
Digiron [165]

32 = 0 \times t +  \frac{1}{2} \times 2.5 \times t^{2}     \\  32 = 0 + 1.25 \times t {}^{2}  \\ 32 = 1.25t {}^{2}   \\   \frac{32}{1.25}  =  \frac{1.25t {}^{2} }{1.25}  \\ t {}^{2}  = 25.6 \\  \sqrt{t {}^{2} }  =  \sqrt{25.6}  \\ t = 5.1seconds \\

7 0
3 years ago
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