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lianna [129]
4 years ago
11

The image shows the electric field lines around two charged particles.

Physics
2 answers:
DENIUS [597]4 years ago
8 0
It's either 3 or 4 I know this becuase I have read a book about electricity
Andrew [12]4 years ago
8 0

Answer:

D. 4

Explanation:

The electric force will be maximum at the position where the number of electric lines are maximum. As we can observe that the maximum number of lines are at position 4.

Thus, electric force is maximum at 4.

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You have a set of calipers that can measure thicknesses of a few inches with an uncertainty of ± 0.005 inches. You measure the t
Bond [772]

Answer:

a) x = (0.0114 ± 0.0001) in , b) the number of decks is 5

Explanation:

a) The thickness of the deck of cards (d) is measured and the thickness of a card (x) is calculated

        x = d / 52

        x = 0.590 / 52

        x = 0.011346 in

Let's look for uncertainty

       Δx = dx /dd Δd

       Δx = 1/52 Δd

       Δx = 1/52  0.005

       Δx = 0.0001 in

The result of the calculation is

        x = (0.0114 ± 0.0001) in

b) You want to reduce the error to Δx = 0.00002, the number of cards to be measured is

           #_cards = n 52

The formula for thickness is

           x = d / n 52

Uncertainty

          Δx = 1 / n 52  Δd

         n = 1/52 Δd / Δx

         n = 1/52 0.005 / 0.00002

         n = 4.8

Since the number of decks must be an integer the number of decks is 5

3 0
4 years ago
An MRI scanner is based on a solenoid magnet that produces a large magnetic field. The magnetic field doesn't stop at the soleno
GaryK [48]

Answer:

The maximum change in  flux is \Delta \o = 0.1404 \ Wb

The average  induced emf     \epsilon =0.11232 V

Explanation:

   From the question we are told that

             The speed of the technician is v = 0.80 m/s

              The distance from the scanner is d = 1.0m

              The  initial magnetic field is  B_i = 0T

               The final magnetic field is B_f = 6.0T

                 The diameter of the loop is  D = 19cm = \frac{19}{100} = 0.19 m

The area of the loop is mathematically represented as

        A  =  \pi [\frac{D}{2} ]^2

             = 3.142 \frac{0.19}{2}

             = 0.02834 m^2

At maximum the change in magnetic field is mathematically represented as

            \Delta \o = (B_f - B_i)A

  =>      \Delta  \o = (6 -0)(0.02834)

                  \Delta \o = 0.1404 \ Wb

The  average induced emf is mathematically represented as

           \epsilon =  \Delta \o v

              = 0.1404 * 0.80

             \epsilon =0.11232 V

7 0
3 years ago
A cubic metal box with sides of 17 cm contains air at a pressure of 1 atm and a temperature of 278 K. The box is sealed so that
const2013 [10]

Answer:

F = 3.98 kN

Explanation:

GIVEN DATA:

sides of box = 17 cm

pressure = 1 atm = 101325 N/m2

T2 = 378K

T1 = 278 K

final pressure can be calculate by using below relation

\frac{P_{1}}{P_{2}}=\frac{T_{1}}{T_{2}}

we know that

force = pressure * area

therefore force is

F =(\frac{T_{1}}{T_{2}}*P_{1})A

F =(\frac{378}{278}*101325)(17*10^{-2})^{2}

F = 3.98 kN

5 0
4 years ago
The right-hand rule predicts the direction of the force on a positively charged object moving in a gravitational field true fals
eimsori [14]
The statement above is FALSE.
The right hand rule is used in physics to predict the direction of the force on a charged object moving in a MAGNETIC FIELD. The right hand rule is used to relate the relationship between the magnetic field and the forces that are exerted on the moving objects in the field. Using the right hand rule, for a positively charged object that is moving in an electric field, the pointer finger will point in the direction the charged object is moving, the middle finger will point in the direction of the magnetic field and the thumb will point in the direction of the magnetic force that is pushing the charged object.
3 0
3 years ago
A person sitting on a pier observes incoming waves that have a sinusoidal form with a distance of 2.5 m between the crests. Of a
Doss [256]

Answer:

Part(a): The frequency is \bf{0.2~Hz}.

Part(b): The speed of the wave is \bf{0.5~m/s}.

Explanation:

Given:

The distance between the crests of the wave, d = 2.5~m.

The time required for the wave to laps against the pier, t = 5.0~s

The distance between any two crests of a wave is known as the wavelength of the wave. So the wavelength of the wave is \lambda = 2.5~m.

Also, the time required for the wave for each laps is the time period of oscillation and it is given by T = 5.0~s.

Part(a):

The relation between the frequency and time period is given by

\nu = \dfrac{1}{T}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(1)

Substituting the value of T in equation (1), we have

\nu &=& \dfrac{1}{5.0~s}\\~~~&=& 0.2~Hz

Part(b):

The relation between the velocity of a wave to its frequency is given by

v = \nu \lambda~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(2)

Substituting the value of \nu and \lambda in equation (2), we have

v &=& (0.2~Hz)(2.5~m)\\~~~&=& 0.5~m/s

5 0
3 years ago
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