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KATRIN_1 [288]
3 years ago
12

Determine the direction of the force (if any) that will act on the charge in each of the following situations. A positive charge

within an electric field that points to the right. negative charge within an electric field that points downward. A positive charge moving downward in a magnetic field that points out of the screen. A negative charge moving to the right in a magnetic field that points into the screen. A positive charge moving upward in an electric field that points into the screen. A negative charge moving to the right in a magnetic field that points to the right. to the right to the left into the screen zero force downward out of the screen upward
Physics
1 answer:
NeTakaya3 years ago
3 0

Answer:

a) F on the right , b)    F up , c) Force to the left , d)  Force down , e)  Force towards the screen , f) Force is zero

Explanation:

The equations for the forces are

Electric

          F = q E

Magnetic

          F = q v x B

The Bold are vectors, the charges are positive

Let's apply these equation to the proposed situations

a) as the load is positive the force goes in the direction of the field

      F on the right

b) as the load is negative the force goes in the opposite direction to the field

     F is up

c) positive charge with speed down and magnetic field out of the screen.

For this part we will use the right hand rule. The thumb is in the direction of the speed, the fingers extended in the direction of the magnetic field and the palm is in the direction of the force for a positive charge

       Force to the left

d) negative charge, speed to the right magnetic field between the screen

   The force has direction

      Force down

e) positive charge, electric field towards the screen.

     Force in the direction of the field

        Force towards the screen

.f) negative charge, speed to the right, magnetic field to the right

    The vector product is zero,

     Force is zero

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A rock with a mass of 540 g in air is found to have an apparent mass of 342 g when submerged in water. (a) What mass of water is
AleksandrR [38]

(a) 198 g

When the rock is submerged into the water, there are two forces acting on the rock:

- its weight, equal to W=mg (m=mass, g=acceleration of gravity), downward

- the buoyant force, equal to B=m_w g (m_w=mass of water displaced), upward

So the resultant force, which is the apparent weight of the rock (W'), is

W'=W-B

which can be rewritten as

m'g = mg-m_w g

where m' is the apparent mass of the rock. Using:

m = 540 g

m' = 342 g

we find the mass of water displaced

m_w = m-m'=540 g-342 g=198 g

(b) 1.98\cdot 10^{-4} m^3

If the rock is completely submerged, the volume of the rock corresponds to the volume of water  displaced.

The volume of water displaced is given by

V_w = \frac{m_w}{\rho_w}

where

m_w = 198 g = 0.198 kg is the mass of the water displaced

\rho_w = 1000 kg/m^3 is the density of the water

Substituting,

V_w = \frac{0.198}{1000}=1.98\cdot 10^{-4} m^3

And so this is also the volume of the rock.

(c) 2727 kg/m^3

The average density of the rock is given by

\rho = \frac{m}{V}

where

m = 540 g = 0.540 kg is the mass of the rock

V=1.98\cdot 10^{-4} m^3 is its volume

Substituting into the equation, we find

\rho = \frac{0.540 kg}{1.98\cdot 10^{-4}}=2727 kg/m^3

3 0
3 years ago
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ch4aika [34]

Answer:

angular range is ( 0.681 rad , 0.35 rad )

Explanation:

given data

wavelength λ = 380 nm = 380 × 10^{-9} m

wavelength λ  = 700 nm =  700 × 10^{-9} m

to find out

angular range of the first-order

solution

we will apply here slit experiment equation that is

d sinθ = m λ    ...........1

here m is 1 for single slit and d is = \frac{1}{900*10^3 m}

so put here value in equation 1 for 380 nm

we get

d sinθ = m λ

\frac{1}{900*10^3} sinθ = 1 × 380 × 10^{-9}

θ = 0.35 rad

and for 700 nm

we get

d sinθ = m λ

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so angular range is ( 0.681 rad , 0.35 rad )

3 0
3 years ago
The distance from earth to teh sun usb approxximately 93 million miles. a scientist would write that number as
erastova [34]

A scientist would write that number as 1.49 x 10⁸ kilometers .

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6 0
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