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creativ13 [48]
2 years ago
7

At time t=0 a proton is a distance of 0.360 m from a very large insulating sheet of charge and is moving parallel to the sheet w

ith speed 960 m/s . The sheet has uniform surface charge density 2.34.×10−9C/m2 . What is the speed of the proton at 5.40
Physics
1 answer:
Yuki888 [10]2 years ago
3 0

Explanation:

Formula for the electric field due to the infinite sheet of charge is as follows.

               E = \frac{\sigma}{2 \epsilon_{o}}

where,   \sigma = surface charge density

Now, formula for electric force acting on the proton is as follows.

             F = eE

where,    e = charge of the proton

According to the Newton's second law of motion, the net force acting on the proton is as follows.

                       F = ma

                 a = \frac{eE}{m}

                    = \frac{e(\frac{\sigma}{2 \epsilon_{o}})}{m}

                    = \frac{e \sigma}{2m \epsilon_{o}}

According to the kinematic equation, speed of the proton in perpendicular direction is as follows.

              v_{f} = v_{i} + at

                     = (0 m/s) + \frac{e \sigma}{2 m \epsilon_{o}}t

                     = \frac{1.6 \times 10^{-19}C \times 2.34 \times 10^{-9} C/m^{2} \times 5.40 \times 10^{-8}s}{2 \times (1.67 \times 10^{-27} kg)(8.85 \times 10^{-12} C^{2}/Nm^{2}}

                     = 683.974 m/s

Hence, total speed of the proton is as follows.

                v' = \sqrt{(960 m/s)^{2} + (683.974 m/s)^{2}}

                    = \sqrt{921600 + 467820.43}

                    = \sqrt{1389420.43}

                    = 1178.73 m/s

Therefore, we can conclude that speed of the proton is 1178.73 m/s.

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A two-resistor voltage divider employing a 2-k? and a 3-k? resistor is connected to a 5-V ground-referenced power supply to prov
vesna_86 [32]

Answer:

circuit sketched in first attached image.

Second attached image is for calculating the equivalent output resistance

Explanation:

For calculating the output voltage with regarding the first image.

Vout = Vin \frac{R_{2}}{R_{2}+R_{1}}

Vout = 5 \frac{2000}{5000}[/[tex][tex]Vout = 5 \frac{2000}{5000}\\Vout = 5 \frac{2}{5} = 2 V

For the calculus of the equivalent output resistance we apply thevenin, the voltage source is short and current sources are open circuit, resulting in the second image.

so.

R_{out} = R_{2} || R_{1}\\R_{out} = 2000||3000 = \frac{2000*3000}{2000+3000} = 1200

Taking into account the %5 tolerance, with the minimal bound for Voltage and resistance.  

if the -5% is applied to both resistors the Voltage is still 5V because the quotient  has 5% / 5% so it cancels. to be more logic it applies the 5% just to one resistor, the resistor in this case we choose 2k but the essential is to show that the resistors usually don't have the same value. applying to the 2k resistor we have:

Vout = 5 \frac{1900}{4900}\\Vout = 5 \frac{19}{49} = 1.93 V

Vout = 5 \frac{2100}{5100}\\Vout = 5 \frac{21}{51} = 2.05 V

R_{out} = R_{2} || R_{1}\\R_{out} = 1900||2850= \frac{1900*2850}{1900+2850} = 1140

R_{out} = R_{2} || R_{1}\\R_{out} = 2100||3150 = \frac{2100*3150 }{2100+3150 } = 1260

so.

V_{out} = {1.93,2.05}V\\R_{1} = {1900,2100}\\R_{2} = {2850,3150}\\R_{out} = {1140,1260}

4 0
2 years ago
Norma kicks a soccer ball with an initial velocity of 10.0 meters per second at an angle of 30.0°. If the ball moves through the
enyata [817]

<u>Answer</u>

27.7


<u>Explanation</u>

The ball was hit at an angle of 30°, with the horizontal at a speed of 10 m/s. We have to find the horizontal component of speed.

cosx = adjacent/hypotenuse

cos 30 = adjacent / 10

adjacent = 10 cos30

             = 8.66 m/s        ⇒ This is the horizontal speed.

Now  find the horizontal distance.

Distance = speed × time

               = 8.66 × 3.2

                = 27.71

Answer to the nearest tenth = 27.7

4 0
2 years ago
Read 2 more answers
A "biconvex" lens is one in which both surfaces of the lens bulge outwards. Suppose you had a biconvex lens with radii of curvat
Stolb23 [73]

Answer: f=150cm in water and f=60cm in air.

Explanation: Focal length is a measurement of how strong light is converged or diverged by a system. To find the variable, it can be used the formula:

\frac{1}{f} = (nglass - ni)(\frac{1}{R1} - \frac{1}{R2}).

nglass is the index of refraction of the glass;

ni is the index of refraction of the medium you want, water in this case;

R1 is the curvature through which light enters the lens;

R2 is the curvature of the surface which it exits the lens;

Substituting and calculating for water (nwater = 1.3):

\frac{1}{f} = (1.5 - 1.3)(\frac{1}{10} - \frac{1}{15})

\frac{1}{f} = 0.2(\frac{1}{30})

f = \frac{30}{0.2} = 150

For air (nair = 1):

\frac{1}{f} = (1.5 - 1)(\frac{1}{10} - \frac{1}{15})

f = \frac{30}{0.5} = 60

In water, the focal length of the lens is f = 150cm.

In air, f = 60cm.

5 0
2 years ago
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A particle in the first excited state of a one-dimensional infinite potential energy well (with U = 0 inside the well) has an en
nataly862011 [7]

Answer:

The energy of this particle in the ground state is E₁=1.5 eV.

Explanation:

The energy E_{n} of a particle of mass <em>m</em> in the <em>n</em>th energy state of an infinite square well potential with width <em>L </em>is:

                                                    E_{n}=\frac{n^{2}h^{2}}{8mL^{2}}

In the ground state (n=1). In the first excited state (n=2) we are told the energy is E₂= 6.0 eV. If we replace in the above equation we get that:

                                                    E_{1}=\frac{h^{2}}{8mL^{2}}            

                                                    E_{2}=\frac{h^{2}}{2mL^{2}}

So we can rewrite the energy in the ground state as:

                                                   E_{1}=\frac{1}{4}(\frac{h^{2}}{2mL^{2}})

                                                      E_{1}=\frac{1}{4} E_{2}

                                                   E_{1}=\frac{1}{4} ( 6.0\ eV)

Finally

                                                    E_{1}=1.5\ eV

                                                   

                                                   

6 0
2 years ago
A helicopter pulls upward by means of a rope on a 250 kg crate to lift it UNIFORMLY. What is the net force on the crate?
Cloud [144]

Answer:

The net force = 0

Explanation:

The given information includes;

The mass of the crate = 250 kg

The way the helicopter lifts the crate = Uniformly (constant rate (speed), no acceleration)

In order to pull the crate upwards, the helicopter has to provide a force equivalent to the weight of the crate keeping the helicopter on the ground.

The weight of the crate = The mass of the crate × The acceleration due gravity acting on the crate

The weight of the crate, F_w↓ = 250 kg × 9.81 m/s² = 2,452.5 N

The force the helicopter should provide to just lift the crate, F_{(helicopter)}↑ = The weight of the crate = 2,452.5 N

The net force, F_{(net)} = F_{(helicopter)}↑ - F_w↓ = 2,452.5 N - 2,452.5 N = 0

The net force = 0.

3 0
2 years ago
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