When the electron is moving in the plane of the page in the direction indicated by the arrow, the force on the electron is directed:_____

a. into the page.
b. toward the left
c. toward the right
d. toward the bottom of the page.
e. toward the top of the page.
f. out of the page.

CHECK COMPLETE QUESTION BELOW

you stood on a planet having a mass four times that of earth mass and a radius two times of earth radius , you would weigh?

A) four times more than you do on Earth.

B) two times less than you do on Earth.

C) the same as you do on Earth

D) two times more than you do on Earth

Answer:

OPTION C is correct

The same as you do on Earth

Explanation :

According to law of gravitation :

F=GMm/R^2......(a)

F= mg.....(b)

M= mass of earth

m = mass of the person

R = radius of the earth

From law of motion

Put equation b into equation a

mg=GMm/R^2

g=GMm/R^2

g=GM/R^2

We know from question a planet having a mass four times that of earth mass and a radius two times of earth radius if we substitute we have

m= 4M

r=(2R)^2=4R^2

g= G4M/4R^2

Then, 4in the denominator will cancel out the numerator we have

g= GM/R^2

Therefore, g remain the same

Given that,

Energy [tex]H=2.7\times10^{31}\ W[/tex]

Surface temperature = 11000 K

Emissivity e =1

(a). We need to calculate the radius of the star

Using formula of energy

[tex]H=Ae\sigma T^4[/tex]

[tex]A=\dfrac{H}{e\sigma T^4}[/tex]

[tex]4\pi R^2=\dfrac{H}{e\sigma T^4}[/tex]

[tex]R^2=\dfrac{H}{e\sigma T^4\times4\pi}[/tex]

Put the value into the formula

[tex]R=\sqrt{\dfrac{2.7\times10^{31}}{1\times5.67\times10^{-8}\times(11000)^4\times 4\pi}}[/tex]

[tex]R=5.0\times10^{10}\ m[/tex]

(b). Given that,

Radiates energy [tex] H=2.1\times10^{23}\ W[/tex]

Temperature T = 10000 K

We need to calculate the radius of the star

Using formula of radius

[tex]R^2=\dfrac{H}{e\sigma T^4\times4\pi}[/tex]

Put the value into the formula

[tex]R=\sqrt{\dfrac{2.1\times10^{23}}{1\times5.67\times10^{-8}\times(10000)^4\times4\pi}}[/tex]

[tex]R=5.42\times10^{6}\ m[/tex]

Hence, (a). The radius of the star is [tex]5.0\times10^{10}\ m[/tex]

(b). The radius of the star is [tex]5.42\times10^{6}\ m[/tex]

Answer:

4.635 *10^12 Neutrinos

Explanation:

Here in this question, we are to determine the number of neutrinos in billions produced, given the power generated by the proton-proton cycle.

We proceed as follows;

In proton-proton cycle generates 26.7 MeV of energy and in this cycle two neutrinos are produced.

From the question, we are given that

Power P = 9.9 watts = 9.9 J/s

Watts is same as J/s

The number of proton-proton cycles required to generate E energy is N = E / E '

Where E ' = Energy generated in proton-proton cycle which is given as 26.7 Mev in the question

Converting Mev to J, we have

= 26.7 x1.6 x10 -13 J

To get the number N which is the number of proton-proton cycle required, we have;

N = 9.9 /(26.7 x1.6 x10^-13) = 2.32 * 10^12

Since we have two proton cycles( proton-proton), it automatically means 2 neutrinos will be produced.

Therefore number of neutrions produced = 2 x Number of proton-proton cycles = 2 * 2.32 * 10^12 = 4.635 * 10^12 neutrinos

Answer:

The water takes longer, because it is the better insulator here.

Explanation:

Conductors and insulators work similarly in "reverse".

If something is a good heat conductor, then it's good at both absorbing heat energy and giving it away. Insulators are good at resisting temperature changes, but also take longer to cool down once they are heated up.

So because copper is the better conductor here, it will cool faster than the water at the same temperature.

Answer:

     I₂ = 143.79

Explanation:

To solve this problem, work them in two parts. A first one where we look for the intensity of the incident light in the set and a second one where we silence the light transmuted by the other set,

Let's start with the set of three curling irons

Beautiful light falls on the first polarized is not polarized, therefore only half the radiation passes

              I₁ = I₀ / 2

this light reaches the second polarized and must comply with the Mule law

             I₂ = I₁ cos² tea

The angle between the first polarized and the second is Tea = 29.0º

             I₂ = I / 2 cos² 29

The light that comes out of the third polarized is

              I₃ = I₂ cos² tea

the angle between the third - second polarizer is

             tea = 58-29

             tea = 29th

               I3 = (I₀ / 2 cos² 29) cos² 29

indicate the output intensity

                I3 = 110

we clear

              I₀ = 2I3 / cos4 29

              I₀ = 2 110 / cos4 29

              I₀ = 375.96 W / cm²

Now we have the incident intensity in the new set of three polarizers

back to the for the first polarizer

                 I₁ = I₀ / 2

when passing the second polarizer

                     I₂ = I1 cos² 29

                    I2 = IO /2 cos²29

let's calculate

                I₂ = 375.96 / 2 cos² 29

               I₂ = 143.79

Answer:   The volume of the block will be [tex]100cm^3[/tex]

Explanation:

Density is defined as the mass contained per unit volume.

[tex]Density=\frac{mass}{volume}[/tex]

Given : Mass of cube = 100 grams

Density of cube = [tex]1g/cm^3[/tex]

Putting in the values we get:

[tex]Volume=\frac{mass}{density}[/tex]

[tex]Volume=\frac{100g}{1g/cm^3}=100cm^3[/tex]

 Thus volume of the block will be [tex]100cm^3[/tex]

Answer:

The correct answer is 500 N

Explanation:

This is an exercise in Newton's third law or law of action and reaction

The Earth exerts a force on the person, which we call a weight of 500 N directed downwards, we can call this action and the person exerts a force on the Earth of equal magnitude 500N and in the opposite direction, that is directed upwards.

Which force we call action does not matter, the analysis and conclusions are the same

The correct answer is 500N

Answer:

True or False

Explanation:

Because.....

easy 50% chance you are right

The image in the attachment describes the situation of the fishing rod.

Answer: F = 10.8 N

Explanation: The image shows a fishing rod attached to an axis. To stay in equilibrium, Torque must be equal for the force of magnitude 18N and for the unknow force.

Torque (τ) is a measure of a force's tendency to cause rotation and, in physics, defined as:

τ = F.r.sin(θ)

F is the force acting on the object;

r is distance between where the torque is measured to where the force is applied;

θ is the angle between F and r;

For the fishing rod:

[tex]\tau_{1} = \tau_{2}[/tex]

[tex]F_{1}.r_{1}.sin(\theta) = F_{2}.r_{2}.sin(\theta)[/tex]

Assuming part (1) is related to unknown force:

[tex]F = \frac{F_{2}.r_{2}.sin(\theta}{r_{1}.sin(\theta) }[/tex]

Replacing the corresponding values:

[tex]F = \frac{18*1.8*sin(30)}{3*sin(30)}[/tex]

[tex]F = \frac{18*1.8}{3}[/tex]

F = 10.8

The fishing line exert on the the rod a force of 10.8N.

Answer:

 object distance  p = 13.33 cm

Explanation:

For this problem of finding the image of an object we must use the constructor equation

         1 / f = 1 / p + 1 / q

where f is the focal length, p and q are the distances to the object and the image, respectively.

In this case they indicate the focal length f = 10 cm, since the person has hyperopia, the image must be formed q = 40 cm, let's find where the object is (p)

        1 / p = 1 / f - 1 / q

        1 / p = 1/10 - 1/40

        1 / p = 0.075

        p = 13.33 cm

Answer:

Static friction is greater than kinetic friction, but they both act opposite the applied force.

Explanation:

Newton's 3rd law states that every action has and equal but opposite reaction.

If an object has static friction, that means it stays in one spot, and it takes a great amount of force to get it moving.

Once the object is moving it has kinetic friction, but it's easier to keep it moving unless you are trying to stop it.

The equal but opposite reaction to something moving it is stopping it, and the equal but opposite reaction to stopping something is moving it.

The same amount of force used to move/stop something is used to stop/move it.

Static friction is greater than kinetic friction, but they both act opposite the applied force.

Friction is the force that opposes motion. Frictional force always acts in opposition to the direction of motion.

There are two kinds of friction;

Since more forces tend to act on a body at rest and prevent it from getting into motion than the forces that tend to stop an already moving body, it follows that static friction is greater than kinetic friction. Both act in opposite direction to the applied force.

Learn more: https://brainly.com/question/18754989

Answer:

41.5 cm to the left of the observer

Explanation:

See attached file

Answer:infrared radiation

Explanation:

Infrared radiation and  microwave radiation  of the electromagnetic spectrum have longer wavelengths than visible light.

EM waves are another name for electromagnetic waves. When an electric field interacts with a magnetic field, electromagnetic waves are created. These electromagnetic waves make up electromagnetic radiations. It is also possible to say that electromagnetic waves are made up of magnetic and electric fields that are oscillating. The basic equations of electrodynamics, Maxwell's equations, have an answer in electromagnetic waves.

If we arrange   electromagnetic wave with decrease in wavelength, we get:

Radio waves > microwave >  Infrared >  Visible light > Ultraviolet > X-rays > Gamma radiation.

Hence,  Infrared  radiation and  microwave radiation  of the electromagnetic spectrum have longer wavelengths than visible light.

Learn more about electromagnetic wave here:

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Given that,

Radius of curvature = 1.4 m

Wavelength = 520 nm

Refraction indexes = 1.6

We know tha,

The condition for constructive interference as,

[tex]t=(m+\dfrac{1}{2})\dfrac{\lambda}{2}[/tex]

Where, [tex]\lambda=wavelength[/tex]

We need to calculate the radius of first bright fringes

Using formula of radius

[tex]r_{1}=\sqrt{2tR}[/tex]

Put the value of t

[tex]r_{1}=\sqrt{2\times(m+\dfrac{1}{2})\dfrac{\lambda}{2}\times R}[/tex]

Put the value into the formula

[tex]r_{1}=\sqrt{2\times(0+\dfrac{1}{2})\dfrac{520\times10^{-9}}{2}\times1.4}[/tex]

[tex]r_{1}=0.603\ mm[/tex]

We need to calculate the radius of second bright fringes

Using formula of radius

[tex]r_{2}=\sqrt{2\times(m+\dfrac{1}{2})\dfrac{\lambda}{2}\times R}[/tex]

Put the value into the formula

[tex]r_{1}=\sqrt{2\times(1+\dfrac{1}{2})\dfrac{520\times10^{-9}}{2}\times1.4}[/tex]

[tex]r_{1}=1.04\ mm[/tex]

Hence, The radius of first bright fringe is 0.603 mm

The radius of second bright fringe is 1.04 mm.

Answer:

 V = - 3.85 V

Explanation:

The electric potential of a continuous charge distribution is

       V = k ∫ dq / r

to find charge differential let's use the concept of linear density

        λ = dq / dx

       dq = λ dx

the distance from a load element to the point of interest

       x₀ = 23 cm = 0.23 m

       r = √ (x-x₀)² = x - x₀

we substitute

        v = k ∫ λ dx / (x-x₀)

we integrate and evaluate between x = 0 and x = l = 0.10 cm

       V = k λ [ln (x-x₀) - ln (-x₀)]

        V = k λ ln ((x-x₀) / x₀)

let's calculate

         V = 9 10⁹  0.74 10⁻⁹ ln ((0.23 - 0.10) / 0.23)

          V = - 3.85 V

Answer:

The vision becomes blurred due to the refractive defects of the eye. There are mainly three common refractive defects of vision. These are (i) myopia or near-sightedness, (ii) Hypermetropia or far – sightedness, and (iii) Presbyopia. These defects can be corrected by the use of suitable spherical lenses.

Answer: Speed = [tex]3.10^{-31}[/tex] m/s

Explanation: Like in classical physics, when external net force is zero, relativistic momentum is conserved, i.e.:

[tex]p_{f} = p_{i}[/tex]

Relativistic momentum is calculated as:

p = [tex]\frac{mu}{\sqrt{1-\frac{u^{2}}{c^{2}} } }[/tex]

where:

m is rest mass

u is velocity relative to an observer

c is light speed, which is constant (c=[tex]3.10^{8}[/tex]m/s)

Initial momentum is zero, then:

[tex]p_{f}[/tex] = 0

[tex]p_{1}-p_{2}[/tex] = 0

[tex]p_{1} = p_{2}[/tex]

To find speed of the heavier fragment:

[tex]\frac{mu_{1}}{\sqrt{1-\frac{u^{2}_{1}}{c^{2}} } }=\frac{mu_{2}}{\sqrt{1-\frac{u^{2}_{2}}{c^{2}} } }[/tex]

[tex]\frac{1.86.10^{-27}u_{1}}{\sqrt{1-\frac{u^{2}_{1}}{(3.10^{8})^{2}} } }=\frac{3.10^{-28}.0.844.3.10^{8}}{\sqrt{1-\frac{(0.844c)^{2}}{c^{2}} } }[/tex]

[tex]\frac{1.86.10^{-27}u_{1}}{\sqrt{1-\frac{u^{2}_{1}}{(3.10^{8})^{2}} } }=1.42.10^{-19}[/tex]

[tex]1.86.10^{-27}u_{1} = 1.42.10^{-19}.{\sqrt{1-\frac{u^{2}_{1}}{(3.10^{8})^{2}} } }[/tex]

[tex](1.86.10^{-27}u_{1})^{2} = (1.42.10^{-19}.{\sqrt{1-\frac{u^{2}_{1}}{(3.10^{8})^{2}} } })^{2}[/tex]

[tex]3.46.10^{-54}.u_{1}^{2} = 2.02.10^{-38}.(1-\frac{u_{1}^{2}}{9.10^{16}} )[/tex]

[tex]3.46.10^{-54}.u_{1}^{2} = 2.02.10^{-38} -[2.02.10^{-38}(\frac{u_{1}^{2}}{9.10^{16}} )][/tex]

[tex]3.46.10^{-54}.u_{1}^{2} = 2.02.10^{-38} -2.24.10^{-23}.u^{2}_{1}[/tex]

[tex]3.46.10^{-54}.u_{1}^{2}+2.24.10^{-23}.u^{2}_{1} = 2.02.10^{-38}[/tex]

[tex]2.24.10^{-23}.u^{2}_{1} = 2.02.10^{-38}[/tex]

[tex]u^{2}_{1} = \frac{2.02.10^{-38}}{2.24.10^{-23}}[/tex]

[tex]u_{1} = \sqrt{9.02.10^{-62}}[/tex]

[tex]u_{1} = 3.10^{-31}[/tex]

The speed of the heavier fragment is [tex]u_{1} = 3.10^{-31}[/tex]m/s.

Answer:

[tex]L = 109.01 db[/tex]

Explanation:

Given

Power, P = 100 W

Distance, d = 10 m

Required

Determine the Sound Level

First, the sound intensity as to be calculated; This is done, as follows;

[tex]I = \frac{P}{4\pi d^2}[/tex]

Substitute for P, d and take π as 3.14

[tex]I = \frac{100}{4 * 3.14 * 10^2}[/tex]

[tex]I = \frac{100}{4 * 3.14 * 100}[/tex]

[tex]I = \frac{100}{1256}[/tex]

[tex]I = 0.0796Wm^{-2}[/tex] --- Approximated

Next is to calculate the Sound Level, as follows

[tex]L = 10 * Log(\frac{I}{I_o})[/tex]

Where [tex]I_o = 10^{-12} Wm^{-2}[/tex]

Substitute for I and Io

[tex]L = 10 * Log(\frac{0.0796}{10^{-12}})[/tex]

[tex]L = 10 * Log(0.0796*10^{12)[/tex]

[tex]L = 10 * Log(0.0796*10^{12)[/tex]

[tex]L = 10 * 10.901[/tex]

[tex]L = 109.01 db[/tex]

Hence, the sound level is 109.01 decibels

Answer:

A. magnetic field

Explanation:

The magnetic field is produced around a wire when an electrical current is in the wire because of the magnetic effect of the electric current therefore the correct answer is option A .

A magnetic field could be understood as an area around a magnet, magnetic material, or an electric charge in which magnetic force is exerted.

As given in the problem statement we have to find out what is produced around a wire when an electrical current is in the wire.

The magnetic field is produced as a result when an electrical current is passed through the conducting wire .

Option A is the appropriate response because a wire's magnetic field is created when an electrical current flows through it due to the magnetic influence of the electric current .

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Answer:

a neutron star

Explanation:

Answer:

d

Explanation:

Answer:

a) The magnitude of the friction force is 55.851 newtons, b) The speed of the crate when it reaches the bottom of the ramp is 2.526 meters per second.

Explanation:

a) This situation can be modelled by the Principle of Energy Conservation and the Work-Energy Theorem, where friction represents the only non-conservative force exerting on the crate in motion. Let consider the bottom of the straight ramp as the zero point. The energy equation for the crate is:

[tex]U_{g,1}+K_{1} = U_{g,2}+K_{2}+ W_{fr}[/tex]

Where:

[tex]U_{g,1}[/tex], [tex]U_{g,2}[/tex] - Initial and final gravitational potential energy, measured in joules.

[tex]K_{1}[/tex], [tex]K_{2}[/tex] - Initial and final translational kinetic energy, measured in joules.

[tex]W_{fr}[/tex] - Work losses due to friction, measured in joules.

By applying the defintions of translational kinetic and gravitational potential energies and work, this expression is now expanded:

[tex]m\cdot g \cdot y_{1} + \frac{1}{2}\cdot m\cdot v_{1}^{2} = m\cdot g \cdot y_{2} + \frac{1}{2}\cdot m\cdot v_{2}^{2} + \mu_{k}\cdot m \cdot g \cdot \cos \theta[/tex]

Where:

[tex]m[/tex] - Mass of the crate, measured in kilograms.

[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.

[tex]y_{1}[/tex], [tex]y_{2}[/tex] - Initial and final height of the crate, measured in meters.

[tex]v_{1}[/tex], [tex]v_{2}[/tex] - Initial and final speeds of the crate, measured in meters per second.

[tex]\mu_{k}[/tex] - Kinetic coefficient of friction, dimensionless.

[tex]\theta[/tex] - Ramp inclination, measured in sexagesimal degrees.

The equation is now simplified and the coefficient of friction is consequently cleared:

[tex]y_{1}-y_{2}+\frac{1}{2\cdot g}\cdot (v_{1}^{2}-v_{2}^{2}) = \mu_{k}\cdot \cos \theta[/tex]

[tex]\mu_{k} = \frac{1}{\cos \theta} \cdot \left[y_{1}-y_{2}+\frac{1}{2\cdot g}\cdot (v_{1}^{2}-v_{2}^{2}) \right][/tex]

The final height of the crate is:

[tex]y_{2} = (1.6\,m)\cdot \sin 30^{\circ}[/tex]

[tex]y_{2} = 0.8\,m[/tex]

If [tex]\theta = 30^{\circ}[/tex], [tex]y_{1} = 0\,m[/tex], [tex]y_{2} = 0.8\,m[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]v_{1} = 5\,\frac{m}{s}[/tex] and [tex]v_{2} = 0\,\frac{m}{s}[/tex], the coefficient of friction is:

[tex]\mu_{k} = \frac{1}{\cos 30^{\circ}}\cdot \left\{0\,m-0.8\,m+\frac{1}{2\cdot \left(9.807\,\frac{m}{s^{2}} \right)}\cdot \left[\left(5\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}\right] \right\}[/tex]

[tex]\mu_{k} \approx 0.548[/tex]

Then, the magnitude of the friction force is:

[tex]f =\mu_{k}\cdot m\cdot g \cdot \cos \theta[/tex]

If [tex]\mu_{k} \approx 0.548[/tex], [tex]m = 12\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex] and [tex]\theta = 30^{\circ}[/tex], the magnitude of the force of friction is:

[tex]f = (0.548)\cdot (12\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot \cos 30^{\circ}[/tex]

[tex]f = 55.851\,N[/tex]

The magnitude of the force of friction is 55.851 newtons.

b) The energy equation of the situation is:

[tex]m\cdot g \cdot y_{1} + \frac{1}{2}\cdot m\cdot v_{1}^{2} = m\cdot g \cdot y_{2} + \frac{1}{2}\cdot m\cdot v_{2}^{2} + \mu_{k}\cdot m \cdot g \cdot \cos \theta[/tex]

[tex]y_{1}+\frac{1}{2\cdot g}\cdot v_{1}^{2} =y_{2} + \frac{1}{2\cdot g}\cdot v_{2}^{2} + \mu_{k}\cdot \cos \theta[/tex]

Now, the final speed is cleared:

[tex]y_{1}-y_{2}+ \frac{1}{2\cdot g}\cdot v_{1}^{2} -\mu_{k}\cdot \cos \theta= \frac{1}{2\cdot g}\cdot v_{2}^{2}[/tex]

[tex]2\cdot g \cdot (y_{1}-y_{2}-\mu_{k}\cdot \cos \theta) + v_{1}^{2} = v_{2}^{2}[/tex]

[tex]v_{2} = \sqrt{2\cdot g \cdot (y_{1}-y_{2}-\mu_{k}\cdot \cos \theta)+v_{1}^{2}}[/tex]

Given that [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]y_{1} = 0.8\,m[/tex], [tex]y_{2} = 0\,m[/tex], [tex]\mu_{k} \approx 0.548[/tex], [tex]\theta = 30^{\circ}[/tex] and [tex]v_{1} = 0\,\frac{m}{s}[/tex], the speed of the crate at the bottom of the ramp is:

[tex]v_{2}=\sqrt{2\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot [0.8\,m-0\,m-(0.548)\cdot \cos 30^{\circ}]+\left(0\,\frac{m}{s} \right)^{2}}[/tex]

[tex]v_{2}\approx 2.526\,\frac{m}{s}[/tex]

The speed of the crate when it reaches the bottom of the ramp is 2.526 meters per second.

Answer:

a) Your player

b) Observer's player

c) Each measures their own first

Explanation:

Because given problem is having relative velocity to each other. The person sitting on the train is moving with a very high speed relative to the person standing next to the track.

In this case, the clock situated in the train will be running slow with respect to the stationary frame of reference

Answer:

Explanation:

This question appears incomplete because of the absence of the data been talked about in the question. However, there is a general ruling here and it can be applied to the data at hand.

If an increase in the distance of charges (let's denote with "d") causes the electric field strength (let's denote with"E") to increase, then the mathematical representation can be illustrated as d ∝ E (meaning distance of charge is directly proportional to electric field strength).

But if an increase in the distance of the charges causes the electric field strength to decrease, then the mathematical representation can be illustrated as d ∝ 1/E (meaning distance of charge is inversely proportional to electric field strength).

A scatterplot can also be used to determine this. If there is a positive correlation (correlation value is greater than zero but less than or equal to 1) on the graph, then it is illustrated as "d ∝ E" BUT if there is a negative correlation (correlation value is less than zero but greater than or equal to -1), then it can be illustrated as "d ∝ 1/E".

Complete question:

a 2.0 kg block slides on the horizontal, frictionless surface until it counters a spring with force constant of  955 N/m. The block comes to rest after compressing the spring a distance of 4.6 cm. Find the initial speed (in m/s) of the block.

Answer:

The initial speed of the block is 1.422 m/s

Explanation:

Given;

mass of the block, m = 2.0 kg

force constant of the spring, K = 955 N/m

compression of the spring, x = 4.6 cm = 0.046 m

Apply Hook's law to determine applied force on the spring;

F = Kx

F = (955 N/m)(0.046 m)

F = 43.93 N

Apply Newton's 2nd law to determine the magnitude of deceleration of the block when it encounters the spring;

F = ma

a = F / m

a = 43.93 / 2

a = 21.965 m/s²

Apply kinematic equation to determine the initial speed of the block;

v² = u² + 2ax

where;

v is the final speed of the block = 0

u is the initial speed of the block

x is the distance traveled by the block = compression of the spring

a is the block deceleration = -21.965 m/s²

0 = u² + 2(-21.965 )(0.046)

0 = u²  - 2.021

u² =  2.021

u = √2.021

u = 1.422 m/s

Therefore, the initial speed of the block is 1.422 m/s

Answer and Explanation:

Data provided as per the question is as follows

Speed at point A = 20 m/s

Acceleration at point C = [tex]5 m/s^2[/tex]

[tex]r_A = 25 m[/tex]

The calculation of the magnitude of the acceleration at A is shown below:-

Centripetal acceleration is

[tex]a_c = \frac{v^2}{r}[/tex]

now we will put the values into the above formula

= [tex]\frac{20^2}{25}[/tex]

After solving the above equation we will get

[tex]= 16 m/s^2[/tex]

Tangential acceleration is

[tex]= \sqrt{ac^2 + at^2} \\\\ = \sqrt{16^2 + 5^2}\\\\ = 16.703 m/s^2[/tex]

Complete question is;

The place you get your hair cut has two nearly parallel mirrors 6.50 m apart. As you sit in the chair, your head is 3.00 m from the nearer mirror. Looking toward this mirror, you first see your face and then, farther away, the back of your head. (The mirrors need to be slightly nonparallel for you to be able to see the back of your head, but you can treat them as parallel in this problem.) How far away does the back of your head appear to be?

Answer:

13 m

Explanation:

We are given;

Distance between two nearly parallel mirrors; d = 6.5 m

Distance between the face and the nearer mirror; x = 3 m

Thus, the distance between the back-head and the mirror = 6.5 - 3 = 3.5m

Now, From the given values above and using the law of reflection, we can find the distance of the first reflection of the back of the head of the person in the rear mirror.

Thus;

Distance of the first reflection of the back of the head in the rear mirror from the object head is;

y' = 2y

y' = 2 × 3.5

y' = 7

The total distance of this image from the front mirror would be calculated as;

z = y' + x

z = 7 + 3

z = 10

Finally, the second reflection of this image will be 10 meters inside in the front mirror.

Thus, the total distance of the image of the back of the head in the front mirror from the person will be:

T.D = x + z

T.D = 3 + 10

T.D = 13m

Answer:

The intensity of light from the first polarizer  is [tex]I_1 = 26 W/m^2[/tex]

Explanation:

  The intensity of the unpolarized light is  [tex]I_o = 52.0 \ W/m^2[/tex]

Generally the intensity of light that emerges from the first polarized light is

            [tex]I_1 = \frac{I_o}{2 }[/tex]

 substituting values

             [tex]I_1 = \frac{52. 0}{2 }[/tex]

             [tex]I_1 = 26 W/m^2[/tex]

Answer:

Upper plate Q/3

Lower plate 2Q/3

Explanation:

See attached file

Answer: E = 39.54 N/C

Explanation: Electric field can be determined using surface charge density:

[tex]E = \frac{\sigma}{2\epsilon_{0}}[/tex]

where:

σ is surface charge density

[tex]\epsilon_{0}[/tex] is permitivitty of free space ([tex]\epsilon_{0} = 8.85.10^{-12}[/tex][tex]C^{2}/N.m^{2}[/tex])

Calculating resulting electric field:

[tex]E=E_{1} - E_{2}[/tex]

[tex]E = \frac{\sigma_{1}-\sigma_{2}}{2\epsilon_{0}}[/tex]

[tex]E = \frac{[0.3-(-0.4)].10^{-9}}{2.8.85.10^{-12}}[/tex]

[tex]E=0.03954.10^{3}[/tex]

E = 39.54

The resulting Electric Field at any point is 39.54N/C.

The magnitude of the resulting electric field at any point should be  28.2 N/C.

Since the Charge of uniform density (0.30 nC/m2) should be allocated over the xy plane, and charge of uniform density (−0.40 nC/m2)should be allocated over the yz plane.

So,

E1

= σ1/2ε0

= 0.30e-9/(2*8.85e-12)

= 16.949 N/C

So, direction of E1 is +z

Now

E2 = σ2/2ε0

= 0.40e-9/(2*8.85e-12)

= 22.6 N/C

So,  direction of E2 is -x

Now

E = √(E1*E1+E2*E2)

= √(16.949*16.949+22.6*22.6)

= 28.2 N/C

Learn more about magnitude here: https://brainly.com/question/14576767

Answer:In the case of two proton beams the protons repel one another because they have the same sign of electrical charge. There is also an attractive magnetic force between the protons, but in the proton frame of reference this force must be zero! Clearly then the attractive magnetic force that reduces the net force between protons in the two beams as seen in our frame of reference is relativistic. In particular the apparent magnetic forces or fields are relativistic modifications of the electrical forces or fields. As such modifications, they cannot be stronger than the electrical forces and fields that produce them. This follows from the fact that switching frames of reference can reduce forces, but it can’t turn what is attractive in one frame into a repulsive force in another frame.

In the case of wires the net charges in two wires are zero everywhere along the wires. That makes the net electrical forces between the wires very nearly zero. Yet the relativistic magnetic forces and fields will be of the same sort as in the case of two beams of charges of a single sign. This is true even in the frame of reference of what we think as the moving charges, that is, the electrons. In the frame of reference moving at the drift velocity of these current-carrying electrons, it is the protons or positively charged ions that are moving in the other direction. Consequently in any frame of reference for current-carrying wires in parallel, the net electrical force will be essentially zero, and there will be a net attractive magnetic force

Explanation:                                                                              

Explanation:

Particles with similar charges (both positive or both negative) will always repel each other, regardless of their speed or direction.