Search Results Web results A car of mass 650 kg is moving at a speed of 0.7

Answer:

a) 2.278 x 10^-5 volts

b) 1.139 x 10^-6 Ampere

c) 2.59 x 10^-11 W

Explanation:

The radius of the wire r = 2 mm = 0.002 m

the number of turns N = 200 turns

direction of the magnetic field ∅ = 25°

magnetic field strength B = 0.02 T

varying time = 2 sec

The cross sectional area of the wire = [tex]\pi r^{2}[/tex]

==> A = 3.142 x [tex]0.002^{2}[/tex] = 1.257 x 10^-5 m^2

Field flux Φ = BA cos ∅ = 0.02 x 1.257 x 10^-5 x cos 25°

==> Φ = 2.278 x 10^-7 Wb

The induced EMF is given as

E = NdΦ/dt

where dΦ/dt = (2.278 x 10^-7)/2 = 1.139 x 10^-7

E = 200 x 1.139 x 10^-7 = 2.278 x 10^-5 volts

b) If the two ends are connected to a resistor of 20 Ω, the current through the resistor is given as

[tex]I[/tex] = E/R

where R is the resistor

[tex]I[/tex] = (2.278 x 10^-5)/20 = 1.139 x 10^-6 Ampere

c) power delivered to the resistor is given as

P = [tex]I[/tex]E

P = (1.139 x 10^-6) x (2.278 x 10^-5) = 2.59 x 10^-11 W

Complete Question

The  complete question is  shown on the first uploaded image  

Answer:

a

  [tex]f= -75 \ cm = - 0.75 \ m[/tex]

b

  [tex]P = -1.33 \ diopters[/tex]

Explanation:

From the question we are told that

    The  image distance is  [tex]d_i = -75 cm[/tex]

The value of the image is negative because it is on the same side with the corrective glasses

    The  object distance is  [tex]d_o = \infty[/tex]

The  reason object distance  is because the object father than it being picture by the eye

General focal length is mathematically represented as

              [tex]\frac{1}{f} = \frac{1}{d_i} - \frac{1}{d_o}[/tex]

substituting values

             [tex]\frac{1}{f} = \frac{1}{-75} - \frac{1}{\infty}[/tex]

=>         [tex]f= -75 \ cm = - 0.75 \ m[/tex]

Generally the power of the corrective lens is  mathematically represented as

        [tex]P = \frac{1}{f}[/tex]

substituting values

       [tex]P = \frac{1}{-0.75}[/tex]

        [tex]P = -1.33 \ diopters[/tex]

Answer:

t = 2.95 min

Explanation:

Given that,

The diameter of flywheeel, d = 1.5 m

Mass of flywheel, m = 250 kg

Initial angular velocity is 0

Final angular velocity, [tex]\omega_f=1200\ rpm = 126\ rad/s[/tex]

We need to find the time taken by the flywheel to each a speed of 1200 rpm if it starts from rest.

Firstly, we will find the angular acceleration of the flywheel.

The moment of inertia of the flywheel,

[tex]I=\dfrac{1}{2}mr^2\\\\I=\dfrac{1}{2}\times 250\times (0.75)^2\\\\I=70.31\ kg-m^2[/tex]

Now,

Let the torque is 50 N-m. So,

[tex]\alpha =\dfrac{\tau}{I}\\\\\alpha =\dfrac{50}{70.31}\\\\\alpha =0.711\ rad/s^2[/tex]

So,

[tex]t=\dfrac{\omega_f-\omega_i}{\alpha }\\\\t=\dfrac{126-0}{0.711}\\\\t=177.21\ s[/tex]

or

t = 2.95 min

Answer:

a) 6.57 m/s

b) 53.75 J

c) 6.37 m/s

d) -98.297 J

Explanation:

mass of player = [tex]m_{p}[/tex] = 117.5 kg

speed of player = [tex]v_{p}[/tex] = 6.5 m/s

mass of ball = [tex]m_{b}[/tex] = 0.43 kg

velocity of ball = [tex]v_{b}[/tex] = 26.5 m/s

Recall that momentum of a body = mass x velocity = mv

initial momentum of the player = mv = 117.5 x 6.5 = 763.75 kg-m/s

initial momentum of the ball = mv = 0.43 x 26.5 = 11.395 kg-m/s

initial kinetic energy of the player = [tex]\frac{1}{2} mv^{2}[/tex] = [tex]\frac{1}{2}[/tex] x 117.5 x [tex]6.5^{2}[/tex] =  2482.187 J

a) according to conservation of momentum, the initial momentum of the system before collision must equate the final momentum of the system.

for this first case that they travel in the same direction, their momenta carry the same sign

[tex]m_{p}[/tex][tex]v_{p}[/tex] + [tex]m_{b}[/tex][tex]v_{b}[/tex] = ([tex]m_{p}[/tex] +[tex]m_{b}[/tex])v

where v is the final velocity of the player.

inserting calculated momenta of ball and player from above, we have

763.75 + 11.395 = (117.5 + 0.43)v

775.145 = 117.93v

v = 775.145/117.93 = 6.57 m/s

b) the player's new kinetic energy = [tex]\frac{1}{2} mv^{2}[/tex] = [tex]\frac{1}{2}[/tex] x 117.5 x [tex]6.57^{2}[/tex] = 2535.94 J

change in kinetic energy = 2535.94 - 2482.187 = 53.75 J  gained

c) if they travel in opposite direction, equation becomes

[tex]m_{p}[/tex][tex]v_{p}[/tex] - [tex]m_{b}[/tex][tex]v_{b}[/tex] = ([tex]m_{p}[/tex] +[tex]m_{b}[/tex])v

763.75 - 11.395 = (117.5 + 0.43)v

752.355 = 117.93v

v = 752.355/117.93 = 6.37 m/s

d) the player's new kinetic energy = [tex]\frac{1}{2} mv^{2}[/tex] = [tex]\frac{1}{2}[/tex] x 117.5 x [tex]6.37^{2}[/tex]  = 2383.89 J

change in kinetic energy = 2383.89 - 2482.187 = -98.297 J

that is 98.297 J  lost

Answer:

The bright fringes will appear much closer together

Explanation:

Because λn = λ/n ,

And the wavelength of light in water is smaller than the wavelength of light in air. Given that the distance between bright fringes is proportional to the wavelength

Explanation:

It is given that,

Speed of bus 1 is 36 km/h and speed of bus 2 is 108 km/h. We need to find the distance between bus 1 and 2 after 20 minutes.

Time = 20 minutes = [tex]\dfrac{20}{60}\ h=\dfrac{1}{3}\ h[/tex]

As the buses are moving in opposite direction, then the concept of relative velocity is used. So,

Distance, [tex]d=v\times t[/tex]

v is relative velocity, v = 108 + 36 = 144 km/h

So,

[tex]d=144\ km/h \times \dfrac{1}{3}\ h\\\\d=48\ km[/tex]

So, the distance between them is 48 km after 20 minutes.

Answer:

It will displace the same weight of fresh water i.e.16000N. The point is the body 'floats'- which is the underlying assumption here, and by Archimedes Principle, for this body or vessel or whatever it may be, to float it should displace an equal weight of water

Explanation:

Answer:

Infrared telescope and camera

Explanation:

An infrared telescope uses infrared light to detect celestial bodies. The infrared radiation is one of the known forms of electromagnetic radiation. Infrared radiation is given off by a body possessing some form of heat. All bodies above the absolute zero temperature in the universe radiates some form of heat, which can then be detected by an infrared telescope, and infrared radiation can be used to study or look into a system that is void of detectable visible light.

Stars are celestial bodies that are constantly radiating heat. In order to see a clearer picture of the these bodies, Infrared images is better used, since they are able to penetrate the surrounding clouds of dust, and have located many more stellar components than any other types of telescope, especially in dusty regions of star clusters like the Trapezium cluster.

Answer:

Explanation:

For this exercise we will use that the potential is created by the charge inside the equinoctial surface and just like in Gauss's law we can consider all the charge concentrated in the center.

Therefore the potential on the ferric surface is

        V = k Q / r

where k is the Coulomb constant, Q the charge of the sphere and r the distance from the center to the point of interest

a) On the surface the potential

        V = 9 10⁹ Q / 0.5

        V = 18 10⁹ Q

Unfortunately you did not write the value of the load, suppose a value to complete the calculations Q = 1 10⁻⁷ C, with this value the potential on the surfaces V = 1800 V

b) The equipotential surfaces are concentric spheres, let's look for the radii for some potentials

for V = 1300V let's find the radius

             r = k Q / V

             r = 9 109 1 10-7 / 1300

             r = 0.69 m

other values ​​are shown in the following table

V (V)      r (m)

1800     0.5

1300     0.69

 800      1,125

 300     3.0

In other words, we draw concentric spheres with these radii and each one has a potential difference of 500V

C) The spacing of the spheres corresponds to lines of radii of the electric field that have the shape

         E = k Q / r²

Answer:

New location at time 3.01 is given by: (7.49, 2.11)

Explanation:

Let's start by understanding what is the particle's velocity (in component form) in that velocity field at time 3:

[tex]V_x=x^2=7^2=49\\V_y=x+y^2=7+2^2=11[/tex]

With such velocities in the x direction and in the y-direction respectively, we can find the displacement in x and y at a time 0.01 units later by using the formula:

[tex]distance=v\,*\, t[/tex]

[tex]distance_x=49\,(0.01)=0.49\\distance_y=11\,(0.01)=0.11[/tex]

Therefore, adding these displacements in component form to the original particle's position, we get:

New position: (7 + 0.49, 2 + 0.11) = (7.49, 2.11)

Answer:

Yes. Something similar occurs here on Earth.

Explanation:

Gravity tends to pull objects perpendicularly to the ground. In space, the absence of this force means there is no compression on the spine due to gravity trying to pull it down. This means that astronauts undergo an increase in height in space.

Here on Earth, we experience gravity pull on our spine during the day. At night when we sleep, we lie down with our spine parallel to the ground, which means that our spine is no longer under compression from gravity force. The result is that we are a few centimetres taller in the morning when we wake up, than we are before going to bed at night. The increase is not much pronounced here on Earth because there is a repeated cycle of compression and decompression of our spine due to gravity, unlike when compared to that of astronauts that spend long duration in space, all the while without gravity forces on their spine

Answer:

a) Height reached before coming to rest is 42.86 m

b) Energy lost to friction is 17902.45 J

Explanation:

mass of the motorcycle = 185 kg

speed of the towards the hill = 29 m/s

The wheels weigh 12 kg each

Wheels are annular rings with an inner radius of 0.280 m and outer radius of 0.330 m

a) To go up the hill, the kinetic energy of motion of the motorcycle will be converted to the potential energy it will gain in going up a given height

the kinetic energy of the motorcycle is given as

[tex]KE[/tex] = [tex]\frac{1}{2}mv^{2}[/tex]

where m is the mass of the motorcycle

v is the velocity of the motorcycle

[tex]KE[/tex]  = [tex]\frac{1}{2}*185*29^{2}[/tex] = 77792.5 J

This will be converted to potential energy

The potential energy up the hill will be

[tex]PE[/tex] = mgh

where m is the mass

g is acceleration due to gravity 9.81 m/s^2

h is the height reached before coming to rest

[tex]PE[/tex] = 185 x 9.81 x m = 1814.85h

equating the  kinetic energy to the potential energy for energy conservation, we'll have

77792.5 = 1814.85h

height reached before coming to rest  = 77792.5/1814.85 = 42.86 m

b) if an altitude of 33 m was reached before coming to rest, then the potential energy at this height is

[tex]PE[/tex] = mgh

[tex]PE[/tex]  = 185 x 9.81 x 33 = 59890.05 J

The energy lost to friction will be the kinetic energy minus this potential energy.

energy lost = 77792.5 - 59890.05 = 17902.45 J

A) The motorcycle can coast up the hill by ; 42.86m  

B) The amount of energy lost to friction :  17902.45 J

A) Determine how high the motorcycle can coast up the hill when friction is neglected

apply the formula for kinetic and potential energies

K.E = 1/2 mv²  ---- ( 1 )

P.E = mgH  ---- ( 2 )

As the motorcycle goes uphiLl the kinetic energy is converted to potential energy.

∴ K.E = P.E

1/2 * mv² = mgH

∴ H = ( 1/2 * mv² ) / mg  ---- ( 3 )

where ; m = 185 kg ,  v = 29 m/s ,  g = 9.81

Insert values into equation ( 3 )

H ( height travelled by motorcycle neglecting friction ) =  42.86m  

B) Determine how much energy is lost to friction if the motorcycle attains 33m before coming to rest  

P.E = mgh = 185 * 9.81 * 33  = 59890.05 J

where : h = 33 m , g = 9.81

K.E = 1/2 * mv² = 77792.5 J   ( question A )

∴ Energy lost ( ΔE ) =  ( 77792.5  - 59890.05 ) = 17902.45 J

Hence we can conclude that The motorcycle can coast up the hill by ; 42.86m , The amount of energy lost to friction :  17902.45 J.

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Answer:

moment of inertia I ≈ 4.0 x 10⁻³ kg.m²

Explanation:

given

point masses = 50g = 0.050kg

note: m₁=m₂=m₃=m₄=50g = 0.050kg

distance, r, from masses to eachother = 20cm = 0.20m

the distance, d, of each mass point from the centre of the mass, using pythagoras theorem is given by

= (20√2)/ 2 = 10√2 cm =14.12 x 10⁻² m  

moment of inertia is a proportion of the opposition of a body to angular acceleration about a given pivot that is equivalent to the entirety of the products of every component of mass in the body and the square of the component's distance from the center

mathematically,

I = ∑m×d²

remember, a square will have 4 equal points

I = ∑m×d² = 4(m×d²)

I = 4 × 0.050 × (14.12 x 10⁻² m)²

I = 0.20 × 1.96 × 10⁻²

I =  3.92 x 10⁻³ kg.m²

I ≈ 4.0 x 10⁻³ kg.m²

attached is the diagram of the equation

Answer:

The  frequency is  [tex]F = 325 Hz[/tex]

Explanation:

From the question we are told that

    The frequency for the first note is  [tex]F_1 = 330 Hz[/tex]

     The  beat frequency of the first note is  [tex]f_b = 5 \ Hz[/tex]

     The  frequency for the second note is  [tex]F_2 = 350 \ H_z[/tex]

      The  beat frequency of the first note is [tex]f_a = 25 \ Hz[/tex]

Generally beat frequency is mathematically represented as

        [tex]F_{beat} = | F_a - F_b |[/tex]

Where [tex]F_a \ and \ F_b[/tex] are frequencies of two sound source

  Now in the case of this question

For the first note

     [tex]f_b = F_1 - F \ \ \ \ \ ...(1)[/tex]

Where  F is the frequency of the string note

For the second note  

      [tex]f_a = F_2 - F \ \ \ \ \ ...(2)[/tex]

Adding  equation 1 from 2

      [tex]f_b + f_a = F_1 + F_2 + ( - F) + (-F) )[/tex]

      [tex]f_b + f_a = F_1 + F_2 -2F[/tex]

substituting values

       [tex]5 +25 = 330 + 350 -2F[/tex]

=>     [tex]F = 325 Hz[/tex]

Answer:

b. Rodger work = 1300 J

Explanation:

Work done: This can be defined as the product of force and distance along the direction of the force.

From the question,

Work is done by Rodger using a force of 100 N  in pushing the skateboard through a distance of 13.0 m.

W = F×d............. Equation 1

Where W = work done, F = force, d = distance.

Given: F = 100 N, d = 13 m

Substitute these values into equation 1

W = 100(13)

W = 1300 J.

Hence the right option is b. Rodger work = 1300 J

Answer:

2.9Ω

Explanation:

Resistors are said to be in parallel when they are arranged side by side such that their corresponding ends are joined together at two common junctions. The combined resistance in such arrangement of resistors is given by;

1/Req= 1/R1 + 1/R2 + 1/R3 .........+ 1/Rn

Where;

Req refers to the equivalent resistance and R1, R2, R3 .......Rn refers to resistance of individual resistors connected in parallel.

Note that;

R1= 6.0Ω

R2 = 9.0Ω

R3= 15.0 Ω

Therefore;

1/Req = 1/6 + 1/9 + 1/15

1/Req= 0.167 + 0.11 + 0.067

1/Req= 0.344

Req= (0.344)^-1

Req= 2.9Ω

The equivalent resistance of this combination of resistors is 2.9Ω.

The combined resistance in such arrangement of resistors is provided by;

1/Req= 1/R1 + 1/R2 + 1/R3 .........+ 1/Rn

here.

Req means  the equivalent resistance and R1, R2, R3

.Rn means the resistance of individual resistors interlinked in parallel.

Also,

R1= 6.0Ω

R2 = 9.0Ω

R3= 15.0 Ω

So,

1/Req = 1/6 + 1/9 + 1/15

1/Req= 0.167 + 0.11 + 0.067

1/Req= 0.344

Req= (0.344)^-1

Req= 2.9Ω

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Answer:

[tex]3.1\times 10^{5}m/s[/tex]

Explanation:

The computation of the speed does the proton gain is shown below:

The potential difference is the difference that reflects the work done as per the unit charged

So, the work done should be

= Potential difference × Charge

Given that

Charge on a proton is

= 1.6 × 10^-19 C

Potential difference = 500 V

[tex]v= \sqrt{\frac{2.q.\Delta V}{m_{p}}} \\\\\\= \sqrt{\frac{2\times 1.6\times 10^{-19}\times 5\times 10^{2}}{1.67\times 10^{-27}}}[/tex]

[tex]v= \sqrt{9.58\times 10^{10}}m/s \\\\= 3.095\times 10^{5}m/s\\\\\approx 3.1\times 10^{5}m/s[/tex]

Simply we applied the above formulas

Answer:

9.6×10⁹ A

Explanation:

From the question above,

P = VI.................... Equation 1

Where P = Electric power, V = Voltage, I = current.

make I the subject of the equation

I = P/V............. Equation 2

Given: P = 200 kW = 200×10³ W, V = 48000 V.

Substitute these vales into equation 2

I = 200×10³×48000

I = 9.6×10⁹ A.

Hence the current in the line is 9.6×10⁹ A.

Answer:

The  percentage of the kinetic energy that is left in the system after collision to that before is 0.174 %

Explanation:

Given;

mass of bullet, m₁ = 4g = 0.004kg

initial velocity of bullet, u₁ = 589 m/s

mass of block of wood, m₂ = 2.3 kg

initial velocity of the block of wood, u₂ = 0

let the final velocity of the system after collision = v

Apply the principle of conservation of linear momentum

m₁u₁ + m₂u₂ = v(m₁+m₂)

0.004(589) + 2.3(0) = v(0.004 + 2.3)

2.356 = 2.304v

v = 2.356 / 2.304

v = 1.0226 m/s

Initial kinetic energy of the system

K.E₁ = ¹/₂m₁u₁² + ¹/₂m₂u₂²

K.E₁ = ¹/₂(0.004)(589)² = 693.842 J

Final kinetic energy of the system

K.E₂ = ¹/₂v²(m₁ + m₂)

K.E₂ = ¹/₂ x 1.0226² x (0.004 + 2.3)

K.E₂ = 1.209 J

The kinetic energy left in the system = final kinetic energy of the system

The percentage of the kinetic energy that is left in the system after collision to that before = (K.E₂ / K.E₁) x 100%

                       = (1.209 / 693.842) x 100%

                        = 0.174 %

Therefore, the  percentage of the kinetic energy that is left in the system after collision to that before is 0.174 %

Answer:

a) 6738.27 J

b) 61.908 J

c)  [tex]\frac{4492.18}{v_{car} ^{2} }[/tex]

Explanation:

The complete question is

A flywheel is a mechanical device used to store rotational kinetic energy for later use. Consider a flywheel in the form of a uniform solid cylinder rotating around its axis, with moment of inertia I = 1/2 mr2.

Part (a) If such a flywheel of radius r1 = 1.1 m and mass m1 = 11 kg can spin at a maximum speed of v = 35 m/s at its rim, calculate the maximum amount of energy, in joules, that this flywheel can store?

Part (b) Consider a scenario in which the flywheel described in part (a) (r1 = 1.1 m, mass m1 = 11 kg, v = 35 m/s at the rim) is spinning freely at its maximum speed, when a second flywheel of radius r2 = 2.8 m and mass m2 = 16 kg is coaxially dropped from rest onto it and sticks to it, so that they then rotate together as a single body. Calculate the energy, in joules, that is now stored in the wheel?

Part (c) Return now to the flywheel of part (a), with mass m1, radius r1, and speed v at its rim. Imagine the flywheel delivers one third of its stored kinetic energy to car, initially at rest, leaving it with a speed vcar. Enter an expression for the mass of the car, in terms of the quantities defined here.

moment of inertia is given as

[tex]I[/tex] = [tex]\frac{1}{2}[/tex][tex]mr^{2}[/tex]

where m is the mass of the flywheel,

and r is the radius of the flywheel

for the flywheel with radius 1.1 m

and mass 11 kg

moment of inertia will be

[tex]I[/tex] =  [tex]\frac{1}{2}[/tex][tex]*11*1.1^{2}[/tex] = 6.655 kg-m^2

The maximum speed of the flywheel = 35 m/s

we know that v = ωr

where v is the linear speed = 35 m/s

ω = angular speed

r = radius

therefore,

ω = v/r = 35/1.1 = 31.82 rad/s

maximum rotational energy of the flywheel will be

E = [tex]Iw^{2}[/tex] = 6.655 x [tex]31.82^{2}[/tex] = 6738.27 J

b) second flywheel  has

radius = 2.8 m

mass = 16 kg

moment of inertia is

[tex]I[/tex] = [tex]\frac{1}{2}[/tex][tex]mr^{2}[/tex] =  [tex]\frac{1}{2}[/tex][tex]*16*2.8^{2}[/tex] = 62.72 kg-m^2

According to conservation of angular momentum, the total initial angular momentum of the first flywheel, must be equal to the total final angular momentum of the combination two flywheels

for the first flywheel, rotational momentum = [tex]Iw[/tex] = 6.655 x 31.82 = 211.76 kg-m^2-rad/s

for their combination, the rotational momentum is

[tex](I_{1} +I_{2} )w[/tex]

where the subscripts 1 and 2 indicates the values first and second  flywheels

[tex](I_{1} +I_{2} )w[/tex] = (6.655 + 62.72)ω

where ω here is their final angular momentum together

==> 69.375ω

Equating the two rotational momenta, we have

211.76 = 69.375ω

ω = 211.76/69.375 = 3.05 rad/s

Therefore, the energy stored in the first flywheel in this situation is

E = [tex]Iw^{2}[/tex] = 6.655 x [tex]3.05^{2}[/tex] = 61.908 J

c) one third of the initial energy of the flywheel is

6738.27/3 = 2246.09 J

For the car, the kinetic energy = [tex]\frac{1}{2}mv_{car} ^{2}[/tex]

where m is the mass of the car

[tex]v_{car}[/tex] is the velocity of the car

Equating the energy

2246.09 =  [tex]\frac{1}{2}mv_{car} ^{2}[/tex]

making m the subject of the formula

mass of the car m = [tex]\frac{4492.18}{v_{car} ^{2} }[/tex]

Answer:

Power=50.17dioptre

Power=50.17D

Explanation:

P=1/f = 1/d₀ + 1/d₁

Where d₀ = the eye's lens and the object distance= 5.70m=

d₁= the eye's lens and the image distance= 0.02m

f= focal length of the lense of the eye

We know that the object can be viewed clearly by the person ,then image and lens of the eye's distance needs to be equal with the retinal and the eye lens distance and this distance is given as 0.02m

Therefore, we can calculate the power using above formula

P= 1/5.70 + 1/0.02

Power=50.17dioptre

Therefore, the power the eye's is using to see the object from distance is 5.70D

Answer:

Pressure is  a scalar quantity defined as per unit area.

Density is the objects ,times its  the acceleration due to gravity.

Explanation:

Pressure is the alternative object increases the area of contact decrease .

Pressure is the force component  to the surface used to calculate pressure.

pressure is that collisions of the gas to container as the per unit time .

pressure is an physical important quantity to play the solid and  fluid .

Pressure is the expressed in a number of units depend the context use, pressure exerted by the liquid alone.

Density is the  objects, times, volume of the object that times acceleration objects.

Density is the used to the system complex objects and materials.

Density  force is the weight of a region or objects static fluid.

Answer:

Explanation:

a )

This type of spectrum is called line emission spectrum . Because it consists of lines . It is emission spectrum because it is due to emission of radiation from a source .

b ) The wavelength of a photon  is inversely proportional to its energy .  Photon  due to transition between n = 1 and n = 3 will have higher energy than

that due to transition between n = 2 and n = 5 . So the later photon ( B)  will have greater wavelength or photon  due to transition between n = 2 and n = 5 will have greater wavelength .

Answer:

B) the particle's momentum.

Explanation:

We know that

The centripetal force  on the particle when its moving in the radius R and velocity V

[tex]F_c=\dfrac{m\times V^2}{R}[/tex]

The magnetic force on the particle when the its moving with velocity V in the magnetic filed B and having charge q

[tex]F_m=q\times V\times B[/tex]

At the equilibrium condition

[tex]F_m=F_c[/tex]

[tex]q\times V\times B=\dfrac{m\times V^2}{R}[/tex]

[tex]R=\dfrac{m\times V}{q\times B}[/tex]

Momentum = m V

Therefore we can say that the radius of curvature is directly proportional to the particle momentum.

B) the particle's momentum.

Answer:

Maximum acceleration is 800m/s^2

Explanation:

See attached file

Answer:

Explanation:

The greatest speed is attained at middle point or equilibrium point or where displacement from equilibrium point is zero .

When the object remains at one of the extreme point it experiences greatest acceleration but at that point velocity is zero . Due to acceleration , its velocity goes on increasing till it come to equilibrium point . At this point acceleration becomes zero . After that its velocity starts decreasing because of negative acceleration . Hence at middle point velocity is maximum .

The greatest acceleration is attained at maximum displacement or at one of the two extreme end .

Greatest restoring force too will be at position where acceleration is maximum because acceleration is produced by restoring force .

Restoring force is proportional to displacement or extension against restoring force . So it will be maximum when displacement is maximum .

Zero restoring force exists at equilibrium position or middle point or at point where displacement is zero . It is so because acceleration at that point is zero .

Answer:

Option A

Explanation:

Acceleration will be obviously zero when Force = 0

That is how:

Force = Mass * Acceleration

So, If force = 0

0 = Mass * Acceleration.

Dividing both sides by Mass

Acceleration = 0/Mass

Acceleration = 0 m/s²

Answer:

[tex]\boxed{\mathrm{A. \: It \: will \: be \: 0 \: meters \: per \: second \: per \: second. }}[/tex]

Explanation:

[tex]\mathrm{force=mass \times acceleration}[/tex]

The force is given 0 newtons.

[tex]\mathrm{force=0 \: N}[/tex]

Plug force as 0.

[tex]\mathrm{0=mass \times acceleration}[/tex]

Divide both sides by mass.

[tex]\mathrm{\frac{0}{mass} =acceleration}[/tex]

[tex]\mathrm{0 =acceleration}[/tex]

[tex]\mathrm{acceleration= 0\: m/s/s}[/tex]

Answer:

Q = 832 kJ

Explanation:

It is given that,

Mass of the water, m = 2.5 kg

The latent heat of fusion, L = 333 kJ/kg

We need to find the heat needed to melt water at its melting point. The formula of heat needed to melt is given by :

Q = mL

[tex]Q=2.5\ kg\times 333\ kJ/kg\\\\Q=832.5\ kJ[/tex]

or

Q = 832 kJ

So, the heat needed to melt the water is 832 kJ.

Answer:

The wave is traveling in the y axis direction

Explanation:

Because the wave will always travel in a direction 90° to the magnetic and electric components