Two identical small charged spheres are a certain distance apart, and each one initially experiences an electrostatic force of magnitude F due to the other. With time, charge gradually leaks off of both spheres. Calculate the magnitude of the electrostatic force, when each of the spheres has lost half of its initial charge. (Your answer will be a function of F, since no values are giving)

Answers

Answer 1

Answer:

1/4F

Explanation:

We already know thatThe electrostatic force is directly proportional to the product of the charge, from Coulomb's law.

So F α Qq

But if it is now half the initial charges, then

F α (1/2)Q *(1/2)q

F α (1/4)Qq

Thus the resultant charges are each halved is (1/4) and the first initial force experienced at full charge.

Thus the answer will be 1/4F


Related Questions

Suppose Young's double-slit experiment is performed in air using red light and then the apparatus is immersed in water. What happens to the interference pattern on the screen?

Answers

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

•• A metal sphere carrying an evenly distributed charge will have spherical equipotential surfaces surrounding it. Suppose the sphere’s radius is 50.0 cm and it carries a total charge of (a) Calculate the potential of the sphere’s surface. (b)You want to draw equipotential surfaces at intervals of 500 V outside the sphere’s surface. Calculate the distance between the first and the second equipotential surfaces, and between the 20th and 21st equipotential surfaces. (c) What does the changing spacing of the surfaces tell you about the electric field?

Answers

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

At what frequency f, in hertz, would you have to move the comb up and down to produce red light, of wavelength 600 nm

Answers

Answer:

f = 500 x 10^12Hz

Explanation:

E=hc/wavelength

E=hf

hc/wavelength =hf

c/wavelength =f

f = 3 x 10^8 / 600 x 10^-9 = 500 x 10^12Hz

A student builds a rocket-propelled cart for a science project. Its acceleration is not quite high enough to win a prize, so he uses a larger rocket engine that provides 39% more thrust, although doing so increases the mass of the cart by 13%. By what percentage does the cart's acceleration increase?

Answers

Answer:

Explanation:

a = F / m

where a is acceleration , F is thrust and m is mass

taking log and differentiating

da / a = dF / F - dm / m

(da / a)x 100 = (dF / F)x100 - (dm / m) x100

percentage increase in a = percentage increase in F - percentage increase in m

= percentage increase in acceleration a   = 39 - 13 = 26 %

required increase = 26 %.

Consider an electromagnetic wave where the electric field of an electromagnetic wave is oscillating along the z-axis and the magnetic field is oscillating along the x-axis.

Required:
In what directions is it possible that the wave is traveling?

Answers

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

An 1300-turn coil of wire that is 2.2 cmcm in diameter is in a magnetic field that drops from 0.14 TT to 0 TT in 9.0 msms . The axis of the coil is parallel to the field.
What is the emf of the coil? (in V)

Answers

Answer:

The induced  emf is  [tex]\epsilon =7.68 \ V[/tex]

Explanation:

From the question we are told that

     The number of turns is  [tex]N = 1300 \ turns[/tex]

    The diameter is  [tex]d = 2.2 \ cm = 2.2*10^{-2}[/tex]

     The initial magnetic field is  [tex]B_i = 0.14 \ T[/tex]

      The final magnetic field is  [tex]B_f = 0 \ T[/tex]

      The  time taken is  [tex]dt = 9.0ms = 9.0*10^{-3} \ s[/tex]

 

The radius is mathematically evaluated as

      [tex]r = \frac{d}{2 }[/tex]

substituting values

     [tex]r = \frac{2.2 *10^{-2}}{2 }[/tex]

     [tex]r = 1.1*10^{-2} \ m[/tex]

The induced emf is mathematically represented as

    [tex]\epsilon =- N * \frac{d\phi }{dt }[/tex]

Where  [tex]d\phi[/tex] is the change in magnetic field which is mathematically represented as

        [tex]d\phi = dB * A * cos\theta[/tex]

=>   [tex]d\phi = [B_f - B_i ] * A * cos\theta[/tex]

Here  [tex]\theta = 0[/tex] given that the axis of the coil is parallel to the field

Also A is the cross-sectional area which is mathematically represented as

       [tex]A = \pi r^2[/tex]

substituting values

      [tex]A = 3.142 * [1.1*10^{-2}]^2[/tex]

       [tex]A = 3.8 *10^{-4] \ m^2[/tex]

So

    [tex]d\phi = [0 - 0.14 ] * 3.8*10^{-4}[/tex]

    [tex]d\phi = -5.32*10^{-5} \ weber[/tex]

So  

     [tex]\epsilon =- 1300 * \frac{-5.32*10^{-5} }{ 9.0*10^{-3} }[/tex]

    [tex]\epsilon =7.68 \ V[/tex]

Tuning a guitar string, you play a pure 330 Hz note using a tuning device, and pluck the string. The combined notes produce a beat frequency of 5 Hz. You then play a pure 350 Hz note and pluck the string, finding a beat frequency of 25 Hz. What is the frequency of the string note?

Answers

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]

       

Calculate the power of the eye in D when viewing an object 5.70 m away. (Assume the lens-to-retina distance is 2.00 cm. Enter your answer to at least one decimal place.)

Answers

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

Two buses are moving in opposite directions with velocities of 36 km/hr and 108
km/hr. Find the distance between them after 20 minutes.

Answers

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.

Describe the orientation of magnetic field lines by drawing a bar magnet, labeling the poles, and drawing several lines indicating the direction of the forces.

Answers

Answer:

A field is a way of mapping forces surrounding any object that can act on another object at a distance without apparent physical connection. The field represents the object generating it. Gravitational fields map gravitational forces, electric fields map electrical forces, and magnetic fields map magnetic forces.

Explanation:

Three resistors, 6.0-W, 9.0-W, 15-W, are connected in parallel in a circuit. What is the equivalent resistance of this combination of resistors?

Answers

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.

Calculation of the equivalent resistance:

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

learn more about resistance here: https://brainly.com/question/15047345

Flywheels are large, massive wheels used to store energy. They can be spun up slowly, then the wheel's energy can be released quickly to accomplish a task that demands high power. An industrial flywheel has a 1.5 m diameter and a mass of 250 kg. Its maximum angular velocity is 1200 rpm.
How long does it take the flywheel to reach top angular speed of 1200 rpm?

Answers

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

A 4g bullet, travelling at 589m/s embeds itself in a 2.3kg block of wood that is initially at rest, and together they travel at the same velocity. Calculate the percentage of the kinetic energy that is left in the system after collision to that before.

Answers

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 %

Determine the position in the oscillation where an object in simple harmonic motion: (Be very specific, and give some reasoning to your answer.) has the greatest speed has the greatest acceleration experiences the greatest restoring force experiences zero restoring force g

Answers

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 .

A skater spins at 3rev/s when she stretches her arms outward. If she keeps her fists on her chest she can spin at 4.5rev/s and her body inertia is 3kg.m2. What is her body inertia when she stretches her arms outward?

Answers

Answer:

Body inertia I = 4.5 kg/m^2

Explanation:

Here, we want to calculate the body inertia when the arms are stretched outwards.

We know from the question that angular momentum is conserved

Thus;

I * 3 = 4.5 * 3

I = 4.5 kg/m^2

g a conductor consists of an infinite number of adjacent wires, each infinitely long. If there are n wires per unit length, what is the magnitude of B~

Answers

Answer:

B=uonI/2

Explanation:

See attached file

Consider the following spectrum where two colorful lines (A and B) are positioned on a dark background. The violet end of the spectrum is on the left and the red end of the spectrum is on the right. A B 5. (1 point) What is the name for this type of spectrum? 6. (1 point) Transition A is associated with an electron moving between the n= 1 and n= 3 levels. Transition B is associated with an electron moving between the n= 2 and n= 5 levels. Which transition is associated with a photon of longer wavelength?

Answers

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 .

"A power of 200 kW is delivered by power lines with 48,000 V difference between them. Calculate the current, in amps, in these lines."

Answers

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.

A magnetic field is entering into a coil of wire with radius of 2(mm) and 200 turns. The direction of magnetic field makes an angle 25° in respect to normal to surface of coil. The magnetic field entering coil varies 0.02 (T) in every 2 seconds. The two ends of coil are connected to a resistor of 20 (Ω).
A) Calculate Emf induced in coil
B) Calculate the current in resistor
C) Calculate the power delivered to resistor by Emf

Answers

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

Based on the graph below, what prediction can we make about the acceleration when the force is 0 newtons? A. It will be 0 meters per second per second. B. It will be 5 meters per second per second. C. It will be 10 meters per second per second. D. It will be 15 meters per second per second.
PLZ HURRY WILL MARK BRAINLIEST IF CORRECT

Answers

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]

The radius of curvature of the path of a charged particle in a uniform magnetic field is directly proportional toA) the particle's charge.B) the particle's momentum.C) the particle's energy.D) the flux density of the field.E)All of these are correct

Answers

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.

A piece of electronic equipment that is surrounded by packing material is dropped so that it hits the ground with a speed of 4 m/s. After impact, the equipment experiences an acceleration of a = 2kx, where k is a constant and x is the compression of the packing material. If the packing material experiences a maximum compression of 20 mm, determine the maximum acceleration of the equipment.

Answers

Answer:

Maximum acceleration is 800m/s^2

Explanation:

See attached file

If an astronomer wants to find and identify as many stars as possible in a star cluster that has recently formed near the surface of a giant molecular cloud (such as the Trapezium cluster in the Orion Nebula), what instrument would be best for her to use

Answers

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.

An object has an acceleration of 6.0 m/s/s. If the net force was doubled and the mass was one-third the original value, then the new acceleration would be _____ m/s/s.

Answers

Hahahahaha. Okay.

So basically , force is equal to mass into acceleration.

F=ma

so when F=ma , we get acceleration=6m/s/s

Force is doubled.

Mass is 1/3 times original.

2F=1/3ma

Now , we rearrange , and we get 6F=ma

So , now for 6 times the original force , we get 6 times the initial acceleration.

So new acceleration = 6*6= 36m/s/s

Two objects are in all respects identical except for the fact that one was coated with a substance that is an excellent reflector of light while the other was coated with a substance that is a perfect absorber of light. You place both objects at the same distance from a powerful light source so they both receive the same amount of energy U from the light. The linear momentum these objects will receive is such that:

Answers

Answer:

absorbent    p = S / c

reflective         p = 2S/c

Explanation:

The moment of radiation on a surface is

          p = U / c

where U is the energy and c is the speed of light.

In the case of a fully absorbent object, the energy is completely absorbed. The energy carried by the light is given by the Poynting vector.

           p = S / c

in the case of a completely reflective surface the energy must be absorbed and remitted, therefore there is a 2-fold change in the process

           p = 2S/c

A proton that is initially at rest is accelerated through an electric potential difference of magnitude 500 V. What speed does the proton gain? (e = 1.60 × 10-19 C , mproton = 1.67 × 10-27 kg)

Answers

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

A particle moves in a velocity field V(x, y) = x2, x + y2 . If it is at position (x, y) = (7, 2) at time t = 3, estimate its location at time t = 3.01.

Answers

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)

two resistors of resistance 10 ohm's and 20 ohm's are connected in parallel to a batery of e.m.f 12V. Calculate the current passing through the 20hm's resister​

Answers

Current through 20 ohm resistor is 0.6 A

A person standing 180m from the foot of a high building claps hi
hand and hears the echo 0.03minutes later. What is the speed
sound in air at that temperature?
A) 331m/s
B) 240m/s C) 200m/s D) 300m/s

Answers

Answer:

C) 200 m/s

Explanation:

The sound travels a total distance of 360 m in 0.03 minutes.

v = (360 m) / (0.03 min × 60 s/min)

v = 200 m/s

Astronauts increased in height by an average of approximately 40 mm (about an inch and a half) during the Apollo-Soyuz missions, due to the absence of gravity compressing their spines during their time in space. Does something similar happen here on Earth

Answers

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

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