which of the following best describes pseudoscience?

Answer:

Time, [tex]t=1.07\times 10^{-23}\ s[/tex]

Explanation:

Given that,

When a high-energy proton or pion traveling near the speed of light collides with a nucleus, it may travel [tex]3.2\times 10^{-15}\ m[/tex] before interacting.

Let t is the time interval required for the strong interaction to occur. It will move with the speed of light. So,

[tex]t=\dfrac{d}{c}\\\\t=\dfrac{3.2\times 10^{-15}}{3\times 10^8}\\\\t=1.07\times 10^{-23}\ s[/tex]

So, the time interval is [tex]1.07\times 10^{-23}\ s[/tex]

Answer:

In the beginning, I was not familiar to assess assessments of the other students. Ifelt a little bit weird that is it possible to check assignments while having an instructor.I was also a bit frustrated, to be honest, that why do we have to assess thoseassessments. It was kind of extra burden for me. But after few weeks assessingmore assignments, my feeling had changed because I was learning lots of thingsthat were changing my perspectives. I was gaining extra knowledge from my peersin the form of assessments. Yes, I am comfortable with assessing assessments,because I got to learn many vocabularies and making structures of the sentencecorrectly by improving grammatically as I am not a native English speaker. Thus, inthis way, I was learning something new in each and every assessment.

Answer:

Explanation:

a )

moment of inertia in the first case will be sum of moment of inertia of two balls + moment of inertia of bar

= 2 x .700 x (2.1 / 2 )² + 3.7 x 2.1² / 12

= 1.5435 + 1.35975

= 2.90325 kg m²

b )

moment of inertia required

= moment of inertia of bar + moment of inertia of the other ball

= 3.70 x (2.1² / 3 )  + .7 x 2.1²

= 5.439 + 3.087

= 8.526 kg m²

c )

In this case moment of inertia of the combination = 0 as distance of masses from given axis is zero .

d )

masses = 3.7 + .7 = 4.4 kg

distance from axis = .5 m  

moment of inertia about given axis

= 4.4 x .5²

= 1.1 kg m².

Explanation:

gravitional force attracts objects with mass toward each other.

Answer:

a

   [tex]\lambda = 1.667 nm[/tex]

b

     [tex]\theta = 0.8681^o[/tex]

Explanation:

From the question we are told that

   The distance of separation is [tex]d = 0.220 \ mm = 0.00022 \ m[/tex]

    The  is distance of the screen from the slit is  [tex]D = 2.60 \ m[/tex]

    The distance between the central bright fringe and either of the adjacent bright   [tex]y = 1.97 cm = 1.97 *10^{-2}\ m[/tex]

Generally  the condition for constructive interference is  

      [tex]d sin \tha(\theta ) = n \lambda[/tex]

From the question we are told that small-angle approximation is valid here.

So    [tex]sin (\theta ) = \theta[/tex]

=>        [tex]d \theta = n \lambda[/tex]

=>        [tex]\theta = \frac{n * \lambda }{d }[/tex]

Here n is the order of maxima and the value is  n =  1 because we are considering the central bright fringe and either of the adjacent bright fringes

Generally the distance between the central bright fringe and either of the adjacent bright  is mathematically represented as

         [tex]y = D * sin (\theta )[/tex]

From the question we are told that small-angle approximation is valid here.

So

       [tex]y = D * \theta[/tex]

=>   [tex]\theta = \frac{ y}{D}[/tex]

So

     [tex]\frac{n * \lambda }{d } = \frac{y}{D}[/tex]

     [tex]\lambda =\frac{d * y }{n * D}[/tex]

substituting values

       [tex]\lambda = \frac{0.00022 * 1.97*10^{-2} }{1 * 2.60 }[/tex]

        [tex]\lambda = 1.667 *10^{-6}[/tex]

        [tex]\lambda = 1.667 nm[/tex]

In the b part of the question we are considering the next set of bright fringe so  n=  2

    Hence

     [tex]dsin (\theta ) = n \lambda[/tex]

    [tex]\theta = sin^{-1}[\frac{ n * \lambda }{d} ][/tex]

    [tex]\theta = sin^{-1}[\frac{ 2 * 1667 *10^{-9}}{ 0.00022} ][/tex]

    [tex]\theta = 0.8681^o[/tex]

Answer:

The width of the slit is 0.4 mm (0.00040 m).

Explanation:

From the Young's interference expression, we have;

(λ ÷ d) = (Δy ÷ D)

where λ is the wavelength of the light, D is the distance of the slit to the screen, d is the width of slit and Δy is the fringe separation.

Thus,

d = (Dλ) ÷ Δy

D = 3.30 m, Δy = 4.7 mm (0.0047 m) and λ = 563 nm (563 ×[tex]10^{-9}[/tex] m)

d = (3.30 × 563 ×[tex]10^{-9}[/tex] ) ÷ (0.0047)

  = 1.8579 × [tex]10^{-6}[/tex] ÷ 0.0047

  = 0.0003951 m

d = 0.00040 m

The width of the slit is 0.4 mm (0.00040 m).

Answer:

The tension on string when the speed was raised is 134.53 N

Explanation:

Given;

Tension on the string, T = 120 N

initial speed of the transverse wave, v₁ = 170 m/s

final speed of the transverse wave, v₂ = 180 m/s

The speed of the wave is given as;

[tex]v = \sqrt{\frac{T}{\mu} }[/tex]

where;

μ is mass per unit length

[tex]v^2 = \frac{T}{\mu} \\\\\mu = \frac{T}{v^2} \\\\\frac{T_1}{v_1^2} = \frac{T_2}{v_2^2}[/tex]

The final tension T₂ will be calculated as;

[tex]T_2 = \frac{T_1 v_2^2}{v_1^2} \\\\T_2 = \frac{120*180^2}{170^2} \\\\T_2 = 134.53 \ N[/tex]

Therefore, the tension on string when the speed was raised is 134.53 N

Answer:

Explanation:

The equation for a traveling wave as expressed as y(x, t) = A cos(kx − [tex]\omega[/tex]t) where An is the amplitude f oscillation, [tex]\omega[/tex] is the angular velocity and x is the horizontal displacement and y is the vertical displacement.

From the formula; [tex]k =\frac{2\pi x}{\lambda} \ and \ \omega = 2 \pi f[/tex] where;

[tex]\lambda \ is\ the \ wavelength \ and\ f \ is\ the\ frequency[/tex]

Before we can get the transverse speed, we need to get the frequency and the wavelength.

frequency = 1/period

Given period = 2/15 s

Frequency = [tex]\frac{1}{(2/15)}[/tex]

frequency = 1 * 15/2

frequency f = 15/2 Hertz

Given wavelength [tex]\lambda[/tex] = 2m

Transverse speed [tex]v = f \lambda[/tex]

[tex]v = 15/2 * 2\\\\v = 30/2\\\\v = 15m/s[/tex]

Hence, the transverse speed at that point is  15m/s

Answer:

Explanation:

Find:

Computation:

F = (9/5)C + 32

3 times that of the Celsius

If C = x

So F = 3x

So,

3x = (9/5)x + 32

15x = 9x +160

6x = 160

x = 26.67

So, C = 26.67° and F = 80°

1/5 times that of the Celsius

If C = x

So F = x/5

So,

x/5 = (9/5)x + 32

x = 9x + 160

x = -20

So, C = -20° and F = -4°

Answer:

a)   I = I₀/2, b)  I = I₀/2 cos² θ

Explanation:

To answer these questions, let's analyze a little the way of working of a polarized

* When a non-polarized light hits a polarizer, the electric field that is not in the direction of the polarizer is absorbed, so the transmitted light is

          i = I₀ / 2

and is polarized in the direction of the polarizer

* when a polarized light reaches the analyzer it must comply with Malus's law

          I = I₁ cos² θ

where the angle is between the polarized light and the analyzer.

With this, let's answer the questions

a) When a polarizer is placed in the non-polarized light path, half of it is absorbed and only the light that has polarization in the direction of the polarizer is transmitted with an intensity of

                  I = I₀/2

b) when a polarizer and an analyzer are fitted, the intensity of the light transmitted by the analyzer is

                I = I₀/2 cos² θ

where the final value depends on the angle between the polarizer and the analyzer.

Let's look at two extreme cases

θ = 0          I = Io / 2

θ = 90º      I = 0

Answer:

This is because the same electron removed from the positively charged plate is what is taken to the negatively charged plate, maintaining the same amount of electron according to the conservation of charge in an electric circuit.

Explanation:

In any circuit, electrons are neither created nor destroyed according to the laws of conservation of charge, but are transferred from one point to another on the circuit. When the plates of a capacitor are connected to battery, the battery pushes the electron to move due to its potential difference. Electrons are then moved from the positive plate, at a steady rate, to the negative plate. The removal of electrons from the positive plate is what leaves it positively charged from deficiency of electrons, and the addition of electrons at the negatively charged plate is what leaves the plate negatively charge from excess of electrons. From this, we can see that the same electrons removed from the positively charged plate are  taken to the negatively charged plate.

Answer:

(a) 10 m/s

(b) 22.4 m/s

Explanation:

(a) Draw a free body diagram of the car when it is at the top of the loop.  There are two forces: weight force mg pulling down, and normal force N pushing down.

Sum of forces in the centripetal direction (towards the center):

∑F = ma

mg + N = mv²/r

At minimum speed, the normal force is 0.

mg = mv²/r

g = v²/r

v = √(gr)

v = √(10 m/s² × 10.0 m)

v = 10 m/s

(b) Energy is conserved.

Initial kinetic energy + initial potential energy = final kinetic energy

½ mv₀² + mgh = ½ mv²

v₀² + 2gh = v²

(10 m/s)² + 2 (10 m/s²) (20.0 m) = v²

v = 22.4 m/s

Explanation:

unit is necessary to communicate values of the physical quantity for example can main to someone a particular length without using some sort of unit is impossible because a length cannot be described without a reference used to make sense of the value given

Answer:

changing the magnetic field more rapidly

Explanation:

According to Faraday's law, whenever there is a change in the magnetic lines of force, it leads the production of induced emf. The magnitude of induced emf is proportional to to the rate of change of flux.

Hence if the magnetic field inside a loop of wire is changed rapidly, the magnitude of induced emf increases in accordance with Faraday's law of electromagnetic induction stated above when the magnetic field is changed more rapidly, hence the answer.

Answer:

The value is  [tex]\mu = 0.76[/tex]

Explanation:

From the question we are told that

    The  acceleration is [tex]a = 7.4 \ m /s^2[/tex]

Generally the force by which the truck bed (truck) is moving with is mathematically represented as

          [tex]F = ma[/tex]

Now for the truck cabinet to slip from the truck bed then the frictional force between the truck cabinet  is equal the force by which the the truck bed is moving with that is  

        [tex]F_f = F[/tex]

Here  [tex]F_f[/tex] is the frictional force which is mathematically represented as

         [tex]F_f = \mu * m * g[/tex]

substituting into above equation

         [tex]\mu * m * g = ma[/tex]

=>        [tex]\mu = \frac{a}{g}[/tex]

substituting values

           [tex]\mu = \frac{ 7.4 }{ 9.8}[/tex]

           [tex]\mu = 0.76[/tex]

Answer:

Increase the distance by a factor of 6.

Explanation:

The intensity at a distance r is given by :

[tex]I=\dfrac{P}{4\pi r^2}[/tex]

Here,

P is power emitted

r is distance from source

It means that the intensity is inversely proportional to the distance from the source.

To decrease the intensity of the sound you are hearing from your speaker system by a factor of 36, we can increase the distance by a factor of 6. Hence, this is the required solution.

Answer:

The power delivered by the battery is 17.04 W

Explanation:

Power through a circuit is given as

P = IV    ....1

where P is the power

I is the current through the circuit

V is the voltage through the circuit

The voltage in a circuit is given as

V = IR    ....2

Let us take the value of each resistor as equal to R

when connected in series, the total resistance will be

[tex]R_{t}[/tex] = R + R = 2R

If we assume constant voltage through the circuit, then from equation 2, the current in this case is

I = V/2R

If the resistors are connected in parallel, then the total resistance will be

[tex]\frac{1}{R_{t} }[/tex] = [tex]\frac{1}{R}[/tex] +

[tex]R_{t}[/tex] = R/2

The current in this case will be increased since the resistance is reduced

I = 2V/R

comparing the two situations, we can see that the current increased when connected in parallel to a ratio of

[tex]\frac{2V}{R}[/tex] ÷  [tex]\frac{V}{2R}[/tex] =  

This means that the current increased 4 times

From equation 1, we can see that electrical power is proportional to the current at a constant voltage, therefore, the power will also increase by four times to

P = 4 x 4.26 = 17.04 W

Answer:

Explanation:

[tex]Distance = 535m\\Time = 17.3s\\\\Velocity = \frac{Distane}{Time} \\\\V = \frac{535m}{17.3s} \\\\Velocity = 30.92m/s[/tex]

[tex]Distance = 535m\\\\535m \:to \: km=0.535km\\\\Time = 17.3s\\\\17.3s = 0.004805556hours\\\\Velocity = \frac{Distance}{Time}\\\\ V= \frac{0.535}{0.004805556} \\\\ V=111.329469472\\\\=111.33km/h[/tex]

Answer:

t = 134834.31 years

Explanation:

First we find the speed of the ship:

v = 0.890 c

where,

v = speed of the ship = ?

c = speed of light = 3 x 10⁸ m/s

Therefore, using the values, we get:

v = (0.89)(3 x 10⁸ m/s)

v = 2.67 x 10⁸ m/s

Now, we find the distance in meters:

Distance = s = (1.2 x 10⁵ light years)(9.461 x 10¹⁵/1 light year)

s = 11.35 x 10²⁰ m

Now, for the time we use the following equation:

s = vt

t = s/v

t = (11.35 x 10²⁰ m)/(2.67 x 10⁸ m/s)

t = (4.25 x 10¹² s)(1 h/3600 s)(1 day/24 h)(1 year/365 days)

t = 134834.31 years

Answer:

The fumes stop getting sucked up by the fume hood once the beaker is pulled out of the hood.

Answer:

Capacitor, is the right answer.

Explanation:

The unknown element is a Capacitor.

Below is the calculation that proves that it is a capacitor.

We know that for the Capacitor

i = Imax × sin(wt+(pi/2)).

i = Imax × sin ((2 × pi/T) × (T/4) + (pi/2))

i = Imax × sin(3.142) = 0 A

at, t = T/2

wt = (2 × pi/T) × (T/2) = pi

wt + (pi/2) = pi + (pi/2) = ( 3 × pi/2) =

i = Imax × sin(3 × pi/2) = -Imax

Which is in a correct agreement with capacitor  therefore, the answer is a Capacitor.

Answer:

I think 9.5

Explanation:

............

Answer:

a) the light is close to normal therefore the reference incidence of medium 1 is less than medium n2 where the ray is transmitted.

b) The ray is far from normal in this case the refractive index of medium 1 is greater than index of medium 2

Explanation:

The expression for the angle of refraction is

          n₁ sin θ₁ = n₂ sin θ₂

refractive index n₁ is for incident light and n₂ is for transmitted light.

We have two cases

a) the light is close to normal therefore the reference incidence of medium 1 is less than medium n2 where the ray is transmitted.

b) The ray is far from normal in this case the refractive index of medium 1 is greater than index of medium 2

Answer:

The induced emf is [tex]\epsilon = 0.0280 \ V[/tex]

Explanation:

From the question we are told

    The diameter of the loop is  [tex]d = 6.7 cm = 0.067 \ m[/tex]

    The magnetic field is  [tex]B = 1.27 \ T[/tex]

    The time taken is  [tex]dt = 0.16 \ s[/tex]

Generally the induced emf is mathematically represented as

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

Where  N =  1 given that it is only a circular loop

            [tex]\Delta \phi = \Delta B * A[/tex]

Where  [tex]\Delta B = B_f - B_i[/tex]

   where [tex]B_i[/tex] is  1.27 T  given that the loop of wire was initially in the magnetic field

    and  [tex]B_f[/tex] is  0 T given that the loop was removed from the magnetic field

Now the area of the of the loop is evaluated as

          [tex]A = \pi r^2[/tex]

Where r is the radius which is mathematically represented as

       [tex]r = \frac{d}{2}[/tex]

substituting values

       [tex]r = \frac{0.067}{2}[/tex]

        [tex]r = 0.0335 \ m[/tex]

So

         [tex]A = 3.142 * (0.0335)^2[/tex]

          [tex]A = 0.00353 \ m^2[/tex]

So

      [tex]\Delta \phi = (0- 127)* (0.00353)[/tex]

      [tex]\Delta \phi = -0.00448 Weber[/tex]

=>    [tex]\epsilon = - 1 * \frac{-0.00448}{0.16}[/tex]

=>   [tex]\epsilon = 0.0280 \ V[/tex]

Answer:

6.13 seconds

Explanation:

At the peak of the fireworks trajectory, the velocity of the firework would be zero. Using equation of motion, we have:

v² = u² + 2gh

0 = 40² - (2)(9.81)(h)

0 = 1600 - 19.62h

19.62h = 1600

h = 1600/19.62

h = 81.55 m

Now during the process of explosion, the two parts gained equal vertical momentum but in opposite directions.

We are told the first piece lands in a time of 2.71 s,

Using 3rd equation of motion, we have;

h = ut + ½gt²

81.55 = u(2.71) + ½(9.81 × 2.71²)

81.55 = 2.71u + 36.0228

2.71u = 81.55 - 36.0228

2.71u = 45.5272

u = 45.5272/2.71

u = 16.8 m/s

The time it takes a projectile to return back to its original launch point assuming the projectile was launched

vertically with speed u = 16.8 m/s is;

t = 2u/g

t = (2 × 16.8)/9.81

t = 3.43 s

Thus total time it takes the second mass to reach the ground = 3.43 + 2.71 = 6.13 seconds

Answer:

When drugs are taken in are body are brain release dopamine: which make us feel so pleasure and good, and for this some people are addicted to drugs which makes them feel good. on other hand damaging their health.

Answer

82.3 wavelengths

Answer:

The blade undergoes 40 revolutions, so neither of the given options is correct!

Explanation:

The revolutions can be found using the following equation:

[tex]\theta_{f} = \theta_{i} + \omega_{i}*t + \frac{1}{2}\alpha*t^{2}[/tex]

Where:

α is the angular acceleration

t is the time = 2.5 s

[tex]\omega_{i}[/tex] is the initial angular velocity = 1500 rev/min                

First, we need to find the angular acceleration:

[tex] \alpha = \frac{\omega_{f} - \omega_{i}}{t} = \frac{400 rev/min*2\pi rad*1 min/60 s - 1500 rev/min *2\pi rad*1 min/60 s}{2.5 s} = -46.08 rad/s^{2} [/tex]

Now, the revolutions that the blade undergo are:

[tex]\theta_{f} - \theta_{i} = \omega_{i}*t + \frac{1}{2}\alpha*t^{2}[/tex]

[tex]\Delta \theta = 1500 rev/min *2\pi rad*1 min/60 s*2.5 s - \frac{1}{2}*(46.08 rad/s^{2})*(2.5)^{2} = 248.7 rad = 39.9 rev[/tex]        

Therefore, the blade undergoes 40 revolutions, so neither of the given options is correct!

I hope it helps you!                              

Answer:

the atom had a dense, positively charged nucleus.​

Explanation:

Ernest Rutherford, based on the experiment carried out by two of his graduate students, established the authenticity of the nuclear model of the atom.

According to the nuclear model, an atom is made up of a dense positive core called the nucleus. Electrons are found to move round this nucleus in orbits. This is akin to the movement of the planets round the sun in the solar system.

Answer:

B. If the container is cooled, the gas particles will lose kinetic energy and temperature will decrease.

C. If the gas particles move more quickly, they will collide more frequently with the walls of the container and pressure will increase.

E. If the gas particles move more quickly, they will collide with the walls of the container more often and with more force, and pressure will increase.

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