A block is released from the top of a frictionless incline plane as pictured above. If the total distance travelled by the block is 1.2 m to get to the bottom, calculate how fast it is moving at the bottom using Conservation of Energy.

Answer:

the high luminosity of an active galactic nucleus primarily consists of light emitted by hot gas in an accretion disk that swirls around the black hole

Answer:

a) q = 4.47 10⁻⁵ C

b)     ΔV = 4.47 10⁴ V

Explanation:

A Leyden bottle works as a condenser that accumulates electrical charge, so we can use the formula of the energy stored in a capacitor

           U = Q² / 2C

         Q = √ (2UC)

let's reduce the magnitudes to the SI system

   c = 1 nF = 1 10⁻⁹ F

let's calculate

         q = √ (2 1 10⁻⁹-9)

         q = 0.447 10⁻⁴ C

         q = 4.47 10⁻⁵ C

b) for the potential difference we use

             C = Q / ΔV

            ΔV = Q / C

            ΔV = 4.47 10⁻⁵ / 1 10⁻⁹

            ΔV = 4.47 10⁴ V

For a wet shirt is put on a clothesline to dry on a sunny day, water molecules gain heat and evaporate.

When a clothe is placed on a line to dry, the idea is to ensure that the water molecules should evaporate.

For the water molecules to evaporate, they must gain more energy that will enable them to transit from liquid to gaseous state.

Recall that he change from liquid to vapor requires energy, this is why water molecules gain energy when they evaporate.

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Answer:

Explanation:

The distance of middle point from centres of spheres will be as follows

From each of 2 cm diameter sphere

R  = 1 + 6.7 / 2 = 4.35 cm = 4.35 x 10⁻² m

Expression for electric field = Q / 4πε R²

Electric field due to positive charge

E₁ = 70  x 10⁻⁹ x 9 x 10⁹ / 4.35² x 10⁻⁴

= 33.3 x 10⁴ N/C

Electric field due to negative  charge

E₂ = 40  x 10⁻⁹ x 9 x 10⁹ / 4.35² x 10⁻⁴

= 19.02 x 10⁴ N/C

E₁ and E₂ act in the same direction so

Total field = (33.3 + 19.02 ) x 10⁴

= 52.32 x 10⁴ N/C .

Answer:

a

 [tex]\theta = 0.0022 rad[/tex]

b

 [tex]I = 0.000304 I_o[/tex]

Explanation:

From the question we are told that  

   The  wavelength of the light is [tex]\lambda = 550 \ nm = 550 *10^{-9} \ m[/tex]

    The  distance of the slit separation is  [tex]d = 0.500 \ mm = 5.0 *10^{-4} \ m[/tex]

Generally the condition for two slit interference  is  

     [tex]dsin \theta = m \lambda[/tex]

Where m is the order which is given from the question as  m = 2

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

 substituting values  

      [tex]\theta = 0.0022 rad[/tex]

Now on the second question  

   The distance of separation of the slit is  

       [tex]d = 0.300 \ mm = 3.0 *10^{-4} \ m[/tex]

The  intensity at the  the angular position in part "a" is mathematically evaluated as

      [tex]I = I_o [\frac{sin \beta}{\beta} ]^2[/tex]

Where  [tex]\beta[/tex] is mathematically evaluated as

       [tex]\beta = \frac{\pi * d * sin(\theta )}{\lambda }[/tex]

  substituting values

     [tex]\beta = \frac{3.142 * 3*10^{-4} * sin(0.0022 )}{550 *10^{-9} }[/tex]

    [tex]\beta = 0.06581[/tex]

So the intensity is  

    [tex]I = I_o [\frac{sin (0.06581)}{0.06581} ]^2[/tex]

   [tex]I = 0.000304 I_o[/tex]

Answer:

Option C - 39.2 J

Explanation:

We are given that;

Mass; m = 2 kg.

Distance moved off the floor;d = 10 m.

Acceleration due to gravity;g = 9.8 m/s².

We want to find the work done.

Now, the Formula for work done is given by;

Work = Force × displacement.

In this case, it's force of gravity to lift up the boots, thus;

Formula for this force is;

Force = mass x acceleration due to gravity

Force = 2 × 9.8 = 19.2 N

∴ Work done = 19.6 × 2

Work done = 39.2 J.

Hence, the Work done to life the boot of 2 kg to a height of 2 m is 39.2 J.

Answer:39.2J

Explanation: I just answered this question and this was the correct answer. 4J is the wrong answer.

Answer:

Final pressure 0.105atm

Explanation:

Let V1 represent the initial volume of dry air at NTP.

under adiabatic condition: no heat is lost or  gained by the system. This does not implies that the constant temperature throughout the system , but rather that no heat gained or loss by the system.

Adiabatic expansion:

[tex]\frac{T_1}{T_2} =(\frac{V_1}{V_2} )^{\gamma -1}[/tex]

273/T2=(5V1/V1)^(1.4−1)

273/T2=5^0.4

Final temperature  T2=143.41 K

Also

P1/P2=(V2/V1)^γ

1/P2=(5V1/V1)^1.4

Final pressure P2=0.105atm

Answer:

n_glass = 1.404

Explanation:

In order to calculate the index of refraction of the light you use the Snell's law, which is given by the following formula:

[tex]n_1sin\theta_1=n_2sin\theta_2[/tex]         (1)

n1: index of refraction of vacuum = 1.00

θ1: angle of the incident light respect to normal of the surface = 38.0°

n2: index of refraction of glass = ?

θ2: angle of the refracted light in the glass respect to normal = 26.0°

You solve the equation (1) for n2 and replace the values of all parameters:

[tex]n_2=n_1\frac{sin\theta_1}{sin\theta_2}=(1.00)\frac{sin(38.0\°)}{sin(26.0\°)}\\\\n_2=1.404[/tex]

The index of refraction of the glass is 1.404

Answer:

The charge stored in the capacitor will stay the same. However, the electric potential across the two plates will increase. (Assuming that the permittivity of the space between the two plates stays the same.)

Explanation:

The two plates of this capacitor are no longer connected to each other. As a result, there's no way for the charge on one plate to move to the other. [tex]Q[/tex], the amount of charge stored in this capacitor, will stay the same.

The formula [tex]\displaystyle Q = C\, V[/tex] relates the electric potential across a capacitor to:

While [tex]Q[/tex] stays the same, moving the two plates apart could affect the potential [tex]V[/tex] by changing the capacitance [tex]C[/tex] of this capacitor. The formula for the capacitance of a parallel-plate capacitor is:

[tex]\displaystyle C = \frac{\epsilon\, A}{d}[/tex],

where

Assume that the two plates are separated with vacuum. Moving the two plates apart will not affect the value of [tex]\epsilon[/tex]. Neither will that change the area of the two plates.

However, as [tex]d[/tex] (the distance between the two plates) increases, the value of [tex]\displaystyle C = \frac{\epsilon\, A}{d}[/tex] will become smaller. In other words, moving the two plates of a parallel-plate capacitor apart would reduce its capacitance.

On the other hand, the formula [tex]\displaystyle Q = C\, V[/tex] can be rewritten as:

[tex]V = \displaystyle \frac{Q}{C}[/tex].

The value of [tex]Q[/tex] (charge stored in this capacitor) stays the same. As the value of [tex]C[/tex] becomes smaller, the value of the fraction will become larger. Hence, the electric potential across this capacitor will become larger as the two plates are moved away from one another.  

Answer:

D). Uranus.

Explanation:

Jovian planets are described as the planets which are giant balls of gases and located farthest from the sun which primarily include Jupiter, Saturn, Uranus, and Neptune.

As per the question, 'Uranus' is the jovian planet that would have the most extreme seasonal changes as its tilted axis leads each season to last for about 1/4 part of its 84 years orbit. The strong tilted axis encourages extreme changes in the season on Uranus. Thus, option D is the correct answer.

Answer:

7.2N/C

Explanation:

Pls see attached file

Answer:

Yes the monkey will get hit and it will not avoid the dart.

Explanation:

Yes, the monkey will be hit anyway because the dart will follow a hyperbolic path and and will thus fall below the branches, so if the monkey jumps it will be hit.

No, the monkey will not avoid the dart because dart velocity doesn't matter. The speed of the bullet doesn’t even matter in this case because a faster bullet will hit the monkey at a higher height and while a slower bullet will simply hit the monkey closer to the ground.

. . . is carrying more energy than a wave in the same medium with a lower amplitude.

Explanation:

The Earth has a solid inner core

Answer:

a) Only the first root is physically reasonable. Therefore, both stones hit the water in 2.866 seconds, b) The initial velocity of the second stone is -16.038 meters per second, c) The speed of the first stone is 30.227 meters per second and the speed of the second stone is 34.338 meters per second.

Explanation:

a) The time after the release after the release of the first stone can be get from the following kinematic formula for the first rock:

[tex]y_{1} = y_{1,o} + v_{1,o} \cdot t +\frac{1}{2}\cdot g \cdot t^{2}[/tex]

Where:

[tex]y_{1}[/tex] - Final height of the first stone, measured in meters.

[tex]y_{1,o}[/tex] - Initial height of the first stone, measured in meters.

[tex]v_{1,o}[/tex] - Initial speed of the first stone, measured in meters per second.

[tex]t[/tex] - Time, measured in seconds.

[tex]g[/tex] - Gravity constant, measured in meters per square second.

Given that [tex]y_{1,o} = 47\,m[/tex], [tex]y_{1} = 0\,m[/tex], [tex]v_{1,o} = -2.12\,\frac{m}{s}[/tex] and [tex]g = -9.807\,\frac{m}{s^{2}}[/tex], the following second-order polynomial is built:

[tex]-4.984\cdot t^{2} - 2.12\cdot t + 47 = 0[/tex]

Roots of the polynomial are, respectively:

[tex]t_{1} \approx 2.866\,s[/tex] and [tex]t_{2}\approx -3.291\,s[/tex]

Only the first root is physically reasonable. Therefore, both stones hit the water in 2.866 seconds.

b) As the second stone is thrown a second later than first one, its height is represented by the following kinematic expression:

[tex]y_{2} = y_{2,o} + v_{2,o}\cdot (t-t_{o}) + \frac{1}{2}\cdot g \cdot (t-t_{o})^{2}[/tex]

[tex]y_{2}[/tex] - Final height of the second stone, measured in meters.

[tex]y_{2,o}[/tex] - Initial height of the second stone, measured in meters.

[tex]v_{2,o}[/tex] - Initial speed of the second stone, measured in meters per second.

[tex]t[/tex] - Time, measured in seconds.

[tex]t_{o}[/tex] - Initial absolute time, measured in seconds.

[tex]g[/tex] - Gravity constant, measured in meters per square second.

Given that [tex]y_{2,o} = 47\,m[/tex], [tex]y_{2} = 0\,m[/tex], [tex]t_{o} = 1\,s[/tex], [tex]t = 2.866\,s[/tex] and [tex]g = -9.807\,\frac{m}{s^{2}}[/tex], the following expression is constructed and the initial speed of the second stone is:

[tex]1.866\cdot v_{2,o}+29.926 = 0[/tex]

[tex]v_{2,o} = -16.038\,\frac{m}{s}[/tex]

The initial velocity of the second stone is -16.038 meters per second.

c) The final speed of each stone is determined by the following expressions:

First stone

[tex]v_{1} = v_{1,o} + g \cdot t[/tex]

Second stone

[tex]v_{2} = v_{2,o} + g\cdot (t-t_{o})[/tex]

Where:

[tex]v_{1,o}, v_{1}[/tex] - Initial and final velocities of the first stone, measured in meters per second.

[tex]v_{2,o}, v_{2}[/tex] - Initial and final velocities of the second stone, measured in meters per second.

If [tex]v_{1,o} = -2.12\,\frac{m}{s}[/tex] and [tex]v_{2,o} = -16.038\,\frac{m}{s}[/tex], the final speeds of both stones are:

First stone

[tex]v_{1} = -2.12\,\frac{m}{s} + \left(-9.807\,\frac{m}{s^{2}} \right)\cdot (2.866\,s)[/tex]

[tex]v_{1} = -30.227\,\frac{m}{s}[/tex]

Second stone

[tex]v_{2} = -16.038\,\frac{m}{s} + \left(-9.807\,\frac{m}{s^{2}} \right) \cdot (2.866\,s-1\,s)[/tex]

[tex]v_{2} = -34.338\,\frac{m}{s}[/tex]

The speed of the first stone is 30.227 meters per second and the speed of the second stone is 34.338 meters per second.

The Tension of the string is going to be less when submerged in water by a value called the buoyancy force, so below in the attached file is explanation on how to calculate the apparent weight and density of the submerged object

Explanation:

KE = q

½ mv² = mCΔT

ΔT = v² / (2C)

ΔT = (200 m/s)² / (2 × 236 J/kg/°C)

ΔT = 84.7°C

This question involves the concepts of the law of conservation of energy.

The temperature change of the bullet is "84.38°C".

According to the law of conservation of energy, total energy of the system must remain constant. Therefore, in this situation.

[tex]Kinetic\ energy\ of\ bullet\ before\ impact=heat\ absorbed\ in\ bullet\\\\\frac{1}{2}mv^2=mC\Delta T\\\\\Delta T = \frac{v^2}{2C}[/tex]

where,

Therefore,

[tex]\Delta T = \frac{(200\ m/s)^2}{2(237\ J/kg.^oC)}[/tex]

ΔT = 84.38°C

Learn more about the law of conservation of energy here:

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Answer:

Net nonzero work is done if the final position of the particle is on options A, C and D

Explanation:

non zero work is done if following will be the final position of the charges :

A) Any point on the line through the charges of the dipole , excluding the midpoint between the two charges.

C) A line that makes an angle 30° with the dipole moment.

D) A line that makes an angle 45°  with the dipole moment.

Answer:

If a key is pressed on a piano, the frequency of the resulting sound will determine the ___PITCH_____, and the amplitude will determine the _____LOUDNESS___ of the perceived musical note.

Explanation:

The frequency of a vibrating string is primarily based on three factors:

The sounding length (longer is lower, shorter is higher)

The tension on the string (more tension is higher, less is lower)

The mass of the string, normally based on a uniform density per unit length (higher mass is lower, lower mass is higher)

To make a shorter string (such as in an upright piano) sound the same fundamental frequency as a longer string (such as in a 9' grand piano), either the thickness of the string must be increased (which increases the density and the mass) or the tension must be decreased, and usually it's a bit of both.

Thicker strings are often stiffer and that creates more inharmonic partials, and lower tension is associated with other problems, so the best way to make a string sound lower is the make it longer, but it is not practical to make a piano from strings that are all the same density and tension, because the lowest strings would have to be ridiculously long. Nine feet is already a great demand on space for a single musical instrument, and of course those pianos are extremely expensive and difficult to move.

And alsoBesides the pitch of a musical note, perhaps the most noticeable feature in how loud the note is. The loudness of a sound wave is determined from its amplitude. While loudness is only associated with sound waves, all types of waves have an amplitude. Waves on a calm ocean may be less than 1 foot high. Good surfing waves might be 10 feet or more in amplitude. During a storm the amplitude might increase to 40 or 50 feet.

Many things can influence the amplitude.

What is producing the sound?

How far are you from the source of the sound? The farther away the smaller the amplitude.

Intervening material. Sound does not travel through walls as well as air.

Depends on what is detecting the wave sound. Ear vs. microphone.

Answer:

The frequency will determine the pitch

the amplitude will determine the loudness

Explanation:

The frequency of a sound refers to the number of vibrations made by the sound wave produced in a unit of time. This usually affects how high or how low a note is perceived in music. High-frequency sounds have higher pitches, while low-frequency sounds have lower pitches.

The amplitude of a sound wave refers to the height between the wave crests and the equilibrium line in a sound wave. It shows how loud a sound will be. High amplitude sounds are loud while low amplitude sounds are quiet.

Answer:

P' = 4 P

Therefore, the power dissipated by the circuit will becomes four times of its initial value.

Explanation:

The power dissipation by an electrical circuit is given by the following formula:

Power Dissipation = (Voltage)(Current)

P = VI

but, from Ohm's Law, we know that:

Voltage = (Current)(Resistance)

V = IR

Substituting this in formula of power:

P = (IR)(I)

P = I²R   ---------------- equation 1

Now, if we double the current , then the power dissipated by that circuit will be:

P' = I'²R

where,

I' = 2 I

Therefore,

P' = (2 I)²R

P' = 4 I²R

using equation 1

P' = 4 P

Therefore, the power dissipated by the circuit will becomes four times of its initial value.

Answer:

Dead lifting uses tho muscle fundamentals

Explanation:

Answer:

Fast twitch

Explanation:

Edmentum

Explanation:

As per Rayleigh criterion, the angular resolution is given as follows:

[tex]\theta=\frac{1.22 \lambda}{D}[/tex]

From this expression larger the size of aperture, smaller will be the value of angular resolution and hence, better will be the device i.e. precision for distinguishing two points at very high angular difference is higher.

Answer:

The acceleration of the crate is [tex]0.3356\,\frac{m}{s^2}[/tex]

Explanation:

Recall the formula that relates force,mass and acceleration from newton's second law;

[tex]F=m\,a[/tex]

Then in our case, we know the force applied and we know the mass of the crate, so we can solve for the acceleration as shown below:

[tex]F=m\,a\\75.5\,N=225\,\,kg\,\,a\\a=\frac{75.5}{225} \,\frac{m}{s^2} \\a=0.3356\,\,\frac{m}{s^2}[/tex]

sorry but I don't understand

Answer:

C - Arteries grow smaller

Explanation:

The option choices are:

A. Faster post-exercise recovery time

B. Lungs expand more easily

C. Arteries grow smaller

D. Diaphragm grows stronger

Explanation:

There are many advantages of cardiorespiratory fitness. It can decrease the risk of heart disease, lung cancer, type 2 diabetes, stroke, and other diseases. Cardiorespiratory health helps develop lung and heart conditions and enhances feelings of well-being.

Answer:

The magnitude of the current in the wire is 5.5A, and the direction of the current is in the positive y direction.

Explanation:

- To find the direction of the conventional current in the wire you use the following formula:

[tex]\vec{F}=i\vec{l}\ X\ \vec{B}[/tex]       (1)

i: current in the wire = ?

F: magnitude of the magnetic force on the wire = 15.1N

B: magnitude of the magnetic field = 6.1T

l: length of the wire that is affected by the magnetic field = 0.45m

The direction of the magnetic force is in the x direction (+^i) and the direction of the magnetic field is in the +z direction (+^k).

The direction of the current must be in the +y direction (+^j). In fact, you have:

^j X ^k = ^i

The current and the magnetic field are perpendicular between them, then, you solve for i in the equation (1):

[tex]F=ilBsin90\°\\\\i=\frac{F}{lB}=\frac{15.1N}{(0.45m)(6.1T)}=5.5A[/tex]

The magnitude of the current in the wire is 5.5A, and the direction of the current is in the positive y direction.

2Q

When a capacitor carries some certain charge, the energy stored in the capacitor is its electric potential energy E. The magnitude of this potential energy is given by;

E  = [tex]\frac{1}{2}qV[/tex]            ------------(i)

Where;

q = charge between the plates of the capacitor

V = potential difference between the plates of the capacitor

From the question;

q = Q

E = Ep

Therefore, equation (i) becomes;

Ep = [tex]\frac{1}{2} QV[/tex]              ----------------(ii)

Make V subject of the formula in equation (ii)

V = [tex]\frac{2E_{p}}{Q}[/tex]

Now, when the energy is doubled i.e E = 2Ep, equation (i) becomes;

2Ep = [tex]\frac{1}{2}qV[/tex]

Substitute the value of V into the equation above;

2Ep = [tex]\frac{1}{2}[/tex]([tex]q *\frac{2E_{p}}{Q}[/tex])

Solve for q;

[tex]2E_{p}[/tex] = [tex]\frac{2qE_p}{2Q}[/tex]

[tex]2E_{p}[/tex] = [tex]\frac{qE_p}{Q}[/tex]

[tex]q = 2Q[/tex]

Therefore, the charge, when the energy stored is twice the originally stored energy, is twice the original charge. i.e 2Q

The question is not complete, the value of z is not given.

Assuming the value of z = 4.0m

Answer:

the magnitude of the electric field at any point having z(4.0 m)  =

E = 5.65 N/C

Explanation:

given

σ(surface density) = 0.20 nC/m² = 0.20 × 10⁻⁹C/m²

z = 4.0 m

Recall

E =F/q (coulumb's law)

E = kQ/r²

σ = Q/A

A = 4πr²

∴ The electric field at point z =

E = σ/zε₀

E = 0.20 × 10⁻⁹C/m²/(4 × 8.85 × 10⁻¹²C²/N.m²)

E = 5.65 N/C

Answer:

The  tension on the rope  is  T  =  900 N

Explanation:

From the question we are told that  

     The mass of the person on the left is  [tex]m_l = 100 \ kg[/tex]

      The force of the person on the left is  [tex]F_l = 1000 \ N[/tex]

       The mass of the person on the right  is  [tex]m_r = 70 \ kg[/tex]

       The force of the person on the right is  [tex]F_r = 830 \ N[/tex]

Generally the net force is  mathematically represented as

         [tex]F_{Net} = F_l - F_r[/tex]

substituting  values

        [tex]F_{Net} = 1000-830[/tex]

       [tex]F_{Net} = 170 \ N[/tex]

Now the acceleration net acceleration of the rope is mathematically evaluated as

        [tex]a = \frac{F_{net}}{m_I + m_r }[/tex]

substituting  values

     [tex]a = \frac{170}{100 + 70 }[/tex]

     [tex]a = 1 \ m/s ^2[/tex]

The  force [tex]m_i * a[/tex]) of the person on the left that caused the rope to accelerate by  a  is  mathematically represented as

        [tex]m_l * a = F_r -T[/tex]

Where T  is  the tension on the rope  

      substituting values

        [tex]100 * 1 = 1000 - T[/tex]

=>      T  =  900 N