A 2kg block is sitting on a hinged ramp such that you can increase the angle of the incline. The coefficient of static friction between the block and the ramp is 0.67 and the coefficient of kinetic friction is 0.25.
a. What angle do you have to tilt the ramp to get the block to slide?
b. What acceleration does the block experience at this angle when kinetic friction takes over?

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]

Answer:

Red light

Explanation:

This because All interference or diffraction patterns depend upon the wavelength of the light (or whatever wave) involved. Red light has the longest wavelength (about 700 nm)

Answer:

Increase of volume (F)  = 8.01%

Explanation:

Given:

Density of fish = 1,080 kg/m³

Density of water = 1,000 kg/m³

density of air = 1.28 kg/m³

Find:

Increase of volume (F)

Computation:

1,080 kg/m³  + [F × 1.28 kg/m³ ] = (1+F) × 1,000 kg/m³  

1,080 + 1.28 F =1,000 F + 1,000

80 = 998.72 F

F = 0.0801 (Approx)

F = 8.01%  (Approx)

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

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:

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.

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.

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:

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.

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

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Answer:

a) The first arrow reaches a maximum height of 56.712 meters, whereas second arrow reaches a maximum height of 342.816 meters, b) Both arrows have a total mechanical energy at their maximum height of 201.720 joules.

Explanation:

a) The first arrow is launch in a parabolic way, that is, horizontal speed remains constant and vertical speed changes due to the effects of gravity. On the other hand, the second is launched vertically, which means that velocity is totally influenced by gravity. Let choose the ground as the reference height for each arrow. Each arrow can be modelled as particles and by means of the Principle of Energy Conservation:

First arrow

[tex]U_{g,1} + K_{x,1} + K_{y,1} = U_{g,2} + K_{x,2} + K_{y,2}[/tex]

Where:

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

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

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

Now, the system is expanded and simplified:

[tex]m \cdot g \cdot (y_{2} - y_{1}) + \frac{1}{2}\cdot m \cdot (v_{y, 2}^{2} -v_{y, 1}^{2}) = 0[/tex]

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

[tex]y_{2}-y_{1} = \frac{1}{2}\cdot \frac{v_{y,1}^{2}-v_{y,2}^{2}}{g}[/tex]

Where:

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

[tex]v_{y,1}[/tex], [tex]v_{y,2}[/tex] - Initial and final vertical speed of the arrow, measured in meters.

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

The initial vertical speed of the arrow is:

[tex]v_{y,1} = v_{1}\cdot \sin \theta[/tex]

Where:

[tex]v_{1}[/tex] - Magnitude of the initial velocity, measured in meters per second.

[tex]\theta[/tex] - Initial angle, measured in sexagesimal degrees.

If [tex]v_{1} = 82\,\frac{m}{s}[/tex] and [tex]\theta = 24^{\circ}[/tex], the initial vertical speed is:

[tex]v_{y,1} = \left(82\,\frac{m}{s} \right)\cdot \sin 24^{\circ}[/tex]

[tex]v_{y,1} \approx 33.352\,\frac{m}{s}[/tex]

If [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]v_{y,1} \approx 33.352\,\frac{m}{s}[/tex] and [tex]v_{y,2} = 0\,\frac{m}{s}[/tex], the maximum height of the first arrow is:

[tex]y_{2} - y_{1} = \frac{1}{2}\cdot \frac{\left(33.352\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}}{9.807\,\frac{m}{s^{2}} }[/tex]

[tex]y_{2} - y_{1} = 56.712\,m[/tex]

Second arrow

[tex]U_{g,1} + K_{y,1} = U_{g,3} + K_{y,3}[/tex]

Where:

[tex]U_{g,1}[/tex], [tex]U_{g,3}[/tex] - Initial and final gravitational potential energy, measured in joules.

[tex]K_{y,1}[/tex], [tex]K_{y,3}[/tex] - Initial and final vertical translational kinetic energy, measured in joules.

[tex]m \cdot g \cdot (y_{3} - y_{1}) + \frac{1}{2}\cdot m \cdot (v_{y, 3}^{2} -v_{y, 1}^{2}) = 0[/tex]

[tex]g \cdot (y_{3}-y_{1}) = \frac{1}{2}\cdot (v_{y,1}^{2}-v_{y,3}^{2})[/tex]

[tex]y_{3}-y_{1} = \frac{1}{2}\cdot \frac{v_{y,1}^{2}-v_{y,3}^{2}}{g}[/tex]

If [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]v_{y,1} = 82\,\frac{m}{s}[/tex] and [tex]v_{y,3} = 0\,\frac{m}{s}[/tex], the maximum height of the first arrow is:

[tex]y_{3} - y_{1} = \frac{1}{2}\cdot \frac{\left(82\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}}{9.807\,\frac{m}{s^{2}} }[/tex]

[tex]y_{3} - y_{1} = 342.816\,m[/tex]

The first arrow reaches a maximum height of 56.712 meters, whereas second arrow reaches a maximum height of 342.816 meters.

b) The total energy of each system is determined hereafter:

First arrow

The total mechanical energy at maximum height is equal to the sum of the potential gravitational energy and horizontal translational kinetic energy. That is to say:

[tex]E = U + K_{x}[/tex]

The expression is now expanded:

[tex]E = m\cdot g \cdot y_{max} + \frac{1}{2}\cdot m \cdot v_{x}^{2}[/tex]

Where [tex]v_{x}[/tex] is the horizontal speed of the arrow, measured in meters per second.

[tex]v_{x} = v_{1}\cdot \cos \theta[/tex]

If [tex]v_{1} = 82\,\frac{m}{s}[/tex] and [tex]\theta = 24^{\circ}[/tex], the horizontal speed is:

[tex]v_{x} = \left(82\,\frac{m}{s} \right)\cdot \cos 24^{\circ}[/tex]

[tex]v_{x} \approx 74.911\,\frac{m}{s}[/tex]

If [tex]m = 0.06\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]y_{max} = 56.712\,m[/tex] and [tex]v_{x} \approx 74.911\,\frac{m}{s}[/tex], the total mechanical energy is:

[tex]E = (0.06\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (56.712\,m)+\frac{1}{2}\cdot (0.06\,kg)\cdot \left(74.911\,\frac{m}{s} \right)^{2}[/tex]

[tex]E = 201.720\,J[/tex]

Second arrow:

The total mechanical energy is equal to the potential gravitational energy. That is:

[tex]E = m\cdot g \cdot y_{max}[/tex]

[tex]m = 0.06\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex] and [tex]y_{max} = 342.816\,m[/tex]

[tex]E = (0.06\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (342.816\,m)[/tex]

[tex]E = 201.720\,J[/tex]

Both arrows have a total mechanical energy at their maximum height of 201.720 joules.

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

Explanation:

The Earth has a solid inner core

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

sorry but I don't understand

Answer:

I₂/I₁ = 0.53

Explanation:

During the motion the angular momentum of the skater remains conserved. Therefore:

Angular Momentum of Skater Before Pulling Arms = Angular Momentum of Skater After Pulling Arms

L₁ = L₂

but, the formula for angular momentum is:

L = Iω

Therefore,

I₁ω₁ = I₂ω₂

I₂/I₁ = ω₁/ω₂

where,

I₁ = Initial Moment of Inertia

I₂ = Final Moment of Inertia

ω₁ = Initial Angular Velocity = 3.14 rad/s

ω₂ = Final Angular velocity = 5.94 rad/s

Therefore,

I₂/I₁ = (3.14 rad/s)/(5.94 rad/s)

I₂/I₁ = 0.53

Answer:

The speed of the arrow after passing through the target is 30.1 meters per second.

Explanation:

The situation can be modelled by means of the Principle of Linear Momentum, let suppose that the arrow and the target are moving on the same axis, where the velocity of the first one is parallel to the velocity of the second one. The Linear Momentum model is presented below:

[tex]m_{a}\cdot v_{a,o} + m_{t}\cdot v_{t,o} = m_{a}\cdot v_{a,f} + m_{t}\cdot v_{t,f}[/tex]

Where:

[tex]m_{a}[/tex], [tex]m_{t}[/tex] - Masses of arrow and target, measured in kilograms.

[tex]v_{a,o}[/tex], [tex]v_{a,f}[/tex] - Initial and final speeds of the arrow, measured in meters per second.

[tex]v_{t,o}[/tex], [tex]v_{t,f}[/tex] - Initial and final speeds of the target, measured in meters per second.

The final speed of the arrow is now cleared:

[tex]m_{a} \cdot v_{a,f} = m_{a} \cdot v_{a,o} + m_{t}\cdot (v_{t,o}-v_{t,f})[/tex]

[tex]v_{a,f} = v_{a,o} + \frac{m_{t}}{m_{a}} \cdot (v_{t,o}-v_{t,f})[/tex]

If [tex]v_{a,o} = 2.1\,\frac{m}{s}[/tex], [tex]m_{t} = 0.3\,kg[/tex], [tex]m_{a} = 0.0225\,kg[/tex], [tex]v_{t,o} = 2.10\,\frac{m}{s}[/tex] and [tex]v_{t,f} = 0\,\frac{m}{s}[/tex], the speed of the arrow after passing through the target is:

[tex]v_{a,f} = 2.1\,\frac{m}{s} + \frac{0.3\,kg}{0.0225\,kg}\cdot (2.10\,\frac{m}{s} - 0\,\frac{m}{s} )[/tex]

[tex]v_{a,f} = 30.1\,\frac{m}{s}[/tex]

The speed of the arrow after passing through the target is 30.1 meters per second.

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

Answer:

8.7 m/s^2

82.15°

Explanation:

Given:-

- The initial speed of the car, vi = 25 m/s

- The radius of track, r = 129 m

- Car makes a circular " quarter turn "

- The constant tangential acceleration, at = 1.2 m/s^2

Solution:-

- We will solve the problem using rotational kinematics. Determine the initial angular velocity of car ( wi ) as follows:

                          [tex]w_i = \frac{v_i}{r} \\\\w_i = \frac{25}{129}\\\\w_i = 0.19379 \frac{rad}{s}[/tex]

- Now use the constant tangential acceleration ( at ) and determine the constant angular acceleration ( α ) for the rotational motion as follows:

                           at = r*α

                           α = ( 1.2 / 129 )

                           α = 0.00930 rad/s^2

- We know that the angular displacement from the initial entry to the exit of the turn is quarter of a turn. The angular displacement would be ( θ = π/2 ).

- Now we will use the third rotational kinematic equation of motion to determine the angular velocity at the exit of the turn (wf) as follows:

                            [tex]w_f^2 = w_i^2 + 2\alpha*theta\\\\w_f = \sqrt{0.19379^2 + 0.00930\pi } \\\\w_f = 0.25840 \frac{rad}{s}[/tex]

- We will use the evaluated final velocity ( wf ) and determine the corresponding velocity ( vf ) as follows:

                            [tex]v_f = r*w_f\\\\v_f = 129*0.2584\\\\v_f = 33.33380 \frac{x}{y}[/tex]

- Now use the formulation to determine the centripetal acceleration ( ac ) at this point as follows:

                            [tex]a_c = \frac{v_f^2}{r} \\\\a_c = \frac{33.3338^2}{129} \\\\a_c = 8.6135 \frac{m}{s^2}[/tex]

- To determine the magnitude of acceleration we will use find the resultant of the constant tangential acceleration ( at ) and the calculated centripetal acceleration at the exit of turn ( ac ) as follows:

                             [tex]|a| = \sqrt{a^2_t + a_c^2} \\\\|a| = \sqrt{1.2^2 + 8.6135^2} \\\\|a| = 8.7 \frac{m}{s^2}[/tex]

- To determine the angle between the velocity vector and the acceleration vector. We need to recall that the velocity vector only has one component and always tangential to the curved path. Hence, the velocity vector is parallel to the tangential acceleration vector ( at ). We can use the tangential acceleration ( at ) component of acceleration ( a ) and the centripetal acceleration ( ac ) component of the acceleration and apply trigonometric ratio as follows:

                          [tex]q = arctan \frac{a_c}{a_t} = arctan \frac{8.7}{1.2} \\\\q = 82.15 ^.[/tex] 

Answer: The angle ( q ) between acceleration vector ( a ) and the velocity vector ( v ) at the exit of the turn is 82.15° .

By  the experiment "The student would have no injuries because he would know hydrochloric acid is dangerous and would be wearing gloves when using it. "

An experiment would be a technique used to confirm or deny a hypothesis, as well as assess the likelihood or effectiveness of something that has never been tried before.

Hydrochloric acid is a kind of compound in which hydrogen and chlorine element is present.

Maintain a safe distance between your hands and your body, mouth, eyes, as well as a face when utilizing lab supplies and chemicals.

By  the experiment "By  the experiment "The student would have no injuries because he would know hydrochloric acid is dangerous and would be wearing gloves when using it. "

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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:

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:

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:

  p₁ = - p₂

the moment value of the two particles is the same, but its direction is opposite

Explanation:

When a nucleus divides spontaneously, the moment of the nucleic must be conserved, for this we form a system formed by the initial nucleus and the two fragments of the fission, in this case the forces during the division are internal and the moment is conserved

initial instant. Before fission

               p₀ = 0

since they indicate that the nucleus is at rest

final moment. After fission

             [tex]p_{f}[/tex] = m₁ v₁ + m₂ v₂

             p₀ = p_{f}

             0 = m₁ v₁ + m₂v₂

             m₁ v₁ = -m₂ v₂

              p₁ = - p₂

this indicates that the moment value of the two particles is the same, but its direction is opposite

Answer:

7.2N/C

Explanation:

Pls see attached file

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:

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:

8.167

Explanation:

Answer:

14.1 m/s

Explanation:

From the question,

μk = a/g...................... Equation 1

Where μk = coefficient of kinetic friction, a= acceleration of the skier, g = acceleration due to gravity.

make a the subject of the equation

a = μk(g).................. Equation 2

Given: μk = 0.160, g = 9.8 m/s²

Substitute into equation 2

a = 0.16(9.8)

a = 1.568 m/s²

Using,

F = ma

Where F = force, m = mass.

Make m the subject of the equation

m = F/a................... Equation 3

m = 160/1.568

m = 102.04 kg.

Note: The work done against air resistance by the skier+ work done against friction is equal to the kinetic energy after cross the patch.

Assuming the initial velocity of the skier to be zero

Fd+mgμ = 1/2mv²........................Equation 4

Where v = speed of the skier after crossing the patch, d = distance/width of the patch.

v = √2(Fd+mgμ)/m)................ Equation 5

Given: F = 160 N, m = 102.04 kg, d = 62 m, g = 9.8 m/s, μk = 0.16

Substitute these values into equation 5

v = √[2[(160×62)+(102.04×9.8×0.16)]/102.04]

v = √197.57

v = 14.1 m/s

v = 9.86 m/s