You walk into an elevator, step onto a scale, and push the "down" button to go directly from the tenth floor to the first floor. You also recall that your normal weight is w= 635 N. If the elevator has an initial acceleration of magnitude 2.45 m/s2, what does the scale read? Express your answer in newtons.

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:

Explanation:

Since B is perpendicular, it does no work on the electron but instead deflects it in a circular path.

q = 1.6 x 10-19 C

v = (17.1j + 12.7k) km/s = square root(17.1² + 12.7²) = 2.13 x 10⁴ m/s

the force acting on electron is

F= qvBsinΦ

F= (1.6 x 10⁻¹⁹C)(2.13.x 10⁴ m/s)(526 x 10⁻⁶ T)(sin90º)

F = 1.793x 10⁻¹⁸ N

The net force acting on electron is

F = e ( E+ ( vXB)

= ( - 1.6 × 10⁻¹⁹) ( E + ( 17.1 × 10³j + 12.7 × 10³ k)X( 529 × 10⁻⁶ ) (i)

= ( -1.6 × 10⁻¹⁹ ) ( E- 6.7k + 9.0j)

a= F/m

1.60 × 10¹² i =  ( -1.6 × 10⁻¹⁹ ) ( E- 6.9 k + 7.56 j)/9.11 × 10⁻³¹

9.11 i = - ( E- 6.7 k + 9.0 j)

E = -9.11i + 6.7k - 9.0j

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:

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:

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:

1.363×10^15 seconds

Explanation:

The spaceship travels an elliptical orbit from a point of 6590km from the earth center to the moon and 38500km from the earth center.

To calculate the time taken from Kepler's third Law :

T^2 = ( 4π^2/GMe ) r^3

Where Me is the mass of the earth

r is the average distance travel

G is the universal gravitational constant. = 6.67×10-11 m3 kg-1 s-2

π = 3.14

Me = mass of earth = 5.972×10^24kg

r =( r minimum + r maximum)/2 ......1

rmin = 6590km

rmax = 385000km

From equation 1

r = (6590+385000)/2

r = 391590/2

r = 195795km

From T^2 = ( 4π^2/GMe ) r^3

T^2 = (4 × 3.14^2/ 6.67×10-11 × 5.972×10^24) × 195795^3

= ( 4×9.8596/ 3.983×10^14 ) × 7.5059×10^15

= 39.4384/ 3.983×10^14 ) × 7.5059×10^15

= (9.901×10^14) × 7.5059×10^15

T^2 = 7.4321× 10^30

T =√7.4321× 10^30

T = 2.726×10^15 seconds

The time for one way trip from Earth to the moon is :

∆T = T/2

= 2.726×10^15 /2

= 1.363×10^15 secs

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:

The permittivity of rubber is  [tex]\epsilon = 8.703 *10^{-11}[/tex]

Explanation:

From the question we are told that

     The  magnitude of the point charge is  [tex]q_1 = 70 \ nC = 70 *10^{-9} \ C[/tex]

      The diameter of the rubber shell is  [tex]d = 32 \ cm = 0.32 \ m[/tex]

       The Electric field inside the rubber shell is  [tex]E = 2500 \ N/ C[/tex]

The radius of the rubber is  mathematically evaluated as

              [tex]r = \frac{d}{2} = \frac{0.32}{2} = 0.16 \ m[/tex]

Generally the electric field for a point  is in an insulator(rubber) is mathematically represented as

         [tex]E = \frac{Q}{ \epsilon } * \frac{1}{4 * \pi r^2}[/tex]

Where [tex]\epsilon[/tex] is the permittivity of rubber

    =>     [tex]E * \epsilon * 4 * \pi * r^2 = Q[/tex]

   =>      [tex]\epsilon = \frac{Q}{E * 4 * \pi * r^2}[/tex]

substituting values

            [tex]\epsilon = \frac{70 *10^{-9}}{2500 * 4 * 3.142 * (0.16)^2}[/tex]

            [tex]\epsilon = 8.703 *10^{-11}[/tex]

sorry but I don't understand

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

Answer:

0.087976 Nm

Explanation:

The magnetic torque (τ) on a current-carrying loop in a magnetic field is given by;

τ = NIAB sinθ     --------- (i)

Where;

N = number of turns of the loop

I = current in the loop

A = area of each of the turns

B = magnetic field

θ = angle the loop makes with the magnetic field

From the question;

N = 200

I = 4.0A

B = 0.70T

θ = 30°

A = π d² / 4        [d = diameter of the coil = 2.0cm = 0.02m]

A = π x 0.02² / 4 = 0.0003142m²         [taking π = 3.142]

Substitute these values into equation (i) as follows;

τ = 200 x 4.0 x 0.0003142 x 0.70 sin30°

τ = 200 x 4.0 x 0.0003142 x 0.70 x 0.5

τ = 200 x 4.0 x 0.0003142 x 0.70      

τ = 0.087976 Nm

Therefore, the torque on the coil is 0.087976 Nm

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:

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:

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:

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:

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)

Answer:

It always contains certain elements

Explanation:

Minerals can be defined as natural inorganic substances which possess an orderly internal structural arrangement as well as a particular, well known chemical composition, crystal structures and physical properties. Minerals include; quartz, dolomite, basalt, etc. Minerals may occur in isolation or in rock formations.

Minerals contain specific, well known chemical elements in certain ratios that can only vary within narrow limits. This is what we mean by a mineral's definite chemical composition. The structure of these minerals are all well known as well as their atom to atom connectivity.

The statement describes one feature of a mineral's definite chemical composition - It always contains certain elements.

A mineral is a naturally occurring chemical compound, usually of a crystalline form.

Thus, The statement describes one feature of a mineral's definite chemical composition - It always contains certain elements.

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Answer:

a) The work required to stretch the spring from 20 centimeters to 25 centimeters is 0.313 joules, b) The area of the region enclosed by one loop of the curve [tex]r(\theta) = 2\cdot \sin 5\theta[/tex] is [tex]4\pi[/tex].

Explanation:

a) The work, measured in joules, is a physical variable represented by the following integral:

[tex]W = \int\limits^{x_{f}}_{x_{o}} {F(x)} \, dx[/tex]

Where

[tex]x_{o}[/tex], [tex]x_{f}[/tex] - Initial and final position, respectively, measured in meters.

[tex]F(x)[/tex] - Force as a function of position, measured in newtons.

Given that [tex]F = k\cdot x[/tex] and the fact that [tex]F = 25\,N[/tex] when [tex]x = 0.3\,m - 0.2\,m[/tex], the spring constant ([tex]k[/tex]), measured in newtons per meter, is:

[tex]k = \frac{F}{x}[/tex]

[tex]k = \frac{25\,N}{0.3\,m-0.2\,m}[/tex]

[tex]k = 250\,\frac{N}{m}[/tex]

Now, the work function is obtained:

[tex]W = \left(250\,\frac{N}{m} \right)\int\limits^{0.05\,m}_{0\,m} {x} \, dx[/tex]

[tex]W = \frac{1}{2}\cdot \left(250\,\frac{N}{m} \right)\cdot [(0.05\,m)^{2}-(0.00\,m)^{2}][/tex]

[tex]W = 0.313\,J[/tex]

The work required to stretch the spring from 20 centimeters to 25 centimeters is 0.313 joules.

b) Let be [tex]r(\theta) = 2\cdot \sin 5\theta[/tex]. The area of the region enclosed by one loop of the curve is given by the following integral:

[tex]A = \int\limits^{2\pi}_0 {[r(\theta)]^{2}} \, d\theta[/tex]

[tex]A = 4\int\limits^{2\pi}_{0} {\sin^{2}5\theta} \, d\theta[/tex]

By using trigonometrical identities, the integral is further simplified:

[tex]A = 4\int\limits^{2\pi}_{0} {\frac{1-\cos 10\theta}{2} } \, d\theta[/tex]

[tex]A = 2 \int\limits^{2\pi}_{0} {(1-\cos 10\theta)} \, d\theta[/tex]

[tex]A = 2\int\limits^{2\pi}_{0}\, d\theta - 2\int\limits^{2\pi}_{0} {\cos10\theta} \, d\theta[/tex]

[tex]A = 2\cdot (2\pi - 0) - \frac{1}{5}\cdot (\sin 20\pi-\sin 0)[/tex]

[tex]A = 4\pi[/tex]

The area of the region enclosed by one loop of the curve [tex]r(\theta) = 2\cdot \sin 5\theta[/tex] is [tex]4\pi[/tex].

Answer:

7.2N/C

Explanation:

Pls see attached file

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° .

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:

d = 229.5 m

Explanation:

It is given that,

Total mass of a ski-plane is 1200 kg

It lands towards the west on a frozen lake at 30.0  m/s.

The coefficient of kinetic friction between the skis and the ice is 0.200.

We need to find the distance covered by the plane before coming to rest. In this case,

[tex]\mu mg=ma\\\\a=\mu g\\\\a=0.2\times 9.8\\\\a=1.96\ m/s^2[/tex]

It is decelerating, a = -1.96 m/s²

Now using the third equation of motion to find the distance covered by the plane such that :

[tex]v^2-u^2=2ad\\\\d=\dfrac{-u^2}{2a}\\\\d=\dfrac{-(30)^2}{2\times -1.96}\\\\d=229.59\ m[/tex]

So, the plane slide a distance of 229.5 m.  

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:

8.167

Explanation:

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

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:

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:

Explanation:

The index of refraction of the liquid can be computed from ...

  [tex]n_i\sin{(\theta_t)}=n_t[/tex]

where ni is the index of refraction of the glass block (1.52) and θt is the angle at which there is total internal refraction. nt is the index of refraction of the liquid.

For the given incidence angles, the computed indices of refraction are ...

  A: n = 1.52sin(52.0°) = 1.198

  B: n = 1.52sin(44.3°) = 1.062

  C: n = 1.52sin(36.3°) = 0.900

Explanation:

The Earth has a solid inner core