The exact ____ to be removed by the patient is determined according to the examination

Explanation:

The Earth has a solid inner core

Answer:

The workdone is [tex]W_v = - 20 \ J[/tex]

Explanation:

From the question we are told that

    The mass of the object is [tex]m = 2.5 \ kg[/tex]

     The speed of fall is [tex]v = 2.5 \ m/s[/tex]

     The depth of fall is  [tex]d = 80\ cm = 0.8 \ m[/tex]

Generally according to the work energy theorem

      [tex]W = \frac{1}{2} mv_2^2 - \frac{1}{2} mv_1^2[/tex]

Now here given the that the velocity is  constant  i.e  [tex]v_1 = v_2 = v[/tex] then

We have that

    [tex]W = \frac{1}{2} mv^2 - \frac{1}{2} mv^2 = 0 \ J[/tex]  

So in terms of workdone by the potential energy of the object and that of the viscous liquid we have

       [tex]W = W_v - W_p[/tex]

Where  [tex]W_v[/tex] is workdone by viscous liquid

             [tex]W_p[/tex] is the workdone by the object which is mathematically represented as

            [tex]W_p = mgd[/tex]

So  

       [tex]0 = W_v + mgd[/tex]

=>    [tex]W_v = - m * g * d[/tex]

substituting values

       [tex]W_v = - (2.5 * 9.8 * 0.8)[/tex]

      [tex]W_v = - 20 \ J[/tex]

Answer:

8.167

Explanation:

Answer:

a. E = 4.62 × 10⁻¹⁹J

b. n = 4.54 × 10¹⁹photons

c. 2.99 × 10²²photons/m²

d. 1.58 ×10¹⁸photons/seconds

e.  n = 1.896 × 10 ⁹ photons

f. the number of photons will be more if it travels slowly

Explanation:

Answer:

11.7 m

Explanation:

The radius of the circle is 13 m.

The central angle of the arc is 0.9 radians

The length of an arc is given as:

L = r θ

where θ = central angle in radians = 0.9

=> L = 0.9 * 13 = 11.7 m

Length of the arc will be 11.7 m ≈ 10 m

Arc length refers to the distance between two points along a curve’s section.

Arc length = radius * theta

where

Arc length  = ? to find

given :

radius = 13 m

theta ( central angle) = 0.9 radians

Arc length = 13 m * 0.9 radians

                = 11.7 m ≈ 10 m

length of the arc will be 11.7 m ≈ 10 m

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Given that,

Weight = 4 pound

[tex]W=4\ lb[/tex]

Stretch = 2 feet

Let the force be F.

The elongation of the spring after the mass attached is

[tex]x=2-1=1\ feet[/tex]

(a). We need to calculate the value of spring constant

Using Hooke's law

[tex]F=kx[/tex]

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

Where, F = force

k = spring constant

x = elongation

Put the value into the formula

[tex]k=\dfrac{4}{1}[/tex]

[tex]k=4[/tex]

(b). We need to calculate the mass

Using the formula

[tex]F=mg[/tex]

[tex]m=\dfrac{F}{g}[/tex]

Where, F = force

g = acceleration due to gravity

Put the value into the formula

[tex]m=\dfrac{4}{32}[/tex]

[tex]m=\dfrac{1}{8}\ lb[/tex]

We need to calculate the natural frequency

Using formula of natural frequency

[tex]\omega=\sqrt{\dfrac{k}{m}}[/tex]

Where, k = spring constant

m = mass

Put the value into the formula

[tex]\omega=\sqrt{\dfrac{4}{\dfrac{1}{8}}}[/tex]

[tex]\omega=\sqrt{32}[/tex]

[tex]\omega=4\sqrt{2}[/tex]

(c). We need to write the differential equation

Using differential equation

[tex]m\dfrac{d^2x}{dt^2}+kx=0[/tex]

Put the value in the equation

[tex]\dfrac{1}{8}\dfrac{d^2x}{dt^2}+4x=0[/tex]

[tex]\dfrac{d^2x}{dt^2}+32x=0[/tex]

(d). We need to find the solution for the position

Using auxiliary equation

[tex]m^2+32=0[/tex]

[tex]m=\pm i\sqrt{32}[/tex]

We know that,

The general equation is

[tex]x(t)=A\cos(\sqrt{32t})+B\sin(\sqrt{32t})[/tex]

Using initial conditions

(I). [tex]x(0)=2[/tex]

Then, [tex]x(0)=A\cos(\sqrt{32\times0})+B\sin(\sqrt{32\times0})[/tex]

Put the value in equation

[tex]2=A+0[/tex]

[tex]A=2[/tex].....(I)

Now, on differentiating of general equation

[tex]x'(t)=-\sqrt{32}A\sin(\sqrt{32t})+\sqrt{32}B\cos(\sqrt{32t})[/tex]

Using condition

(II). [tex]x'(0)=0[/tex]

Then, [tex]x'(0)=-\sqrt{32}A\sin(\sqrt{32\times0})+\sqrt{32}B\cos(\sqrt{32\times0})[/tex]

Put the value in the equation

[tex]0=0+\sqrt{32}B[/tex]

So, B = 0

Now, put the value in general equation from equation (I) and (II)

So, The general solution is

[tex] x(t)=2\cos\sqrt{32t}[/tex]

(e). We need to calculate the  time

Using formula of time

[tex]T=\dfrac{2\pi}{\omega}[/tex]

Put the value into the formula

[tex]T=\dfrac{2\pi}{4\sqrt{2}}[/tex]

[tex]T=1.11\ sec[/tex]

Hence, (a). The value of spring constant is 4.

(b). The natural frequency is 4√2.

(c). The differential equation is [tex]\dfrac{d^2x}{dt^2}+32x=0[/tex]

(d). The solution for the position is [tex] x(t)=2\cos\sqrt{32t}[/tex]

(e). The time period is 1.11 sec.

Answer:

The  self-inductance is  [tex]L = 0.0053 \ H[/tex]

Explanation:

From the question we are told that  

      The number of loops is  [tex]N = 900[/tex]

      The  length of the rod is  [tex]l =6 \ cm = 0.06 \ m[/tex]

      The radius of the rod is  [tex]r = 1 \ cm = 0.01 \ m[/tex]

The  self-inductance for the solenoid is mathematically represented as

        [tex]L = \frac{\mu_o * A * N^2 }{l}[/tex]

Now the cross-sectional of the solenoid is mathematically evaluated as

        [tex]A = \pi r^2[/tex]

substituting values  

         [tex]A =3.142 * 0.01 ^2[/tex]

        [tex]A = 3.142 *10^{-4} \ m^2[/tex]

and  [tex]\mu_o[/tex] is the permeability of free space with a value  [tex]\mu_o = 4\pi * 10^{-7} N/A^2[/tex]

    substituting values into above equation

          [tex]L = \frac{ 4\pi * 10^{-7} ^2* 3.142*10^{-4} * 900^2 }{0.06}[/tex]

          [tex]L = 0.0053 \ H[/tex]

sorry but I don't understand

Answer:

Speed = 20 m/sec at 75 deg South of East = 20 m/sec at 15 deg East of South

Explanation:

given data

truck moves = 25 m/s toward the east.

throws a baseball = 28 m/s southwest

solution

first we take here Speed of truck w.r.to ground i.e. V(p/g) = 25 m/sec toward the east  so we can say

V(p/g) = (25 i) m/sec     ........................1

and

Speed of baseball w.r.t. pickup i.e. V(b/p) = 28 m/sec toward the South West  and   we know that south west direction is in third quadrant

and here both component (x and y) are negative

So  that we can say it

V(b/p) = -28 × cos(45) i - 28 × sin(45) j     =  -19.8 i - 19.8 j  

and

now we use here relative motion  velocity for ball w.r.t ground

V(b/g) = V(b/p) + V(p/g )      ..........................2

put here value and we get

V(b/g) = (-19.8 i - 19.8 j) + 25 i     = 5.2 i - 19.8 j

so

Magnitude of that velocity

| V(b/g) | = [tex]\sqrt{(5.2^2 + 19.8^2)}[/tex]  

| V(b/g) | = 20.47 m/sec

so that  Direction will be here

Direction = arctan (19.8 ÷ 5.2)

Direction = 75.3° South of East

so that

Speed = 20.47 m/sec at 75.3 deg South of East

and 2 significant  

Speed = 20 m/sec at 75 deg South of East = 20 m/sec at 15 deg East of South

Answer:

The arrangement with the greatest resistance is the light bulb of option C. 4 W, 4.5 V

Explanation:

The equation for electric power is

power P = IV

also,  I = V/R,

substituting into the equation, we have

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

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

a)  [tex]R = \frac{1.5^{2} }{0.8}[/tex] = 2.8 Ω

b) [tex]R = \frac{3^{2} }{6}[/tex] = 1.5 Ω

c) [tex]R = \frac{4.5^{2} }{4}[/tex] 5.06 Ω

d) [tex]R = \frac{6^{2} }{8}[/tex] = 4.5 Ω

from the calculations, one can see that the lightbulb with te greates resistance is

C. 4 W, 4.5 V

Answer:

The rotational inertia of the pendulum around its pivot point is [tex]0.280\,kg\cdot m^{2}[/tex].

Explanation:

The angular frequency of a physical pendulum is measured by the following expression:

[tex]\omega = \sqrt{\frac{m\cdot g \cdot d}{I_{o}} }[/tex]

Where:

[tex]\omega[/tex] - Angular frequency, measured in radians per second.

[tex]m[/tex] - Mass of the physical pendulum, measured in kilograms.

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

[tex]d[/tex] - Straight line distance between the center of mass and the pivot point of the pendulum, measured in meters.

[tex]I_{O}[/tex] - Moment of inertia with respect to pivot point, measured in [tex]kg\cdot m^{2}[/tex].

In addition, frequency and angular frequency are both related by the following formula:

[tex]\omega =2\pi\cdot f[/tex]

Where:

[tex]f[/tex] - Frequency, measured in hertz.

If [tex]f = 0.658\,hz[/tex], then angular frequency of the physical pendulum is:

[tex]\omega = 2\pi \cdot (0.658\,hz)[/tex]

[tex]\omega = 4.134\,\frac{rad}{s}[/tex]

From the formula for the physical pendulum's angular frequency, the moment of inertia is therefore cleared:

[tex]\omega^{2} = \frac{m\cdot g \cdot d}{I_{o}}[/tex]

[tex]I_{o} = \frac{m\cdot g \cdot d}{\omega^{2}}[/tex]

Given that [tex]m = 1.15\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]d = 0.425\,m[/tex] and [tex]\omega = 4.134\,\frac{rad}{s}[/tex], the moment of inertia associated with the physical pendulum is:

[tex]I_{o} = \frac{(1.15\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (0.425\,m)}{\left(4.134\,\frac{rad}{s} \right)^{2}}[/tex]

[tex]I_{o} = 0.280\,kg\cdot m^{2}[/tex]

The rotational inertia of the pendulum around its pivot point is [tex]0.280\,kg\cdot m^{2}[/tex].

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 value of flux will be "744.1 N.m²/C".

Explanation:

The given values are:

Magnitude,

E = 21.5 N/C

Radius,

R = 1.66 m

As we know,

⇒  [tex]Flux = Area\times E[/tex]

On putting the estimated values, we get

⇒           [tex]=-21.5\tines (4\times \pi\times 1.66^2 )[/tex]

⇒           [tex]=744.1 \ N.m^2/C[/tex]

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 maximum number of bright spot is [tex]n_{max} =5001[/tex]

Explanation:

From the question we are told that

     The  slit distance is [tex]d = 1 \ mm = 0.001 \ m[/tex]

      The  wavelength is  [tex]\lambda = 400 \ nm = 400*10^{-9 } \ m[/tex]

Generally the condition for interference is  

        [tex]n * \lambda = d * sin \theta[/tex]

Where n is the number of fringe(bright spots) for the number of bright spots to be maximum  [tex]\theta = 90[/tex]

=>     [tex]sin( 90 )= 1[/tex]

So

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

substituting values

     [tex]n = \frac{ 1 *10^{-3} }{ 400 *10^{-9} }[/tex]

     [tex]n = 2500[/tex]

given there are two sides when it comes to the double slit apparatus which implies that the fringe would appear on two sides so the maximum number of bright spots is mathematically evaluated as

        [tex]n_{max} = 2 * n + 1[/tex]

The  1  here represented the central bright spot

So  

      [tex]n_{max} = 2 * 2500 + 1[/tex]

     [tex]n_{max} =5001[/tex]      

Answer:

8 Hz

Explanation:

Given that

Standing wave at one end is 24 Hz

Standing wave at the other end is 32 Hz.

Then the frequency of the standing wave mode of a string having a length, l, is usually given as

f(m) = m(v/2L), where in this case, m could be 1. 2. 3. 4 etc

Also, another formula is given as

f(m) = m.f(1), where f(1) is the fundamental frequency..

Thus, we could say that

f(m+1) - f(m) = (m + 1).f(1) - m.f(1) = f(1)

And as such,

f(1) = 32 - 24

f(1) = 8 Hz

Then, the fundamental frequency needed is 8 Hz

Answer:

Work done = 96.15 KJ

Heat transferred = 24 KJ

Explanation:

We are given;

Mass of air;m = 1.5 kg

Initial pressure;P1 = 100 KPa

Initial temperature;T1 = 17°C = 273 + 17 K = 290 K

Final volume;V2 = 0.5V1

Since the polytropic exponent is 1.3,thus it means;

P2 = P1[V1/V2]^(1.3)

So,P2 = 100(V1/0.5V1)^(1.3)

P2 = 100(2)^(1.3)

P2 = 246.2 KPa

To find the final temperature, we will make use of combined gas law.

So,

(P1×V1)/T1 = (P2×V2)/T2

T2 = (P2×V2×T1)/(P1×V1)

Plugging in the known values;

T2 = (246.2×0.5V1×290)/(100×V1)

V1 cancels out to give;

T2 = (246.2×0.5×290)/100

T2 = 356.99 K

The boundary work for this polytropic process is given by;

W_b = - ∫P. dv between the initial boundary and final boundary.

Thus,

W_b = - (P2.V2 - P1.V1)/(1 - n) = -mR(T2 - T1)/(1 - n)

R is gas constant for air = 0.28705 KPa.m³/Kg.K

n is the polytropic exponent which is 1.3

Thus;

W_b = -1.5 × 0.28705(356.99 - 290)/(1 - 1.3)

W_b = 96.15 KJ

The formula for the heat transfer is given as;

Q_out = W_b - m.c_v(T2 - T1)

c_v for air = 0.718 KJ/Kg.k

Q_out = 96.15 - (1.5×0.718(356.99 - 290))

Q_out = 24 KJ

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:

The wavelength becomes twice the original wavelength

Explanation:

Recall that for regular waves, the relationship between wavelength, velocity (i.e speed) and frequency is given by

v = fλ

where,

v = velocity,

f = frequency

λ = wavelength

Before a change was made to the frequency, we have: v₁ = f₁ λ₁

After a change was made to the frequency, we have: v₂ = f₂ λ₂

We are told that the speed remains the same, so

v₁ = v₂

f₁ λ₁ = f₂ λ₂ (rearranging this)

f₁ / f₂ = λ₂/λ₁ --------(1)

we are given that the frequency is cut in half.

f₂ = (1/2) f₁     (rearranging this)

f₁/f₂ = 2 -------------(2)

if we substitute equation (2) into equation (1):

f₁ / f₂ = λ₂/λ₁

2 = λ₂/λ₁

λ₂ = 2λ₁

Hence we can see that the wavelength after the change becomes twice (i.e doubles) the initial wavelength.

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

The container has the greatest total force acting on its base is the conatainer with the greatest total weight.

For each container, there is a pressure, P acting on its base given by

P = F/A where

Now, since the force on the base is the weight of water acting on the base, W, we have that

P = W/A where w = total weight of water acting on base

Now, for each container, the area of base is the same. So, we have that

P ∝ W

So, we see that the pressure is directly proportional to the weight on the container.

Since pressure is directly proportional to force, we see that the force on the base is directly proportional to the weight of water acting on the base.

So, for the greatest force acting on the base, we would have the greatest total weight of water acting on the base.

So, the container has the greatest total force acting on its base is the conatainer with the greatest total weight.

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Answer:

B = 0.046T

Explanation:

given

size of the screen = 51.2cm

distance from center = 11.1cm

region of magnetic field = 1.00cm

V= 22000V= 22kV

Answer:

all of the above

Explanation:

because it is.

Answer:

non of the above

Explanation:

Quantity of heat = mass× specific heat× change in temperature

m= 7kg c= 4.18 temp= 46-25=21°

.......H= 7×4.18×21= 614.46kJ

Answer:610 KJ

Explanation:A P E X answers

Answer:

Planet A = planet B > planet D > planet A = planet E

Explanation:

Gravitational force obeys the inverse square law. That is,

Gravitation force F is inversely proportional to the square of the distance between the two masses.

The larger the distance, the weaker the gravitational force F.

From largest to smallest, the rank of gravitational force that each planet exerts on the star with mass M. Distances are as follows:

Planet B, planet C, planet D, planet planet A, planet E.

Planet B and C may experience different the same gravitational force depending on their masses. This is also applicable to planet A and E.

Therefore,

Planet A = planet B > planet D > planet A = planet E

Answer:

Gas 1

Explanation:

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:

dV/dt = 150.79 m^3/s

Explanation:

In order to calculate the rate of change of the volume, you calculate the derivative, respect to the radius of the sphere, of the volume of the sphere, as follow:

[tex]\frac{dV}{dt}=\frac{d}{dt}(\frac{4}{3}\pi r^3)[/tex]                     (1)

r: radius of the sphere

You calculate the derivative of the equation (1):

[tex]\frac{dV}{dt}=\frac{d}{dt}(\frac{4}{3}\pi r^3)=3\frac{4}{3}\pi r^2\frac{dr}{dt}=4\pi r^2\frac{dr}{dt}\\\\\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}[/tex](2)

where dr/dt = 3m/s

You replace the values of dr/dt and r=2m in the equation (2):

[tex]\frac{dV}{dt}=4\pi (2m)^2(3\frac{m}{s})=150.79\frac{m^3}{s}[/tex]

The rate of change of the sphere, when it has a radius of 2m, is 150.79m^3/s

Answer:

The  gauge pressure is  [tex]P = 4.2*10^{5} \ N/m^2[/tex]

Explanation:

From the question we are told that

     The height of the 14th floor from the point where the water entered the building  is  h = 43 m

The  gauge pressure is  mathematically represented as

           [tex]P = mgh[/tex]

Where  the m is the mass of the water which is  mathematically represented as  

      [tex]m = \frac{\rho}{V}[/tex]

Where  [tex]\rho[/tex] is the density of the water which has a constant value of  [tex]\rho = 1000 \ kg/m^3[/tex] and this standard value of density the volume is  [tex]1 m^3[/tex] so

     [tex]m = \frac{1000}{1}[/tex]

     [tex]m = 1000 \ kg[/tex]

Thus  

     [tex]P = 1000 * 9.8 * 43[/tex]

     [tex]P = 4.2*10^{5} \ N/m^2[/tex]

Answer:

Electric dipole

Explanation:

the dipole axis makes an angle with the electric field. depending on direction (clockwise/aniclockwise) you get the torque

Hope this helps