The pressure of sea water increases by 1.0atm for each 10m increase in the depth, by what percentage is the density of water increased in the deepest ocean of water of 12km. Compressibility is 5.0×10^-5 atm

Answer

A)184.9J

B)=3.63m/s

C) Zero

Explanation:

A)potential energy of the child at the initial position, measured relative the her potential energy at the bottom of the motion, is

U=Mgh

Where m=28kg

g= 9.8m/s

h= difference in height between the initial position and the bottom position

We are told that the rope is L = 2.30 m long and inclined at 45.0° from the vertical

h=L-Lcos(x)= L(1-cosx)=2.30(1-cos45)

=0.674m

Her Potential Energy will now

= 28× 9.8×0.674

=184.9J

B)we can see that at the bottom of the motion, all the initial potential energy of the child has been converted into kinetic energy:

E= 0.5mv^2

where

m = 28.0 kg is the mass of the child

v is the speed of the child at the bottom position

Solving the equation for v, we find

V=√2k/m

V=√(2×184.9/28

=3.63m/s

C)we can find work done by the tension in the rope is given using expresion below

W= Tdcosx

where W= work done

T is the tension

d = displacement of the child

x= angle between the directions of T and d

In this situation, we have that the tension in the rope, T, is always perpendicular to the displacement of the child, d. x= 90∘ and cos90∘=0 hence, the work done is zero.

Complete question:

Two 10-cm-diameter charged rings face each other, 25.0cm apart. Both rings are charged to +20.0nC. What is the electric field strength at:

a) the midpoint between the two rings?

b) the center of the left ring?

Answer:

a) the electric field strength at the midpoint between the two rings is 0

b) the electric field strength at the center of the left ring is 2712.44 N/C

Explanation:

Given;

distance between the two rings, d = 25 cm = 0.25 m

diameter of each ring, d = 10 cm = 0.1 m

radius of each ring, r = [tex]\frac{0.1}{2} = 0.05 \ m[/tex]

the charge on each ring, q = 20 nC

Electric field strength for a ring with radius r and distance x from the center of the ring is given as;

[tex]E = \frac{kxQ}{(x^2 +r^2)^{3/2}}[/tex]

The electric field strength at the midpoint;

the distance from the left ring to the mid point , x = 0.25 m / 2 = 0.125 m

[tex]E = \frac{kxQ}{(x^2 +r^2)^{3/2}} \\\\E = \frac{8.99*10^{9}*0.125*20*10^{-9}}{(0.125^2 + 0.05^2)^{3/2}} \\\\E = 9210.5 \ N/C[/tex]

[tex]E_{left} = 9210.5 \ N/C[/tex]

The electric field strength due to right ring is equal in magnitude to left ring but opposite in direction;

[tex]E_{right} = -9210.5 \ N/C[/tex]

The electric field strength at the midpoint;

[tex]E_{mid} = E_{left} + E_{right}\\\\E_{mid} = 9210.5 \ N/C - 9210.5 \ N/C\\\\E_{mid} = 0[/tex]

(b)

The distance from the right ring to center of the left ring, x = 0.25 m.

[tex]E = \frac{KxQ}{(x^2 +r^2)^{3/2}} \\\\E = \frac{8.99*10^{9} *0.25*20*10^{-9}}{(0.25^2 + 0.05^2)^{3/2}} \\\\E = 2712.44 \ N/C[/tex]

Answer:

Amount of charge

Explanation:

Mark me brainliest

Answer:

Explanation:

Suppose initially the plane was horizontal and light was reflected back at some angle θ from the normal .

Now the reflecting surface is twisted so that is becomes inclined at angle alpha .

The reflected light will be deviated from its original direction by angle

2 x alpha .

Similarly when the reflecting surface is further twisted so that it becomes inclined at angle beta then again the reflected beam will deviated by angle

2 x beta

Hence angle between these two reflected beam

= 2 beta - 2 alpha

= 2 ( β - α )

So, angular separation between the rays reflected from the two surfaces

= 2 ( β - α ) .

Answer:

The answer is A

Explanation:

Answer:

implausible or untestable scientific claims

Answer:

a) m = 0.626 kg , b) T = 2.09 s , c)   a = 1.0544 m / s²

Explanation:

In a spring mass system the equation of motion is

        x = A cos (wt + Ф)

with      w = √(k / m)

a) velocity is defined by

        v = dx / dt

        v = - A w sin (wt + Ф)          (1)

give us that the speed is

        v = 26.1 m / s

for the point

        x = a / 2

the range of motion is a = 11.0 cm

       x = 11.0 / 2

       x = 5.5 cm

Let's find the time it takes to get to this distance

       wt + Ф = cos⁻¹ (x / A)

       wt + Ф = cos 0.5

        wt + Ф = 0.877

In the exercise they do not indicate that the body started its movement with any speed, therefore we assume that for the maximum elongation the body was released, therefore the phase is zero f

       Ф = 0

       wt = 0.877

       t = 0.877 / w

we substitute in equation 1

       26.1 = -11.0 w sin (w 0.877 / w)

        w = 26.1 / (11 sin 0.877))

        w = 3.096 rad / s

from the angular velocity equation

       w² = k / m

       m = k / w²

       m = 6 / 3,096²

       m = 0.626 kg

b) angular velocity and frequency are related

       w = 2π f

frequency and period are related

        f = 1 / T

we substitute

        w = 2π / T

        T = 2π / w

        T = 2π / 3,096

        T = 2.09 s

c) maximum acceleration

 the acceleration of defined by

        a = dv / dt

        a = - Aw² cos (wt)

the acceleration is maximum when the cosine is ±1

         a = A w²

          a = 11  3,096²

        a = 105.44 cm / s²

we reduce to m / s

        a = 1.0544 m / s²

a. I've attached a plot of the surface. Each face is parameterized by

• [tex]\mathbf s_1(x,y)=x\,\mathbf i+y\,\mathbf j[/tex] with [tex]0\le x\le2[/tex] and [tex]0\le y\le6-x[/tex];

• [tex]\mathbf s_2(u,v)=u\cos v\,\mathbf i+u\sin v\,\mathbf k[/tex] with [tex]0\le u\le2[/tex] and [tex]0\le v\le\frac\pi2[/tex];

• [tex]\mathbf s_3(y,z)=y\,\mathbf j+z\,\mathbf k[/tex] with [tex]0\le y\le 6[/tex] and [tex]0\le z\le2[/tex];

• [tex]\mathbf s_4(u,v)=u\cos v\,\mathbf i+(6-u\cos v)\,\mathbf j+u\sin v\,\mathbf k[/tex] with [tex]0\le u\le2[/tex] and [tex]0\le v\le\frac\pi2[/tex]; and

• [tex]\mathbf s_5(u,y)=2\cos u\,\mathbf i+y\,\mathbf j+2\sin u\,\mathbf k[/tex] with [tex]0\le u\le\frac\pi2[/tex] and [tex]0\le y\le6-2\cos u[/tex].

b. Assuming you want outward flux, first compute the outward-facing normal vectors for each face.

[tex]\mathbf n_1=\dfrac{\partial\mathbf s_1}{\partial y}\times\dfrac{\partial\mathbf s_1}{\partial x}=-\mathbf k[/tex]

[tex]\mathbf n_2=\dfrac{\partial\mathbf s_2}{\partial u}\times\dfrac{\partial\mathbf s_2}{\partial v}=-u\,\mathbf j[/tex]

[tex]\mathbf n_3=\dfrac{\partial\mathbf s_3}{\partial z}\times\dfrac{\partial\mathbf s_3}{\partial y}=-\mathbf i[/tex]

[tex]\mathbf n_4=\dfrac{\partial\mathbf s_4}{\partial v}\times\dfrac{\partial\mathbf s_4}{\partial u}=u\,\mathbf i+u\,\mathbf j[/tex]

[tex]\mathbf n_5=\dfrac{\partial\mathbf s_5}{\partial y}\times\dfrac{\partial\mathbf s_5}{\partial u}=2\cos u\,\mathbf i+2\sin u\,\mathbf k[/tex]

Then integrate the dot product of f with each normal vector over the corresponding face.

[tex]\displaystyle\iint_{S_1}\mathbf f(x,y,z)\cdot\mathrm d\mathbf S=\int_0^2\int_0^{6-x}f(x,y,0)\cdot\mathbf n_1\,\mathrm dy\,\mathrm dx[/tex]

[tex]=\displaystyle\int_0^2\int_0^{6-x}0\,\mathrm dy\,\mathrm dx=0[/tex]

[tex]\displaystyle\iint_{S_2}\mathbf f(x,y,z)\cdot\mathrm d\mathbf S=\int_0^2\int_0^{\frac\pi2}\mathbf f(u\cos v,0,u\sin v)\cdot\mathbf n_2\,\mathrm dv\,\mathrm du[/tex]

[tex]\displaystyle=\int_0^2\int_0^{\frac\pi2}-u^2(2\sin v+\cos v)\,\mathrm dv\,\mathrm du=-8[/tex]

[tex]\displaystyle\iint_{S_3}\mathbf f(x,y,z)\cdot\mathrm d\mathbf S=\int_0^2\int_0^6\mathbf f(0,y,z)\cdot\mathbf n_3\,\mathrm dy\,\mathrm dz[/tex]

[tex]=\displaystyle\int_0^2\int_0^60\,\mathrm dy\,\mathrm dz=0[/tex]

[tex]\displaystyle\iint_{S_4}\mathbf f(x,y,z)\cdot\mathrm d\mathbf S=\int_0^2\int_0^{\frac\pi2}\mathbf f(u\cos v,6-u\cos v,u\sin v)\cdot\mathbf n_4\,\mathrm dv\,\mathrm du[/tex]

[tex]=\displaystyle\int_0^2\int_0^{\frac\pi2}-u^2(2\sin v+\cos v)\,\mathrm dv\,\mathrm du=\frac{40}3+6\pi[/tex]

[tex]\displaystyle\iint_{S_5}\mathbf f(x,y,z)\cdot\mathrm d\mathbf S=\int_0^{\frac\pi2}\int_0^{6-2\cos u}\mathbf f(2\cos u,y,2\sin u)\cdot\mathbf n_5\,\mathrm dy\,\mathrm du[/tex]

[tex]=\displaystyle\int_0^{\frac\pi2}\int_0^{6-2\cos u}12\,\mathrm dy\,\mathrm du=36\pi-24[/tex]

c. You can get the total flux by summing all the fluxes found in part b; you end up with 42π - 56/3.

Alternatively, since S is closed, we can find the total flux by applying the divergence theorem.

[tex]\displaystyle\iint_S\mathbf f(x,y,z)\cdot\mathrm d\mathbf S=\iiint_R\mathrm{div}\mathbf f(x,y,z)\,\mathrm dV[/tex]

where R is the interior of S. We have

[tex]\mathrm{div}\mathbf f(x,y,z)=\dfrac{\partial(3x)}{\partial x}+\dfrac{\partial(x+y+2z)}{\partial y}+\dfrac{\partial(3z)}{\partial z}=7[/tex]

The integral is easily computed in cylindrical coordinates:

[tex]\begin{cases}x(r,t)=r\cos t\\y(r,t)=6-r\cos t\\z(r,t)=r\sin t\end{cases},0\le r\le 2,0\le t\le\dfrac\pi2[/tex]

[tex]\displaystyle\int_0^2\int_0^{\frac\pi2}\int_0^{6-r\cos t}7r\,\mathrm dy\,\mathrm dt\,\mathrm dr=42\pi-\frac{56}3[/tex]

as expected.

Answer:

[tex]N = 208 \ turns[/tex]

Explanation:

From the question we are told that

    The  rate of  current change is  [tex]\frac{di }{dt} = 0.0200 \ A/s[/tex]

    The  magnitude of the induced emf is  [tex]\epsilon = 12.7 \ mV = 12.7 *10^{-3} \ V[/tex]

     The  current is  [tex]I = 1.50 \ A[/tex]

      The  average  flux is  [tex]\phi = 0.00458 \ Wb[/tex]

Generally the number of  turns the number of turn the solenoid has is mathematically represented as  

            [tex]N = \frac{\epsilon_o * I}{ \phi * \frac{di}{dt} }[/tex]

substituting values

           [tex]N = \frac{ 12.7*10^{-3} * 1.50 }{ 0.00458 * 0.0200 }[/tex]

            [tex]N = 208 \ turns[/tex]

Answer:

Not slow down or speed up.

Explanation:

Hitting the puck accelerates the speed of the puck from zero to the speed with which it leaves at the instance they lose contact. Since there is no friction between the puck and the ice, there will be no force decelerating or accelerating the hockey puck, allowing the puck to move away and remain in motion without speeding up or slowing down indefinitely theoretically.

Answer:

The speed of the electron is 1.371 x 10⁶ m/s.

Explanation:

Given;

wavelength of the ultraviolet light beam, λ = 130 nm = 130 x 10⁻⁹ m

the work function of the molybdenum surface, W₀ = 4.2 eV = 6.728 x 10⁻¹⁹ J

The energy of the incident light is given by;

E = hf

where;

h is Planck's constant = 6.626 x 10⁻³⁴ J/s

f = c / λ

[tex]E = \frac{hc}{\lambda} \\\\E = \frac{6.626*10^{-34} *3*10^{8}}{130*10^{-9}} \\\\E = 15.291*10^{-19} \ J[/tex]

Photo electric effect equation is given by;

E = W₀ + K.E

Where;

K.E is the kinetic energy of the emitted electron

K.E = E - W₀

K.E = 15.291 x 10⁻¹⁹ J - 6.728 x 10⁻¹⁹ J

K.E = 8.563 x 10⁻¹⁹ J

Kinetic energy of the emitted electron is given by;

K.E = ¹/₂mv²

where;

m is mass of the electron = 9.11 x 10⁻³¹ kg

v is the speed of the electron

[tex]v = \sqrt{\frac{2K.E}{m} } \\\\v = \sqrt{\frac{2*8.563*10^{-19}}{9.11*10^{-31}}}\\\\v = 1.371 *10^{6} \ m/s[/tex]

Therefore, the speed of the electron is 1.371 x 10⁶ m/s.

Answer:

298rad/s and 116.96 ohms

Explanation:

Given an L-R-C series circuit where

L = 0.450 H,

C=2.50×10^−5F, and resistance R= 0

In this situation we have a simple LC circuit with angular frequency

Wo = 1√LC

= 1/√(0.450)(2.50×10^-5)

= 1/√0.00001125

= 1/0.003354

= 298rad/s

B) Now we need to find the value of R such that it gives a 10% decrease in angular frequency.

Wi/W° = (100-10)/100

Wi/W° = 90/100

Wi/W° = 0.90 ............... 1

Angular frequency of oscillation

The complete aspect of the solution is attached, please check.

a. The angular frequency of the circuit when R = 0 Ohms is 294.12 rad/s.

b. The value R must have to give a decrease in angular frequency of 10.0 % compared to the initial value is equal to 116.96 Ohms.

Given the following data:

a. To determine the angular frequency of the circuit when R = 0 Ohms:

Mathematically, the angular frequency of a LC circuit is given by the formula:

[tex]\omega = \frac{1}{\sqrt{LC} } \\\\\omega =\frac{1}{\sqrt{0.450 \times 2.50\times 10^{-5}}} \\\\\omega =\frac{1}{\sqrt{1.125 \times 10^{-5}}} \\\\\omega = \frac{1}{0.0034} \\\\\omega = 294.12\;rad/s[/tex]

b. To find the value R must have to give a decrease in angular frequency of 10.0 % compared to the value calculated above:

The mathematical expression is given as follows:

[tex]\frac{\omega_f}{\omega_i} = \frac{100-10}{100} \\\\\frac{\omega_f}{\omega_i} =\frac{90}{100} \\\\\frac{\omega_f}{\omega_i} =0.9[/tex]

[tex](\frac{\omega_f}{\omega_i})^2 = 1 - \frac{R^2C}{4L} \\\\0.90^2=1 - \frac{R^2C}{4L}\\\\R=\sqrt{\frac{4L(1-0.81)}{C}} \\\\R=\sqrt{\frac{4\times 0.450 \times (0.19)}{2.50\times 10^{-5}}}\\\\R = \sqrt{\frac{0.342}{2.50\times 10^{-5}} }\\\\R =\sqrt{13680}[/tex]

R = 116.96 Ohms.

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Answer:

The amplitude of the resultant wave is 12.93 cm.

Explanation:

The amplitude of resultant of two waves, y₁ and y₂, is given as;

Y = y₁ + y₂

Let y₁ = A sin(kx - ωt)

Since the wave is out phase by φ, y₂ is given as;

y₂ = A sin(kx - ωt + φ)

Y = y₁ + y₂ = 2A Cos (φ / 2)sin(kx - ωt + φ/2 )

Given;

phase difference, φ = 45°

Amplitude, A = 7.00 cm

Y = 2(7) Cos (45 /2) sin(kx - ωt + 22.5° )

Y = 12.93 cm

Therefore, the amplitude of the resultant wave is 12.93 cm.

Answer:

[tex]v_B/v_A=\sqrt{3}[/tex]

Explanation:

Consider the two kinematic equations for velocity and position of an object falling due to the action of gravity:

[tex]v=-g\,t\\ \\position=-\frac{1}{2} g\,t^2[/tex]

Therefore, if we consider [tex]t_A[/tex] the time for the object to reach point A, and [tex]t_B[/tex] the time for it to reach point B, then:

[tex]v_A=-g\,t_A\\v_B=-g\,t_B\\\frac{v_B}{v_A}= \frac{-g\,t_B}{-g\,t_A} =\frac{t_B}{t_A}[/tex]

Let's work in a similar way with the two different positions at those different times, and for which we have some information;

[tex]x_A=-x=-\frac{1}{2}\, g\,t_A^2\\x_B=-3\,x=-\frac{1}{2}\, g\,t_B^2\\ \\\frac{x_B}{x_A} =\frac{t_B^2}{t_A^2} \\\frac{t_B^2}{t_A^2}=\frac{-3\,x}{-x} \\\frac{t_B^2}{t_A^2}=3\\(\frac{t_B}{t_A})^2=3[/tex]

Notice that this quotient is exactly the square of the quotient of velocities we are looking for, therefore:

[tex](\frac{t_B}{t_A})^2=3\\(\frac{v_B}{v_A})^2=3\\ \frac{v_B}{v_A}=\sqrt{3}[/tex]

Answer:

d' = 75.1 cm

Explanation:

It is given that,

The actual depth of a shallow pool is, d = 1 m

We need to find the apparent depth of the water in the pool. Let it is equal to d'.

We know that the refractive index is also defined as the ratio of real depth to the apparent depth. Let the refractive index of water is 1.33. So,

[tex]n=\dfrac{d}{d'}\\\\d'=\dfrac{d}{n}\\\\d'=\dfrac{1\ m}{1.33}\\\\d'=0.751\ m[/tex]

or

d' = 75.1 cm

So, the apparent depth is 75.1 cm.

Answer:

m = 22,877 kg / s

Explanation:

Let's solve this exercise in parts, first look for the amount of heat generated by the plant and then the amount of water to dissipate this heat

The plant generates a power of 300 MW at a rate of 40%, let's use a direct ratio rule to find the heat. If the power is 400 MW it corresponds to 40%, what heat (Q) corresponds to the other 60%

           Q = 300 60% / 40%

           Q = 450 MW

having the amount of heat generated we can use the calorimeter equation,

           Q = m [tex]c_{e}[/tex] [tex](T_{f} - T_{o})[/tex]

            m = Q / c_{e} (T_{f} - T_{o})

let's use the maximum temperature change allowed

           (T_{f} - T_{o}) = 5

the specific heat of sea water is 3934 J / kg ºC, note that it is less than that of pure water, due to the salts dissolved in sea water

power and energy are related

              W = Q / t

               Q = W t

let's calculate

             m = 450 10⁶ / (3934 5)

             m = 22,877 kg / s

Answer:

B. 1.6 x 10⁻³ m

Explanation:

The magnetic field at the center of the loop is given by;

[tex]B = \frac{\mu_o I }{2R}[/tex]

Where;

μ₀ is the permeability of free space

I is the current in the loop

R is the radius of the circular loop

B is the magnetic field

Given;

diameter of the loop = 1cm

radius of the loop, r = 0.5 cm = 0.005 m

magnetic field, B = 2.5mT = 2.5 x 10⁻³ T

The current in the loop is calculated as;

[tex]I = \frac{2BR}{\mu_o} \\\\I = \frac{2*2.5*10^{-3}*0.005}{4\pi*10^{-7}} \\\\I = 19.89 \ A[/tex]

The magnetic at a distance from the long straight wire is calculated as;

[tex]B = \frac{\mu_o I}{2\pi d}[/tex]

where;

d is the distance from the wire;

[tex]d = \frac{\mu_o I}{2\pi B} \\\\d = \frac{4\pi *10^{-7} * 19.89}{2\pi *2.5*10^{-3}} \\\\d = 1.6 *10^{-3} \ m[/tex]

Therefore, the distance from the wire where the magnetic field is 2.5 mT is 1.6 x 10⁻³ m.

B. 1.6 x 10⁻³ m

This question involves the concepts of the magnetic field due to a loop and a  current-carrying wire and current.

A long straight wire carrying the same current as the loop of wire has a magnetic field of 2.5 mT at a distance of b "B. 1.5 x 10⁻³ m".

The magnetic field at the center of a loop of wire is given by the following formula:

[tex]B=\frac{\mu_o I}{2r}[/tex]

where,

B = Magnetic Field = 2.5 mT = 2.5 x 10⁻³ T

μ₀ = permeability of free space = 4π x 10⁻⁷ N/A²

I = current = ?

r = radius = diameter/2 = 1 cm/2 = 0.5 cm = 0.005 m

Therefore,

[tex]I = \frac{(2.5\ x\ 10^{-3}\ T)(2)(0.005\ m)}{4\pi\ x\ 10^{-7}\ N/A^2}[/tex]

I = 19.9 A

Now, the magnetic field at a distance from the straight wire is given by the following formula:

[tex]B=\frac{\mu_o I}{2\pi R}[/tex]

where,

R = distance from wire = ?

Therefore,

[tex]R = \frac{(4\pi \ x \ 10^{-7}\ N/A^2)(19.9\ A)}{2\pi(2.5\ x\ 10^{-3}\ T)}[/tex]

R = 1.6 x 10⁻³ m

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Answer:

E = 2.48 eV

Explanation:

The energy of a photon is given by the following formula:

E = hυ

where,

E = Energy of Photon = ?

h = Plank's Constant = 6.626 x 10⁻³⁴ J.s

υ = frequency of photon = c/λ

Therefore,

E = hc/λ

where,

c = speed of light = 3 x 10⁸ m/s

λ = wavelength of light = 500 nm = 5 x 10⁻⁷ m

Therefore,

E = (6.626 x 10⁻³⁴ J.s)(3 x 10⁸ m/s)/(5 x 10⁻⁷ m)

E = (3.97 x 10⁻¹⁹ J)(1 eV/1.6 x 10⁻¹⁹ J)

E = 2.48 eV

A photon of visible light that has a wavelength of 500 nm, has an energy of 2.48 eV.

We can calculate the energy (E) of a photon with a wavelength (λ) of 500 nm using the Planck's-Einstein relation.

[tex]E = \frac{h \times c}{\lambda } = \frac{(6.63 \times 10^{-34}J.s ) \times (3.00 \times 10^{8}m/s )}{500 \times 10^{-9}m } = 3.98 \times 10^{-19} J[/tex]

where,

We can convert 3.98 × 10⁻¹⁹ J to eV using the conversion factor 1 J = 6.24 × 10¹⁸ eV.

[tex]3.98 \times 10^{-19} J \times \frac{6.24 \times 10^{18} eV }{1J} = 2.48 eV[/tex]

A photon of visible light that has a wavelength of 500 nm, has an energy of 2.48 eV.

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Answer:

The  period is [tex]T = 0.00255 \ s[/tex]

Explanation:

From the question we are told that

  The  frequency is  [tex]f = 392 \ Hz[/tex]

Generally the period is mathematically represented as  

           [tex]T = \frac{1}{f}[/tex]

=>       [tex]T = \frac{1}{ 392}[/tex]

=>       [tex]T = 0.00255 \ s[/tex]

Answer:

The magnitude of the magnetic flux through the loop is 0.0982 T.m²

Explanation:

Given;

magnitude of magnetic field, B = 0.5 T

radius of the loop, r = 0.25 m

Area of the loop is given by;

A = πr²

A = 3.142 x (0.25)²

A = 0.1964 m²

The magnitude of the magnetic flux through the loop is given by;

Ф = BA

Where;

B is the magnitude of the magnetic field

A is area of the field

Ф = 0.5 x 0.1964

Ф = 0.0982 T.m²

Therefore, the magnitude of the magnetic flux through the loop is 0.0982 T.m²

Answer:

Magnitude of momentum = q × B0 × [d^2 + 2L^2] / 2d.

Explanation:

So, from the question, we are given that the charge = q, the momentum = p.

=> From the question We are also given that, "initially, there is movement along the x-axis which then enters a region where a uniform magnetic field* B = (B0)(k) which then extends over a width x = L, the distance = d in the +y direction as it traverses the field."

Momentum,P = mass × Velocity, v -----(1).

We know that for a free particle the magnetic field is equal to the centrepetal force. Thus, we have the magnetic field = mass,.m × (velocity,v)^2 / radius, r.

Radius,r = P × v / B0 -----------------------------(2).

Centrepetal force = q × B0 × v. ----------(3).

(If X = L and distance = d)Therefore, the radius after solving binomially, radius = (d^2 + 2 L^2) / 2d.

Equating Equation (2) and (3) gives;

P = B0 × q × r.

Hence, the Magnitude of momentum = q × B0 × [d^2 + 2L^2] / 2d.

Answer: Wavelength can be defined as the distance between two successive crests or troughs of a wave. It is measured in the direction of the wave.

Explanation:

Wavelength refers to the length or distance between two identical points of neighboring cycles of a wave signal traveling in space or in any physical medium. ... The wavelength of a signal is inversely proportional to its frequency, that is, the higher the frequency, the shorter the wavelength.

Explanation:

Using Equations of Motion :

[tex]s = ut + \frac{1}{2} g {t}^{2} [/tex]

Height = 0 * 4 + 4.9 * 16

Height = 78.4 m

Explanation:

Given that,

Separation between two long parallel conductors, r = 11 cm = 0.11 m

Current in first wire, [tex]I_1=3\ A[/tex]

Current in second wire, [tex]I_2=8\ A[/tex]

(a) The magnetic field created by I₁ at the location of I₂ is given by :

[tex]B_{12}=\dfrac{\mu_o I_1}{2\pi r}\\\\B_{12}=\dfrac{4\pi \times 10^{-7}\times 3I_1}{2\pi \times 0.11}\\\\B_{12}=5.45\times 10^{-6}\ T[/tex]

(b) Magnetic force per unit length exerted by [tex]I_1[/tex] on [tex]I_2[/tex] is given by :

[tex]\dfrac{F}{l}=\dfrac{\mu_o I_1I_2}{2\pi r}\\\\\dfrac{F}{l}=\dfrac{4\pi \times 10^{-7}\times 3\times 8}{2\pi \times 0.11}\\\\\dfrac{F}{l}=4.36\times 10^{-5}\ N/m[/tex]

(c) The magnetic field created by I₂ at the location of I₁ is given by :

[tex]B_{21}=\dfrac{\mu_o I_2} {2\pi r}\\\\B_{12}=\dfrac{4\pi \times 10^{-7}\times 8}{2\pi \times 0.11}\\\\B_{12}=1.45\times 10^{-5}\ T[/tex]

Hence, this is the required solution.

Answer:

Explanation:

The force experienced by the moving electron in the magnetic field is expressed as F = qvBsinθ where;

q is the charge on the electron

v is the velocity of the electron

B is the magnetic field strength

θ is the angle that the velocity of the electron make with the magnetic field.

Given parameters

F =  1.40*10⁻¹⁶ N

q = 1.6*10⁻¹⁹C

v = 3.94*10³m/s

B = 1.23T

Required

Angle that the velocity of the electron make with the magnetic field

Substituting the given parameters into the formula:

1.40*10⁻¹⁶ =  1.6*10⁻¹⁹ * 3.94*10³ * 1.23 * sinθ

1.40*10⁻¹⁶ = 7.75392 * 10⁻¹⁹⁺³sinθ

1.40*10⁻¹⁶ = 7.75392 * 10⁻¹⁶sinθ

sinθ = 1.40*10⁻¹⁶/7.75392 * 10⁻¹⁶

sinθ = 1.40/7.75392

sinθ = 0.1806

θ = sin⁻¹0.1806

θ₁ = 10.4⁰

Since sinθ is positive in the 1st and 2nd quadrant, θ₂ = 180-θ₁

θ₂ = 180-10.4

θ₂ = 169.6⁰

Hence, the angle that the velocity of the electron make with the magnetic field are 10.4⁰ and 169.6⁰

Answer:

B.

Explanation:

The water collects in the ocean; it is then evaporated by the sun. After evaporation the water turns into water vapor, it then condenses to form clouds.

The ocean water prior to the part of the water cycle should be option B.

The ocean water should be collected back in the ocean prior to the part of the water cycle.

Because this should be done when it is evaporated by the sun.  When the evaporation is done so the water should be transformed into water vapor.

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Answer:

b. approaches infinity

Explanation:

Because Capacitive reactance is given as Xc = 1/ωC

​

So we can see that the value of capacitive reactance and therefore its overall impedance (in Ohms) decreases to zero as the frequency increases acting like a short circuit.

Same as the frequency approaches zero or DC, the capacitors reactance increases to infinity, acting like an open circuit which is why capacitors block DC

Answer:

The induced current is clockwise

Answer:

covalent

Explanation:

covalent bonds share electrons

a) y(max)  = 337.76 m

b) t₁ = 5.30 s  the time for y maximum

c)t₂ =  13.60 s  time for y = 0 time when the fly finish

d) vₓ = 30 m/s        vy = - 81.32 m/s

e)x = 408 m

Equations for projectile motion:

v₀ₓ = v₀ * cosα          v₀ₓ = 60*(1/2)     v₀ₓ = 30 m/s   ( constant )

v₀y = v₀ * sinα           v₀y = 60*(√3/2)     v₀y = 30*√3  m/s

a) Maximum height:

The following equation describes the motion in y coordinates

y  =  y₀ + v₀y*t - (1/2)*g*t²      (1)

To find h(max), we need to calculate t₁ ( time for h maximum)

we take derivative on both sides of the equation

dy/dt  = v₀y  - g*t

dy/dt  = 0           v₀y  - g*t₁  = 0    t₁ = v₀y/g

v₀y = 60*sin60°  = 60*√3/2  = 30*√3

g = 9.8 m/s²

t₁ = 5.30 s  the time for y maximum

And y maximum is obtained from the substitution of t₁  in equation (1)

y (max) = 200 + 30*√3 * (5.30)  - (1/2)*9.8*(5.3)²

y (max) = 200 + 275.40 - 137.64

y(max)  = 337.76 m

Total time of flying (t₂)  is when coordinate y = 0

y = 0 = y₀  + v₀y*t₂ - (1/2)* g*t₂²

0 = 200 + 30*√3*t₂  - 4.9*t₂²            4.9 t₂² - 51.96*t₂ - 200 = 0

The above equation is a second-degree equation, solving for  t₂

t =  [51.96 ±√ (51.96)² + 4*4.9*200]/9.8

t =  [51.96 ±√2700 + 3920]/9.8

t =  [51.96 ± 81.36]/9.8

t = 51.96 - 81.36)/9.8         we dismiss this solution ( negative time)

t₂ =  13.60 s  time for y = 0 time when the fly finish

The components of the velocity just before striking the ground are:

vₓ = v₀ *cos60°       vₓ = 30 m/s  as we said before v₀ₓ is constant

vy = v₀y - g *t        vy = 30*√3  - 9.8 * (13.60)

vy = 51.96 - 133.28         vy = - 81.32 m/s

The sign minus means that vy  change direction

Finally the horizontal distance is:

x = vₓ * t

x = 30 * 13.60  m

x = 408 m

Explanation:

We know that rate of flow through a cross section :

[tex]v1 \times a1 = v2 \times a2[/tex]

5 m/s * 1.76cm^2 = v2 * 7.06cm^2

[tex]v2 = 1.24 \: m {s}^{ - 1} [/tex]

At a point where the pipe diameter is 3.00 cm, the velocity is 1.25 m/s.

Fluid Flow, a branch of fluid dynamics, is concerned with fluids. It involves the movement of a fluid under the influence of uneven forces. As long as unbalanced pressures are applied, this motion will persist.

Given parameters:

Initial velocity of the water: u = 5.00 m/s

Initial diameter of the pipe: d = 1.50 cm.

Final diameter of the pipe: D = 3.00 cm.

Final velocity of the water: v = ?

In fluid motion:

velocity×(diameter)² = constant

Hence, initial velocity × ( initial diameter)² = final velocity × ( final diameter)²

ud² = vD²

v = u (d/D)²

v= 5 × (1.50/3.0)²

v= 5/2²

v= 5/4

v= 1.25 m/s.

Hence, at a point where the pipe diameter is 3.00 cm, the velocity is 1.25 m/s.

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