A rock weighing 20 N (mass = 2 kg) is swung in a horizontal circle of radius 2 m at a constant speed of 6 m/s. What is the tension in the cord?

Answer:

Explanation:

If the river moves towards the south at 3m/s and the both moves towards the east at 4.0m/s, the speed of the boat relative to me will be the resulting displacement of both velocities of the river and that of the boat. This can be gotten using pythagoras theorem.

Let Vr be the relative speed. According to the theorem;

[tex]V_r^2 = V_s^2 + V_e^2\\\\V_r^2 = 3.0^2 + 4.0^2\\\\V_r^2 = 9+16\\\\V_r^2 = \sqrt{25}\\ \\V_r = 5.0m/s[/tex]

Hence this relative speed is 5.0 m/s

Answer:

[tex]density \: = \frac{mass}{volume} [/tex]

1 / 5.587 is equal to 0.179 kg/m³

Hope it helps:)

Answer:

The answer is

Explanation:

Density of a substance is given by

[tex]Density \: = \frac{mass}{volume} [/tex]

From the

mass = 1 kg

volume = 5.583 m³

Substitute the values into the above formula

We have

[tex]Density \: = \frac{1 \: kg}{5.583 \: {m}^{3} } [/tex]

We have the final answer as

Hope this helps you

Answer:

E = 0    r <R₁

Explanation:

If we use Gauss's law

      Ф = ∫ E. dA = [tex]q_{int}[/tex] / ε₀

in this case the charge is distributed throughout the spherical shell and as we are asked for the field for a radius smaller than the radius of the spherical shell, therefore, THERE ARE NO CHARGES INSIDE this surface.

Consequently by Gauss's law the electric field is ZERO

           E = 0    r <R₁

Answer:

I will explain the concept of magnetic field and how it can be calculated.

Explanation:

The formula for magnetic field at the center of a loop is given as

B = μ[tex]_{o}[/tex]I / 2R

where B is the magnetic field

R is the radius of the loop

I is the current

and μ[tex]_{o}[/tex] is the magnetic permeability of free space which is a constant 4π × [tex]10^{-7}[/tex] newtons/ampere²

If the magnetic field at the center of the loop is 0, then μ[tex]_{o}[/tex]I = 0

I = 0 which means there will be no current flow in the loop.

Answer:

The position on the y-axis where the magnetic field is zero is at y = 10 cm

Explanation:

The magnetic field B due to a long straight wire carrying a current, i at a distance R from the wire is given by

B = μ₀i/2πR

Now, let y be the point where the magnetic fields of both wires are equal.

So, the magnetic field due to wire 1 carrying current i₁ = 73 A is

B₁ = μ₀i₁/2π(y - 3) and

the magnetic field due to wire 2 carrying current i₂ = 31 A is

B₂ = μ₀i₂/2π(13 - y)

At the point where the magnetic field is zero, B₁ = B₂. So,

μ₀i₁/2π(y - 3) = μ₀i₂/2π(13 - y)

cancelling out μ₀ and 2π, we have

i₁/(x - y) = i₂/(13 - y)

cross-multiplying, we have

(13 - y)i₁ = (y - 3)i₂

Substituting the values of i₁ and i₂, we have

(13 - y)73 = (y - 3)31

949 - 73y = 31y - 93

Collecting like terms, we have

949 + 93 = 73y + 31y

1042 = 104y

dividing through by 104, we have

y = 1042/104

y = 10.02 cm

y ≅ 10 cm

So, the position on the y-axis where the magnetic field is zero is at y = 10 cm

Answer:

The resultant vector is [tex]\vec R = \vec A + \vec B = 7.196\,i + 6.464\,j[/tex].

Explanation:

First, each vector is determined in terms of absolute coordinates:

6-meter vector with direction: 30º north of east.

[tex]\vec A = (6\,m)\cdot (\cos30^{\circ} \,i + \sin 30^{\circ}\,j)[/tex]

[tex]\vec A = 5.196\,i + 3\,j[/tex]

4-meter vector with direction: 30º east of north.

[tex]\vec B = (4\,m)\cdot (\cos 60^{\circ}\,i + \sin 60^{\circ}\,j)[/tex]

[tex]\vec B = 2\,i + 3.464\,j[/tex]

The resultant vector is obtaining by sum of components:

[tex]\vec R = \vec A + \vec B = 7.196\,i + 6.464\,j[/tex]

The resultant vector is [tex]\vec R = \vec A + \vec B = 7.196\,i + 6.464\,j[/tex].

Answer:

206.8965517 n

Explanation:

First, we need to see that 60:29 is 2.078965517:1. Then we need to multiply the energy put 29 cm from the fulcrum by 2.078965517, giving us the end result of our answer.

Answer:

Using a process of testing ideas and gathering evidence

Explanation:

Answer:

95.7v

Explanation

Using Faraday's law of electromagnetic induction we know that rate of change in magnetic flux will induce EMF in closed loop

So it is given as

E= Ndစ/dt

E= N BA-0/ deta t

Given that

N = 58turns

B = 1.10T

A = 0.150m^²

Deta t= 0.1s

now we have

E = 58(1.10x0.150)/0.1

= 95.7v

Magnetic flux is decreasing, so the direction of the current will be to aid the decreasing flux $decrease= CLOCKWISE

Explanation:

Answer:

10m

Explanation:

Since Given frequency f= 1/t

and velocity ν=10 m/s

We know ν=λf

λ= ν/f

​ = 10/1/0.5

=5m

Since both the waves are similar but moves in opposite direction its total wavelength of the wave will be 10 m

Answer:

the force decreases by a factor of 4

Explanation:

The expression for the law of universal gravitation is

          F = [tex]G \frac{m_1m_2}{r^2}[/tex]

let's call the force Fo for the distance r

          F₀ = [tex]G \frac{m_1m_2}{r^2}[/tex]

They indicate that the distance doubles

          r ’= 2 r

we substitute

          F = [tex]G \frac{m_1m_2}{(r')^2}[/tex]

          F = [tex]G \frac{m_1m_2}{r^2} \ \frac{1}{4}[/tex]

          F = ¼ F₀

consequently the correct answer is that the force decreases by a factor of 4

If the distance between two large masses are doubled, the gravitational force between them weakens by a factor of 4.

Let the initial force be F

Let the initial distance apart be r

Thus, we can obtain the final force as follow:

Initial force (F₁) = F

Initial distance apart (r₁) = r

Final distance apart (r₂) = 2r

F = GM₁M₂ / r²

Fr² = GM₁M₂ (constant)

Thus,

F₁r₁² = F₂r₂²

Fr² = F₂(2r)²

Fr² = F₂4r²

Divide both side by 4r²

F₂ = Fr² /4r²

From the illustration above, we can see that when the distance (r) is doubled, the force (F) is decreased (i.e weakens) by a factor of 4

Learn more: https://brainly.com/question/975812

Answer:

 V = - 3.85 V

Explanation:

The electric potential of a continuous charge distribution is

       V = k ∫ dq / r

to find charge differential let's use the concept of linear density

        λ = dq / dx

       dq = λ dx

the distance from a load element to the point of interest

       x₀ = 23 cm = 0.23 m

       r = √ (x-x₀)² = x - x₀

we substitute

        v = k ∫ λ dx / (x-x₀)

we integrate and evaluate between x = 0 and x = l = 0.10 cm

       V = k λ [ln (x-x₀) - ln (-x₀)]

        V = k λ ln ((x-x₀) / x₀)

let's calculate

         V = 9 10⁹  0.74 10⁻⁹ ln ((0.23 - 0.10) / 0.23)

          V = - 3.85 V

Explanation:

Mary had 21 plants when she went on vacation.

When she got back, she only had 14 left alive.

We need to find the percent decrease in the number of plants.

Decrease in plants = 21 - 14 = 7

Percent decrease is given by :

[tex]\%=\dfrac{7}{21}\times 100\\\\\%=33.33\%[/tex]

So, there is 33% pf decrease in the number of plants.

Explanation:

EE = ½ kx²

EE = ½ (800 N/m) (6 m)²

EE = 14,400 J

Answer:

14,400 J

Explanation:

Its the answer

Answer:

The width of Center bright fringe is 10.2mm

Explanation:

Given that if

Y/ L << 1 then

Sin theta will be approx Y/L

So sin theta approx Y/L = lamda/a

Y= a x lambda/a

By substituting

1.9x 10^ -3m x 550*10^-9/ 0.2 x 10^-3m

= 5.2mm

But

Change in y = 2y = 10.4mm

Answer:

a. 2143 turns/m

b. 111.5 m

Explanation:

a. The minimum number of turns per unit length (N/L) can be found using the following equation:

[tex] B = \frac{\mu_{0}NI}{L} [/tex]

[tex] \frac{N}{L} = \frac{B}{\mu_{0}I} = \frac{3.50 \cdot 10^{-2} T}{4\pi \cdot 10^{-7} Tm/A*13.0 A} = 2143 turns/m [/tex]

Hence, the minimum number of turns per unit length is 2143 turns/m.

b. The total length of wire is the following:

[tex] N = 2143 turns/m*L = 2143 turns/m*46.0 \cdot 10^{-2} m = 986 turns [/tex]

Since each turn has length 2πr of wire, the total length is:

[tex] L_{T} = N*2\pi r = 986 turn*2*\pi*1.80 \cdot 10^{-2} m = 111.5 m [/tex]

Therefore, the total length of wire required is 111.5 m.

I hope it helps you!

Explanation:

In Single Slit Experiment:

The width of the central diffraction maximum is inversely proportional to the width of the slit.

Therefore, if we make the slit width smaller, the angle T(representing the angle between the wave ray to a point on the screen and the normal line between the slit and the screen) increases, giving a wider central band.

Answer:

prove that Sin^6 ϴ-cos^6ϴ=(2Sin^2ϴ-1)(cos^2ϴ+sin^4ϴ)

please sove step by step with language it is opt maths question

Answer:

1190 N

Explanation:

Force: This can be defined as the product of mass and velocity. The unit of force is Newton(N).

From the question,

F = ma................. Equation 1

Where F = average force, m = mass, a = acceleration.

But,

a = (v-u)/t................ Equation 2

Where v = final velocity, u = initial velocity, t = time.

Substitute equation 2 into equation 1

F = m(v-u)/t.............. Equation 3

Given: m = 70 kg, v = 1.7 m/s, u = 0 m/s (from rest), t = 0.1 s.

Substitute into equation 3

F = 70(1.7-0)/0.1

F = 1190 N.

Explanation:

The De-Broglie wavelength is given by :

[tex]\lambda=\dfrac{h}{p}[/tex]

h is Planck's constant

p is momentum

In this case, an electron and a proton both moving at nonrelativistic speeds have the same de Broglie wavelength. Mass of electron and proton is different. It means their velocity and energy are different.

Only momentum is the factor that remains same for both particles i.e. momentum.

Answer:

Explanation:

The fundamental frequency of the string  f₀ is expressed as f₀ = V/4L where V is the speed experienced by the string.

[tex]V = \sqrt{\frac{T}{\mu} }[/tex] where T is the tension in the string and  [tex]\mu[/tex] is the density of the string

Given T = 600N and [tex]\mu[/tex] = 0.015 g/cm  = 0.0015kg/m

[tex]V = \sqrt{\frac{600}{0.0015} }\\ \\V = \sqrt{400,000}\\ \\V = 632.46m/s[/tex]

The next is to get the length L of the string. Since the string is stretched and fixed at both ends, 200 cm apart, then the length of the string in metres is 2m.

L = 2m

Substituting the derived values into the formula f₀ = V/2L

f₀ = 632.46/2(2)

f₀ = 632.46/4

f₀ = 158.12 Hertz

Hence the fundamental frequency of the string is 158.12 Hertz

Answer:

first order date and most recent order date

Explanation:

it was switched. column 5 should be most recent order date because it's 2020 while column 6 should be first order date because it was in 2019

Answer:

Δx = 4.68 x 10⁻³ m = 4.68 mm

Explanation:

The distance between the consecutive maxima, in Young's Double Slit Experiment is given bu the following formula:

Δx = λD/d

So, the distance between the eighth order maximum and the fourth order maximum on the screen will be given as:

Δx = 4λD/d

where,

Δx = distance between eighth order maximum and fourth order maximum=?

λ = wavelength = 487 nm = 4.87 x 10⁻⁷ m

d = slit separation = 0.2 mm = 2 x 10⁻⁴ m

D = Distance between slits and screen = 48 cm = 0.48 m

Therefore,

Δx = (4)(4.87 x 10⁻⁷ m)(0.48 m)/(2 x 10⁻⁴ m)

Δx = 4.68 x 10⁻³ m = 4.68 mm

Answer:

a) k = 95.54 N / m,   c =   19.55 , b)      m₃ = 0.9078 kg

Explanation:

In a simple harmonic movement with friction, we can assume that this is provided by the speed

          fr = -c v

when solving the system the angular value remains

          w² = w₀² + (c / 2m)²

They give two conditions

1) m₁ = 1 kg

     f₁ = 1.1 Hz

the angular velocity is related to frequency

         w = 2π f₁

Let's find the angular velocity without friction is

         w₂ = k / m₁

we substitute

        (2π f₁)² = k / m₁ + (c / 2m₁)²

2) m₂ = 2 kg

    f₂ = 0.8 Hz

        (2π f₂)² = k / m₂ + (c / 2m₂)²

we have a system of two equations with two unknowns, so we can solve it

we solve (c / 2m)² is we equalize the expression

           (2π f₁)² - k / m₁ = (2π f₂²) 2 - k / m₁

           k (1 / m₂ - 1 / m₁) = 4π² (f₂² - f₁²)

           k = 4π² (f₂² -f₁²) / (1 / m₂ - 1 / m₁)

a) Let's calculate

           k = 4 π² (0.8² -1.1²) / (½ -1/1)

           k = 39.4784 (1.21) / (-0.5)

           k = 95.54 N / m

now we can find the constant of friction

              (2π f₁) 2 = k / m₁ + (c / 2m₁)²

           c2 = ((2π f₁)² - k / m₁) 4m₁²

           c2 = (4ππ² f₁² - k / m₁) 4 m₁²

let's calculate

           c² = (4π² 1,1² - 95,54 / 1) 4 1²

           c² = (47.768885 - 95.54) 8

           c² = -382.1689

           c =   19.55    

b) f₃ = 0.2 Hz

   m₃ =?

              (2πf₃)² = k / m₃ + (c / 2m₃) 2

we substitute the values

              (4π² 0.2²) = 95.54 / m₃ + 382.1689 2/4 m₃²

              1.579 = 95.54 / m₃ + 95.542225 / m₃²

let's call

              x = 1 / m₃

              x² = 1 / m₃²

- 1.579 + 95.54 x + 95.542225 x² = 0

              60.5080 x² + 60.5080 x -1 = 0

                x² + x - 1.65 10⁻² = 0

                  x = [1 ±√ (1- 4 (-1.65 10⁻²)] / 2

                  x = [1 ± 1.03] / 2

                  x₁ = 1.015 kg

                  x₂ = -0.015 kg

Since the mass must be positive we eliminate the second results

                  x₁ = 1 / m₃

                 m₃ = 1 / x₁

                  m₃ = 1 / 1.1015

Answer:

The rms current is 0.3112 A.

Explanation:

Given that,

Suppose, The capacitance is 170 μF and the inductance is 2.94 mH. The resistance in the top branch is 278 Ohms, and in the bottom branch is 151 Ohms. The potential of the power supply is 47 V .

We know that,

When the frequency is very large then the capacitance can be treated as a short circuit and inductance as open circuit.

So,

We need to calculate the rms current

Using formula of current

[tex]I=\dfrac{V}{R}[/tex]

Where, V = voltage

R = resistance

Put the value into the formula

[tex]I=\dfrac{47}{151}[/tex]

[tex]I= 0.3112 \ A[/tex]

Hence, The rms current is 0.3112 A.

Complete question is;

A force stretches a wire by 0.60 mm. A second wire of the same material has the same cross section and twice the length.

a) How far will it be stretched by the same force?

b) A third wire of the same material has the same length and twice the diameter as the first. How far will it be stretched by the same force?

Answer:

0.15 mm

Explanation:

According to Hooke's Law,

E = Stress(σ)/Strain(ε)

Where E is youngs modulus

Formula for stress is;

Stress(σ) = Force(F)/Area(A)

Formula for strain is;

Strain(ε) = Change in length/original length = (Lf - Li)/Li

We are also told that a second wire of the same material has the same cross section and twice the length.

Thus;

Rearranging Hooke's Law to get the constants on one side, we have;

F/(AE) = ε

Thus from the conditions given;

ε1 = 0.6/Li

ε2 = (Change in length)/(2*Li)

And ε1 = ε2

Thus;

0.6/Li = Change in length/(2*Li)

Li will cancel out and we now have;

Change in length = 2 × 0.6 = 1.2 mm

Finally, we are told A third wire of the same material has the same length and twice the diameter as the first.

Area of a circle;A1 = πd²/4

Now, we are told d is doubled.

Thus, new area of the new circle is;

A2 = π(2d)²/4 = πd²

Rearranging Hooke's Law,we have;

F/A = εE

Since F and E are now constants, we have;

F/E = constant = Aε

Thus;

A1(ε1) = A2(ε2)

A1 = πd²/4

e1 = 0.60/Li

A2 = πd²

e2 = Change in length/Li

Thus;

((πd²/4) × 0.6)/Li = (πd² × Change in length)/ Li

Rearranging, Li and πd² will cancel out to give;

0.6/4 = Change in length

Change in length = 0.15 mm

Given that,

A charged particle enters a magnetic field with an angle theta .

If [tex]\theta=90^{\circ}[/tex]

We know that,

If the angle is 90° then the charged particle enters perpendicular to the B.

B is magnetic field.

The charged particle will be follow of the circular path.

If the angle is greater than 0 and less than 90° then the charged particle will be show the helical path.

Hence, This is required answer.

Answer:

Current =,charge / time

Charge = e = 1.6E-19 coulombs

t = T time for 1 revolution  (period)

v = S / T = distance traveled in 1 revolution / time for 1 revolution

T = S / v = 2 pi * 4.76E-10 / 2.43E5 = 1.23E-14

I = Q / T = 1.6E-19 / 1.23E-14 = 1.30E-5

Answer:

Explanation:

When plastic balloon is rubbed with wool , charges are created on both balloon and silk in equal amount . Rubber balloon will acquire negative charge and silk will acquire positive charge .

Now when balloon is brought near a wall , there is induction of charge on the wall due to charge on the balloon . On the near surface of wall positive charge is produced and on the surface deep inside the wall negative charge is produced . The charge deep inside goes inside the earth but the positive charge near the surface of wall can not escape . It remains trapped by negative charge on the balloon .

hence there is mutual attraction between balloon and surface of wall is just like attraction between opposite charges . But once the ballon due to mutual attraction comes in contact with the wall , the charge on balloon and on wall neutralises each other and hence after some time the balloon falls off from the wall on the ground . It does not remain attracted to wall for ever . It happens due to neutralisation of charges on balloon and wall .