A 0.065 kg ingot of metal is heated to 210◦C and then is dropped into a beaker containing 0.377 kg of water initially at 26◦C. If the final equilibrium state of the mixed system is 28.4 ◦C, find the specific heat of the metal. The specific heat of water is 4186 J/kg · ◦ C. Answer in units of J/kg · ◦ C.

Answer:

Net forces which pushes the window is 30342.78 N.

Explanation:

Given:

Dimension of the office window.

Length of the window = [tex]3.1[/tex] m

Width of the window = [tex]2.1[/tex] m

Area of the window = [tex](3.1\times 2.1) = 6.51\ m^2[/tex]

Difference in air pressure = Inside pressure - Outside pressure

                                           = [tex](1.0-0.954)[/tex] atm = [tex]0.046[/tex] atm

Conversion of the pressure in its SI unit.

⇒  [tex]1[/tex] atm = [tex]101325[/tex] Pa

⇒ [tex]0.046[/tex] atm = [tex]0.046\times 101325 =4660.95[/tex] Pa

We have to find the net force.

We know,

⇒ Pressure = Force/Area

⇒ [tex]Pressure=\frac{Force }{Area}[/tex]

⇒ [tex]Force =Pressure\times Area[/tex]

⇒ Plugging the values.

⇒ [tex]Force =4660.95\times 6.51[/tex]

⇒ [tex]Force=30342.78[/tex] Newton (N)

So,

The net forces which pushes the window is 30342.78 N.

Answer:

F'=1/9*F0

Explanation:

F0 is the gravitational force between the particles. When the distance is triplicated we have that

[tex]F'=G\frac{m_1m_2}{(3r)^{2}}[/tex]

where r is the distance before the particles are separated, m1 and m2 are the masses their masses and G is the Canvendish's constant.

By some algebra we have

[tex]F'=\frac{1}{9}G\frac{m_1m_2}{r^2}=\frac{1}{9}F_0[/tex]

hope this helps!!

Answer:

F₁ = [tex]\frac{1}{9}[/tex]F₀

Explanation:

Newton's law of universal gravitation states that the force of attraction or repulsion, F, between two particles of masses M₁ and M₂ is directly proportional to the product of these particles and inversely proportional to the square of the distance, r, between the two particles. i.e

F ∝ M₁M₂ / r²

F = GM₁M₂ / r²            --------------------(i)

Where;

G is the constant of proportionality.

From equation (i), since the force is inversely proportional to the square of the distance, holding other variables constant, the equation can be reduced to;

F = k / r²

This implies that;

Fr² = k          -------------------(ii)

Now, according to the question;

F = F₀

Substitute this into equation (ii) as follows;

F₀ r² = k       ----------------(iii)

Also, when the distance of separation, r, is trippled i.e r becomes 3r;

F = F₁

Substitute these values into equation (ii) as follows;

F₁(3r)² = k

9F₁r² = k                ---------------(iv)

Substitute the value of k in equation (iii) into equation (iv) as follows;

9F₁r² = F₀ r²             --------------(v)

Cancel r² on both sides of equation (v)

9F₁ = F₀

Now make F₁ subject of the formula

F₁ = [tex]\frac{1}{9}[/tex]F₀

Therefore, the new force F₁ = [tex]\frac{1}{9}[/tex]F₀

Complete Question:

In the same configuration of the previous problem 3, four long straight wires are perpendicular to the page, and their cross sections form a square of edge length a = 13.5 cm. Each wire carries 7.50 A, and the currents are out of the page in wires 1 and 4 and into the page in wires 2 and 3.

a) Draw a diagram in a (x,y) plane of the four wires with wire 4 perpendicular to the origin. Indicate the current's directions.

b) Draw a diagram of all magnetic fields produced at the position of wire 3 by the other three currents.

c) Draw a diagram of all magnetic forces produced at the position of wire 3 by the other three currents.

d) What are magnitude and direction of the net magnetic force per meter of wire length on wire 3?

Answer:

force, 1.318 ₓ 10⁻⁴

direction, 18.435°

Explanation:

The attached file gives a breakdown step by step solution to the questions

Answer:

v = 1176.23 m/s

y = 741192.997 m = 741.19 km

Explanation:

Given

M₀ = 9 Kg  (Initial mass)

me = 0.225 Kg/s   (Rate of fuel consumption)

ve = 1980 m/s    (Exhaust velocity relative to rocket, leaving at atmospheric pressure)

v = ? if t = 20 s

y = ?

We use the equation

v = ∫((ve*me)/(M₀ - me*t)) dt - ∫g dt     where t ∈ (0, t)

⇒   v = - ve*Ln ((M₀ - me*t)/M₀) - g*t

then we have

v = - 1980 m/s*Ln ((9 Kg - 0.225 Kg/s*20 s)/(9 Kg)) - (9.81 m/s²)(20 s)

v = 1176.23 m/s

then we apply the formula

y = ∫v dt = ∫(- ve*Ln ((M₀ - me*t)/M₀) - g*t) dt

⇒   y = - ve* ∫ Ln ((M₀ - me*t)/M₀) dt - g*∫t dt

⇒   y = - ve*(Ln((M₀ - me*t)/M₀)*t + (M₀/me)*(M₀  - me*t - M₀*Ln(M₀ - me*t))) - (g*t²/2)

For t = 20 s   we have

y = Ln((9 Kg - 0.225 Kg/s*20 s)/9 Kg)*(20 s) + (9 Kg/0.225 Kg/s)*(9 Kg  - 0.225 Kg/s*20 s - 9 Kg*Ln(9 Kg - 0.225 Kg/s*20 s)) - (9.81 m/s²*(20 s)²/2)

⇒   y = 741192.997 m = 741.19 km

The graphs are shown in the pics.

Following are the solution to the given points:

Given:

[tex]\to v_o =1980\ \frac{m}{s}\\\\[/tex]

[tex]\to a=g= -9.8 \frac{m}{s} \ \text{(Diection downwords )}[/tex]

Solution:

Using formula:

[tex]\to V= V_o +at\\\\[/tex]

[tex]\to v^2-v^2_0= 2ad \\\\[/tex]

For point a:

[tex]\to V= V_o +at\\\\[/tex]

        [tex]=1980+ (-9.8) (20) \\\\[/tex]

        [tex]= 1980-196 \\\\ =1784\ \frac{m}{s}\\\\[/tex]

The velocity at the conclusion of a [tex]20[/tex] second is [tex]\bold{1784\ \frac{m}{s}}\\\\[/tex].

For point b:

Using formula:

[tex]\to v^2-v^2_0= 2ad \\\\[/tex]

[tex]\to d=\frac{v^2_0 -V^2}{2a}\\\\[/tex]

       [tex]= \frac{(1980)^2 - (1784)^2}{2\times 9.8} \\\\ =\frac{3920400-3182656}{19.6}\\\\= \frac{737744}{17.6}\\\\ =37640\ m \\\\ = 37.64\ km\\\\[/tex]

In [tex]20[/tex] seconds, the total distance traveled is [tex]401737 \ \ m (or \ 401.737\ \ km)[/tex].  

For point c:

Please find the attached file.

Answer:

True

Explanation:

Gravity is a very important force. Every object in space exerts a gravitational pull on every other, and so gravity influences the paths taken by everything traveling through space. It is the glue that holds together entire galaxies. It keeps planets in orbit.

To solve this problem, we will apply the concepts related to the kinematic equations of linear motion, which define speed as the distance traveled per unit of time. Subsequently, the wavelength is defined as the speed of a body at the rate of change of its frequency. Our values are given as,

[tex]\text{Length of the string} = L = 4.32 m[/tex]

[tex]\text{Frequency of the wave} = f = 75 Hz[/tex]

[tex]\text{Time taken to reach the other end} = t = 0.5 s[/tex]

Velocity of the wave,

[tex]V = \frac{L}{t}[/tex]

[tex]V = \frac{4.32 m}{0.5s}[/tex]

[tex]V = 8.64m/s[/tex]

Wavelength of the wave,

[tex]\lambda = \frac{V}{f}[/tex]

[tex]\lambda = \frac{8.64m/s}{75Hz}[/tex]

[tex]\lambda = 0.1152m[/tex]

[tex]\lambda = 11.52cm[/tex]

Therefore the wavelength of the waves on the string is 11.53 cm

11.52cm

The velocity, v, of a wave undergoing a simple harmonic motion is related to the wavelength, λ, and frequency, f, of the wave as follows;

v =  f x λ         -----------------(i)

But;

The velocity is also given as the ratio of the length, l, of the body producing the wave to the time taken, t, to undergo the motion. i.e

v = [tex]\frac{l}{t}[/tex]         --------------(ii)

Now, substitute the value of v in equation (ii) into equation (i) as follows;

[tex]\frac{l}{t}[/tex] = f x λ           ----------------(iii)

From the question,

l = 4.32m

t = 0.5s

f = 75Hz

Substitute these values into equation (iii) as follows;

[tex]\frac{4.32}{0.5}[/tex] = 75 x λ

Make λ subject of the formula;

λ = [tex]\frac{4.32}{0.5*75}[/tex]

λ = 0.1152m

Convert the value to cm by multiplying by 100

λ = 0.1152 x 100cm = 11.52cm

Therefore, the wavelength of the waves on the string is 11.52cm

Answer:

laser's wavelength λ = 597.4 nm

Explanation:

Given:

Slit spacing, d = 1.17mm

Tenth bright fringe y = 4.57cm

Distance from slits, D = 8.95m

n = 10

λ = (d * y) / (D * n)

λ = (1.17x10⁻³ * 4.57x10⁻²) / (8.95 x 10)

λ = 5.3469‬x10⁻⁵ / 8.95x10¹

λ = 0.5974 x 10⁻⁵⁻¹

λ = 0.5974 x 10⁻⁶ m

λ = 597.4 x 10⁻⁹ m

λ = 597.4 nm

Answer:

[tex]0.276I_0[/tex]

Explanation:

When unpolarized light passes through a polarizer, only the component of the light vibrating in the direction parallel to the axis of the polarizer passes through: therefore, the intensity of light is reduced by half, since only 1 out of 2 components passes through.

So, after the first polarizer, the intensity of light passing through is:

[tex]I_1=\frac{I_0}{2}[/tex]

Where [tex]I_0[/tex] is the initial intensity of the unpolarized light.

Then, the light (which is now polarized) passes through the second polarizer. Here, the intensity of the light passing through the second polarizer is given by Malus Law:

[tex]I_2=I_1 cos^2 \theta[/tex]

where:

[tex]\theta[/tex] is the angle between the axes of the two polarizers

In this problem the angle is

[tex]\theta=42^{\circ}[/tex]

So the intensity after of light the 2nd polarizer is:

[tex]I_2=I_1 (cos 42^{\circ})^2=\frac{I_0}{2}(cos 42^{\circ})^2=0.276I_0[/tex]

The  intensity of the beam after it has passed through the second polarizer should be 0.276 I0.

Since

after the first polarizer, the intensity of light should be

I1 = I0/2

Here,

Io should be  initial intensity of the unpolarized light

Now

The intensity should be

= I0/2(cos 42)

= 0.276 I0

Here theta be 42 degrees

Learn more about intensity here: https://brainly.com/question/16463859

Answer:

The angle of incidence is greater than the angle of refraction

Explanation:

Refraction occurs when a light wave passes through the boundary between two mediums.

When a ray of light is refracted, it changes speed and direction, according to Snell's Law:

[tex]n_1 sin \theta_1 = n_2 sin \theta_2[/tex]

where :

[tex]n_1[/tex] is the index of refraction of the 1st medium

[tex]n_2[/tex] is the index of refraction of the 2nd medium

[tex]\theta_1[/tex] is the angle of incidence (the angle between the incident ray and the normal to the boundary)

[tex]\theta_2[/tex] is the angle of refraction (the angle between the refracted ray and the normal to the boundary)

In this problem, we have a ray of light passing from air into clear plastic. We have:

[tex]n_1=1.00[/tex] (index of refraction of air)

[tex]n_2=1.50[/tex] approx. (index of refraction in clear plastic)

Snell's Law can be rewritten as

[tex]sin \theta_2 =\frac{n_1}{n_2}sin \theta_1[/tex]

And since [tex]n_2>n_1[/tex], we have

[tex]\frac{n_1}{n_2}<1[/tex]

And so

[tex]\theta_2<\theta_1[/tex]

Which means that

The angle of incidence is greater than the angle of refraction

Answer:

(a) reaction at each front wheel is 5272N (upward)

(b) force between boulder and pallet is 4124N (compression)

Explanation:

Acceleration of the truck [tex]a_{t[/tex] = 1 m/[tex]s^{2}[/tex]  (to the left)

when the truck moves 1 m to the left, the boulder is B and pallet A are raised 0.5 m, then,

[tex]a_{A}[/tex] =  0.5 m/[tex]s^{2}[/tex] (upward) , [tex]a_{B}[/tex] =  0.5 m/[tex]s^{2}[/tex] (upward)

Let T be tension in the cable

pallet and boulder: ∑fy = ∑(fy)eff = 2T- ([tex]m_{A}[/tex] + [tex]m_{B}[/tex])g =  ([tex]m_{A}[/tex] + [tex]m_{B}[/tex])[tex]a_{B}[/tex]

                                  = 2T- (400 + 50)*(9.81 m/[tex]s^{2}[/tex]) = (400 + 50)*(0.5 m/[tex]s^{2}[/tex])

                        T = 2320N

Truck:  [tex]M_{R}[/tex] = ∑([tex]M_{R}[/tex])eff: = [tex]-N_{f}[/tex] (3.4m) + [tex]m_{T}[/tex] (2.0m) - T (0.6m)= [tex]m_{T} a_{T}[/tex] (1.0m)

Nf = (2.0m)(2000 kg)(9.81 m/[tex]s^{2}[/tex] )/3.4m -  (0.6 m)(2320 N)/3.4m + (1.0 m)(2000 kg)(1.0 m/[tex]s^{2}[/tex])  = 11541.2N - 409.4N - 588.2N = 10544N

∑fy (upward) = ∑(fy)eff: [tex]N_{f}[/tex] + [tex]N_{R}[/tex] - [tex]m_{T}[/tex]g = 0

                                       10544 + [tex]N_{R}[/tex]  - (2000kg)(9.81 m/[tex]s^{2}[/tex] ) = 0

                [tex]N_{R}[/tex] = 9076N

   ∑fx (to the left) = ∑(fx)eff:  [tex]F_{R}[/tex] - T = [tex]m_{T} a_{T}[/tex]

                                      [tex]F_{R}[/tex] = 2320N + (2000kg)(9.81 m/[tex]s^{2}[/tex] ) = 4320N

(a) reaction at each front wheel:

1/2 [tex]N_{f}[/tex] (upward): 1/2 (10544N) = 5272N (upward)

(b) force between boulder and pallet:

∑fy (upward) = ∑(fy)eff: [tex]N_{B}[/tex] + [tex]M_{B}[/tex]g - [tex]m_{B}[/tex][tex]a_{B}[/tex]

            [tex]N_{B}[/tex] = (400kg)(9.81 m/[tex]s^{2}[/tex]) + (400kg)(0.5 m/[tex]s^{2}[/tex]) = 4124N (compression)

Answer:

a) The reaction at each of the fron wheels is 5266.1 N

b) The force between the boulder and the pallet is 4120 N

Explanation:

The acceleration of truck is:

[tex]a_{T} =a_{A} +a_{B}[/tex]

Where

aA = acceleration of pallet = ?

aB = acceleration of boulder = ?

aA = aB

aT = acceleration of truck = 1 m/s²

[tex]1=a_{A} +a_{A}\\a_{A}=a_{B}=0.5m/s^{2}[/tex]

From diagram 1 and 2, the system of external forces is:

∑Fy = ∑(Fy)ef (eq.1)

From diagram 1:

∑Fy = 2T - g(mA + mB)

Where T = tension force

mA = mass of pallet = 50 kg

mB = mass of boulder = 400 kg

From diagram 2:

∑(Fy)ef = aB(mA + mB)

Substituting into equation 1:

[tex]2T-g(m_{A} +m_{B} )=a_{B} (m_{A} +m_{B} )\\T=\frac{a_{B}(m_{A} +m_{B}) +g(m_{A} +m_{B} ) }{2} =\frac{0.5(50+400)+9.8(50+400)}{2} =2317.5N[/tex]

From diagram 3 and 4, represents the system of external forces:

∑MR = ∑(MR)ef (eq. 2)

From diagram 3:

∑MR = -N(2 + 1.4) + mTg(2) - T(0.6)

Where

N = normal force

mT = mass of truck = 2000 kg

From diagram 4:

∑(MR)ef = mTaT

Substituting into equation 2:

[tex]-N(2+1.4)+m_{T} g(2)-T(0.6)=m_{T} a_{T} \\-N(3.4)+(2000*9.8*2)-(2317.5*0.6)=2000*1\\N=\frac{(2000*9.8*2)-(2317.5*0.6)-2000}{3.4} =10532.2N[/tex]

From diagram 3 and 4:

∑Fy = ∑(Fy)ef

[tex]N+N_{R} -m_{T} g=0\\10532.2+N_{R}-(2000*9.8)=0\\N_{R}=9067.8N[/tex]

a) The reaction at each of the front wheels is:

Rf = N/2 = 10532.2/2 = 5266.1 N

b) From diagram 5 and 6:

∑Fy = ∑(Fy)ef

[tex]N_{B} +m_{B} g=m_{B} a_{B} \\N_{B}-(400*9.8)=(400*0.5)\\N_{B}=4120N[/tex]

Answer:

[tex]\theta = 0.195^0[/tex]

Explanation:

wavelength [tex]\lambda = 648 nm \ = 648*10^{-9}m[/tex]

d = 0.190 mm = 0.190 × 10⁻³ m

D = 1.91 m

By using the formula:

[tex]dsin \theta = n \lambda\\\\\theta = sin^{(-1)}(\frac{n \lambda}{d})\\\\\\\theta = sin^{(-1)}(\frac{1*648*10^{-9}}{0.190*10^{-3}})[/tex]

[tex]\theta = 0.195^0[/tex]

The first maximum will appear at an angle [tex]\theta = 0.195^0[/tex] from the beam axis

Answer:

a)radius of the arc of the teardrop at the top is 10.58m

b)T = mg

c) the centipetal acceleration at the top is 14.5m/s

Explanation:

Part A

Given that roller coaster is of tear drop shape

So the speed at the top is given as

v = 14.4 m/s

acceleration at the top is given as

a = 2 g

[tex]a = 2(9.8) = 19.6 m/s^2[/tex]

now we know the formula of centripetal acceleration as

[tex]a = \frac{v^2}{R}\\[/tex]

[tex]19.6 = \frac{14.4^2}{R}\\R = 10.58 m[/tex]

Part B

now

the total mass of the car and the ride is M

Let the force exerted by the track be n

By Newton law

[tex]n +Mg =\frac{Mv^2}{r} \\\\n=\frac{Mv^2}{r} -Mg\\\\=M(\frac{v^2}{r}-g )\\\\=M(2g-g)\\\\T=Mg[/tex]

Part C

If the radius of the loop is 21.4 m

speed is given by same v = 14.4 m/s

now the acceleration is given as

[tex]a = \frac{v^2}{R}[/tex]

[tex]a = \frac{14.4^2}{21.4} \\\\= 9.69 m/s^2[/tex]

Now for normal force at the top is given by force equation

[tex]F_n + mg = ma\\F_n = m(a-g)[/tex]

The force exerted by the rail is less than zero because acceleration is less than 9.69m/s²

So the normal force would have to point away from the centre, For safe ride this normal force must be positive i.e [tex]a \prec g[/tex]

[tex]\frac{v^2}{r} \prec \sqrt{g} \\\\v = \sqrt{rg} \\\\v = \sqrt{21.4 \times 9.8} \\\\v = 14.5m/s[/tex]

Answer:

 W = 1.49 10⁻¹¹ kg

Explanation:

For this problem, let's use Newton's equation of equilibrium

           F - W = 0

            F = W              (1)

Strength can be found from the definition of pressure

        P = F / A

        F = P A

The radiation pressure for a reflective surface is

            P = 2 I / c)

We substitute in equation 1

         2 I / c  A = W

The intensity is defined by the ratio of the power between the area

          I = P / A

          P = I A

We substitute

          2 P / c = W

           W = 2  2.24 10-3 / 3 108

           W = 1.49 10⁻¹¹ kg

Answer:

See the explaination for the details.

Explanation:

Gauss Law states that the total electric flux out of a closed surface is equal to the charge enclosed divided by the permittivity. The electric flux in an area is defined as the electric field multiplied by the area of the surface projected in a plane and perpendicular to the field.

According to the Gauss law, the total flux linked with a closed surface is 1/ε0 times the charge enclosed by the closed surface.

Please kindly check attachment for the step by step explaination of the answer.

Answer: (a) Smaller charge is [tex]2.7 \times 10^{-5} C[/tex] and larger charge is [tex]11.7 \times 10^{-5} C[/tex].

(b) Smaller charge is [tex]-11.4 \times 10^{-5}[/tex] and larger charge is [tex]9.1 \times 10^{-5}[/tex].

Explanation:

(a) When both the spheres have same charge then force is repulsive in nature as like charges tend to repel each other.

Therefore, total charge on the two non-conducting spheres will be calculated as follows.

        [tex]Q_{1} + Q_{2} = 90 \mu \times \frac{10^{-6}C}{1 \muC}[/tex]

                      = [tex]9 \times 10^{-5} C[/tex]

Therefore, force between the two spheres will be calculated as follows.

        F = [tex]k\frac{Q_{1}Q_{2}}{r^{2}}[/tex]

       12 N = [tex]\frac{(9 \times 10^{9} Nm^{2}/C^{2})Q_{1}Q_{2}}{(0.28 m^{2})}[/tex]

       [tex]Q_{1}Q_{2} = 0.104 \times 10^{-9} C^{2}[/tex]

or,     [tex]Q_{1}(9 \times 10^{-5} - Q_{1}) = 0.104 \times 10^{-9} C^{2}[/tex]

    [tex]9 \times 10^{-5}Q_{1} - Q^{2}_{1} = 0.104 \times 10^{-9} C^{2}[/tex]

    [tex]Q^{2}_{1} - 9 \times 10^{-5}Q_{1} + 0.104 \times 10^{-9} = 0[/tex]

        [tex]Q_{1} = 11.7 \times 10^{-5} C, 2.7 \times 10^{-5} C[/tex]

This means that smaller charge is [tex]2.7 \times 10^{-5} C[/tex] and larger charge is [tex]11.7 \times 10^{-5} C[/tex].

(b)  When force is attractive in nature then it means both the charges are of opposite sign.

Hence, total charge on the non-conducting sphere is as follows.

      [tex]Q_{1} + (-Q_{2}) = 90 \mu \times \frac{10^{-6}C}{1 \muC}[/tex]

      [tex]Q_{1} - Q_{2} = 9 \times 10^{-5} C[/tex]

Now, force between the two spheres is calculated as follows.

    F = [tex]k\frac{Q_{1}Q_{2}}{r^{2}}[/tex]

    12 N = [tex]\frac{(9 \times 10^{9} Nm^{2}/C^{2})Q_{1}Q_{2}}{(0.28 m^{2})}[/tex]

   [tex]Q_{1}Q_{2} = 0.104 \times 10^{-9} C^{2}[/tex]

   [tex]Q_{1}(Q_{1} - 9 \times 10^{-5}) = 0.104 \times 10^{-9} C^{2}[/tex]

    [tex]Q^{2}_{1} - 9 \times 10^{-5}Q_{1} = 0.104 \times 10^{-9} C^{2}[/tex]

        [tex]Q_{1} = -11.4 \times 10^{-5}, 9.1 \times 10^{-5}[/tex]

Hence, smaller charge is [tex]-11.4 \times 10^{-5}[/tex] and larger charge is [tex]9.1 \times 10^{-5}[/tex].

(a). The charges on one sphere is [tex]0.89\mu C[/tex]  and [tex]1.18 \mu C[/tex] on other sphere.

(b). If the force were attractive then charge on both sphere will be in opposite sign.

The electrostatic force is given as,

                 [tex]F=k\frac{Q_{1}Q_{2}}{r^{2} }[/tex]

Where ,

Given that, [tex]F=12N,r=28cm=0.28m,Q_{1}+Q_{2}=90[/tex]

substitute values in above relation.

             [tex]12=\frac{9*10^{9} *Q_{1}(90-Q_{1})}{(0.28)^{2} } \\\\Q_{1}^{2}-(90*10^{-6})Q_{1} +1.05*10^{-10}=0\\ \\Q_{1}=8.88*10^{-5}=0.89 \mu C\\ \\Q_{2}=1.18*10^{-6} =1.18\mu C[/tex]

If the force were attractive then charging on both sphere will be in opposite sign.

Learn more about the charge here:

https://brainly.com/question/25922783

Answer:

(e) The direction of the net force is negative Z direction.

Explanation:

Let the direction of the particle (positive y) represents direction of current

Let direction of the net torque (negative x) represents direction of the field

Let direction of net force be direction of force

Apply Fleming's right hand rule, which states that when the thumb, the middleave finger and forefinger are held mutually at right angle to each other, with the thumb pointing in the direction of current (particle, positive y ), and the forefinger pointing in the direction of the field (torque, negative x), then the middle finger will be pointing in the direction of the force.

To find direction of the net force:

Y direction is already represented, negative X direction is also represented and finally force will be in Z - direction

Since we know that torque = force x position of particle

Torque is negative, thus force will be negative.

Therefore, the direction of the net force is negative Z direction

Answer:

HEAT FLOWS INTO THE SURROUNDING FROM THE SYSTEM AND THE SIGN OF THE SYSTEM WILL BE POSITIVE.

Explanation:

This is so of an endothermic reaction and the contraction of the gas is an endothermic one because the total heat content of the products formed is more than that of the reactants and the gas therefore absorbs heat from the surrounding during the contraction. An endothermic reaction is one in which heat is absorbed from the surroundings and the change in heat content ΔH of is positive.

The reaction for the contraction of the gas produces + 10 J of heat.

Answer:the wave is an electromagnetic wave.

Explanation:

Answer:

A: The wave is a mechanical wave.

C: The wave moves energy through matter.

F: The wave transfers energy perpendicular to the motion of the wave.

Hope it works!

Explanation:

Explanation:

Let l is the length of the wire. such that, side of square is l/4. Area of square loop is, [tex]A_s=\dfrac{l^2}{16}[/tex]

Radius of the circular loop,

[tex]l=2\pi r\\\\r=\dfrac{l}{2\pi}[/tex]

Area of the circular loop,

[tex]A_a=\pi r^2\\\\A_a=\pi \times (\dfrac{l}{2\pi})^2\\\\A=\dfrac{l^2}{4\pi}[/tex]

Torque in magnetic field is given by :

[tex]\tau=NIAB\sin\theta[/tex]

It is clear that, [tex]\tau\propto A[/tex]

So,

[tex]\dfrac{\tau_s}{\tau_a}=\dfrac{A_s}{A_a}\\\\\dfrac{\tau_s}{\tau_a}=\dfrac{\dfrac{l^2}{16}}{\dfrac{l^2}{4\pi}}\\\\\dfrac{\tau_s}{\tau_a}=\dfrac{\pi}{4}[/tex]

So, the ratio of maximum torque on the square loop to the maximum torque on the circular loop is [tex]\pi :4[/tex].

Answer:

Explanation:

solution is found below

Answer:

This demonstrates Newton's Third Law because for every action there is an equal and opposite force.

Explanation:

The cat's paw hitting her toy string would be an action therefore when is swings to and from her it it creating and equal and opposite force. In other words, the cat hits it and when the string comes back to her it's coming back the same force just opposite of the direction she hit.

Answer:

The value of the constant force is [tex]\bf{296.88~N}[/tex].

Explanation:

Given:

Mass of the merry-go-round, [tex]m = 210~Kg[/tex]

Radius of the horizontal disk, [tex]r = 1.5~m[/tex]

Time required, [tex]t = 2.00~s[/tex]

Angular speed, [tex]\omega = 0.600~rev/s[/tex]

Torque on an object is given by

[tex]\tau = F.r = I.\alpha~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(1)[/tex]

where [tex]I[/tex] is the moment of inertia of the object, [tex]\alpha[/tex] is the  angular acceleration and [tex]F[/tex] is the force on the disk.

The moment of inertia of the horizontal disk is given by

[tex]I = \dfrac{1}{2}mr^{2}[/tex]

and the angular acceleration is given by

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

Substituting all these values in equation (1), we have

[tex]F &=& \dfrac{I\alpha}{r}\\&=& \dfrac{\pi m r \omega}{t}\\&=& \dfarc{\pi(210~Kg)(1.5~m)(0.600~rev/s)}{2.00~s}\\&=& 296.88~N[/tex]

Answer:the mass Of The object,and there area Of The bottom surface

Explanation:

There mass Of The object, and the area Of The bottom surface

The amount of pressure exerted by a solid is based on the mass of the object and the area of the bottom surface. therefore the correct answer is option B.

The total applied force per unit of area is known as the pressure.

Both the external force being applied and the area to which it is being applied affect the pressure.

The mathematical expression for the pressure

Pressure = Force /Area

the pressure is expressed by the unit pascal or N /m²

A specific force produces less pressure when its region of influence is expanded. It is also true that increasing pressure is produced as a force's region of influence is reduced. For instance, because the force is acting across a smaller area at the pointy end of a high-heeled shoe, it applies more pressure there than it does at the flat end.

Thus, The mass of the object and the size of the bottom surface determine how much pressure a solid will exert. Therefore, choice B is the appropriate answer.

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Answer:

Refractive index = 1.33

Explanation:

Refractive index is of a material explains how light passes through a medium. It is the ratio of speed of light in vacuum to the speed of light in the medium.

We know the speed of light in water is [tex] 2.25* 10^8 [/tex] and there is a change iñ speed of light in every medium it passes (except when it passes through vacuum). The change in speed of light is known to be = [tex] 2.99*10^8[/tex]

Given speed of light = 2.25*10^8 m/s

Formula for refractive index:

[tex]n = \frac{c}{v}[/tex]

Where,

change in speed of light, c= [tex] 2.99*10^8[/tex]

speed of light, v = 2.25*10^8 m/s

[tex] Refractive index, n = \frac{2.99*10^8}{2.55*10^8} [/tex]

= 1.328

= 1.33

The liquid is water, and the index of refraction of water is 1.33

As we know,

The refractive index will be:

→ [tex]n = \frac{c}{v}[/tex]

By substituting the values,

      [tex]= \frac{2.99\times 10^8}{2.55\times 10^8}[/tex]

      [tex]= 1.33[/tex]

Thus the approach above is right.

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Answer:

Explanation:

a )  In this case A rests on B and both rest on horizontal surface

Both moves together .

Total reaction due to weight of both of them

R = 1.2 + 3.6 = 4.8 N

friction force by horizontal surface = μ R , μ is coefficient of friction .

= .3 x 4.8 = 1.44 N

Force equal to frictional force will be required to put both of them in uniform motion .

b ) If A is held stationary , friction force will arise at both , the upper and lower surface of B .

At upper surface friction force =  μ x weight of A

= .3 x 1.2 = .36 N

At lower surface friction force = μ x weight of A +B

= .3 x 4.8

= 1.44

Total frictional force on B

= 1.8 N  N .

So 1.8 N force will be required to put B in uniform keeping A stationary.

To find the magnitude of the horizontal force F necessary to drag block B to the left at constant speed, we need to consider two scenarios: a) when block A rests on block B and moves with it, and b) when block A is held at rest. In both scenarios, the friction force between block B and the surface is given by the equation F_friction = μ * N, where μ is the coefficient of friction and N is the normal force. The normal force is the sum of the weights of the blocks in scenario a) and the weight of block B in scenario b).

To find the magnitude of the horizontal force F necessary to drag block B to the left at constant speed, we need to consider two scenarios: a) when block A rests on block B and moves with it, and b) when block A is held at rest.

a) When block A rests on block B and moves with it, the friction force between the blocks and the surface is the force needed to overcome the friction. The friction force is given by the equation:
F_friction = μ * N

Here, μ is the coefficient of kinetic friction and N is the normal force between the blocks and the surface. In this case, the normal force N is the sum of the weights of both blocks, so N = m_A * g + m_B * g. Substituting this value into the equation, we get:
F_friction = μ * (m_A * g + m_B * g)

b) When block A is held at rest, the friction force between block B and the surface is the force needed to overcome the friction. The friction force is the product of the coefficient of static friction and the normal force:
F_friction = μ * N

Here, μ is the coefficient of static friction and N is the normal force between block B and the surface. The normal force N is equal to the weight of block B, so N = m_B * g. Substituting this value into the equation, we get:
F_friction = μ * (m_B * g)

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To solve this problem, we will apply the concepts related to Faraday's law that describes the behavior of the emf induced in the loop. Remember that this can be expressed as the product between the number of loops and the variation of the magnetic flux per unit of time. At the same time the magnetic flux through a loop of cross sectional area is,

[tex]\Phi = BA Cos \theta[/tex]

Here,

[tex]\theta[/tex] = Angle between areal vector and magnetic field direction.

According to Faraday's law, induced emf in the loop is,

[tex]\epsilon= -N \frac{d\Phi }{dt}[/tex]

[tex]\epsilon = -N \frac{(BAcos\theta)}{dt}[/tex]

[tex]\epsilon = -NAcos\theta \frac{dB}{dt}[/tex]

[tex]\epsilon = -N\pi r^2 cos\theta \frac{d}{dt} ( ( 3.75 T ) + ( 3.05T/s ) t + ( -6.95 T/s^2 ) t^2)[/tex]

[tex]\epsilon = -N\pi r^2 cos\theta( (3.05T/s)-(13.9T/s)t )[/tex]

At time [tex]t = 5.71s[/tex],  Induced emf is,

[tex]\epsilon = -(1) \pi (0.220m)^2 cos(19.5\°)( (3.05T/s)-(13.9T/s)(5.71s))[/tex]

[tex]\epsilon = 10.9V[/tex]

Therefore the magnitude of the induced emf is 10.9V

10.90V

Faraday's law of induction states that the electromotive force (E) produced in a magnetic field is directly proportional to the change in flux, ΔΦ, inversely proportional to change in time, Δt. i.e

E = - N ΔΦ / Δt       ----------------(i)

Where;

N = proportionality constant called the number of coils in the wire.

With a small change in time, equation (i) could be re-written as follows;

E = - N δΦ / δt              --------------(ii)

Also, the magnetic flux, Φ, is given as follows;

Φ = BA cos θ          --------------------(iii)

Where;

B = magnetic field

A = cross sectional area of the wire

θ = angle between the field and the cross-section of the wire

Substitute equation (iii) into equation (ii) as follows;

E = - N δ(BAcosθ) / δt

E = - NAcosθ δ(B) / δt               -----------------------(iv)

From the question;

B(t) = (3.75 T) + (3.05 T/s)t + (− 6.95 T/s²) t²           -------------------(v)

Substitute equation (v) into equation (iv) as follows;

E = - NAcosθ δ[(3.75 T) + (3.05 T/s)t + (− 6.95 T/s²) t²] / δt         -----(vi)

Solve equation (vi) by taking derivative as follows;

E = - NAcosθ [3.05 − (2)6.95 t]

E = - NAcosθ [3.05 − 13.9t]            ----------------(vii)

Solve for A using the following relation;

A = πr²    -----------------(viii)

Where;

r = radius of the wire loop = 0.220m

π = 3.142

Substitute these values into equation (viii) as follows;

A = 3.142 x 0.220²

A = 0.152m²

Now substitute A = 0.152m², N = 1 (a single coil), θ = 19.5° and t = 5.71s into equation (vii)

E = - (1) (0.152)cos(19.5)° [3.05 − 13.9(5.71)]

E = - (1) (0.152)(0.94) [3.05 − 13.9(5.71)]

E = 10.90V

Therefore, the induced EMF in the loop is 10.90V

For the NACA 44XX series airfoils, using thin airfoil theory, the angle of zero lift is determined to be -1.6 degrees. The design angle of attack, which eliminates the suction peak, coincides with this angle. The design lift coefficient is 0.04, equivalent to the maximum camber of the airfoil.

The question provided revolves around the aerodynamic properties of the NACA 44XX series airfoils, specifically in terms of their performance at cruise without the need for leading edge flaps. Given the NACA numbering system, where the second digit represents the location of the maximum camber (p) as a fraction of the chord, we can determine several characteristics for the 44XX airfoils.

(a) Angle of zero lift: 0 degrees. (b) Design angle: ~1.376°. (c) Design lift coefficient: ~0.047. Utilized thin airfoil theory.

To address these questions, we can utilize thin airfoil theory, which provides approximate solutions for the aerodynamic characteristics of airfoils at small angles of attack. Let's go through each question:

(a) Determine the angle of zero lift for the NACA 44XX airfoils:

The angle of zero lift, also known as the zero-lift angle of attack [tex](\(\alpha_{L=0}\))[/tex], can be calculated using thin airfoil theory. For symmetrical airfoils like the NACA 44XX series, the zero-lift angle of attack is simply 0 degrees.

(b) Determine the angle at which the suction peak is eliminated. We will call this the design angle of attack:

The suction peak is eliminated when the airfoil is operating at its design lift coefficient [tex](\(C_{L_{\text{design}}}\))[/tex]. In thin airfoil theory, the lift coefficient is directly proportional to the angle of attack [tex](\(\alpha\))[/tex]. For cambered airfoils, this design lift coefficient is achieved at an angle of attack where the camber is aligned with the oncoming flow, resulting in no suction peak.

For NACA 44XX airfoils, the maximum camber is located at [tex]\(p = \frac{4}{10} = 0.4\)[/tex]. The angle of attack at which the camber aligns with the flow is called the camber line slope angle [tex](\(\alpha_{camber}\))[/tex]. For NACA 44XX series airfoils, the camber line slope angle can be approximated by:

[tex]\[\alpha_{camber} \approx 0.06 \times p = 0.06 \times 0.4 = 0.024 \text{ radians}\][/tex]

To convert this to degrees:

[tex]\[\alpha_{camber} \approx 0.024 \times \frac{180}{\pi} \approx 1.376^\circ\][/tex]

So, the design angle of attack where the suction peak is eliminated is approximately [tex]\(1.376^\circ\).[/tex]

(c) What is the design lift coefficient for the NACA 44XX airfoils?

At the design angle of attack, the lift coefficient [tex](\(C_L\))[/tex] can be calculated using thin airfoil theory. For symmetrical airfoils, [tex]\(C_L\)[/tex] is given by:

[tex]\[C_L = 2\pi \alpha\][/tex]

Substituting the design angle of attack [tex]\(\alpha_{camber}\)[/tex], we get:

[tex]\[C_{L_{\text{design}}} = 2\pi \times 1.376^\circ = 0.047\][/tex]

So, the design lift coefficient for the NACA 44XX airfoils is approximately [tex]\(0.047\).[/tex]

To solve this problem we will apply the concepts related to the double slit experiment. Here we test a relationship between the sine of the deviation angle and the distance between slit versus wavelength and the bright fringe order. Mathematically it can be described as,

[tex]dsin\theta = m\lambda[/tex]

Here,

d = Distance between slits

m = Any integer which represent the order number or the number of repetition of the spectrum

[tex]\lambda[/tex] = Wavelength

[tex]\theta[/tex] = Angular deviation

Replacing with our values we have,

[tex](6.93*10^{-6}) sin\theta = (3)(491*10^{-9})[/tex]

[tex]\theta = sin^{-1} (\frac{(3)(491*10^{-9}}{6.93*10^{-6}) })[/tex]

Part A)

[tex]\theta = 0.2141rad[/tex]

PART B)

[tex]\theta = 0.2141rad(\frac{360\°}{2\pi rad})[/tex]

[tex]\theta = 12.27\°[/tex]

Answer:

a) If the currents are in the same direction, the magnetic field is zero at x = 0.298 m = 29.8 cm

That is, in between the wires, 29.8 cm from the 73.0 A wire and 10.2 cm from the 25.0 A wire.

b) If the currents are in opposite directions, the magnetic field is zero at x = 0.608 m = 60.8 cm

That is, along the positive x-axis, 60.8 cm from the 73.0 A wire and 20.8 cm from the 25.0 A wire.

Explanation:

The origin is at the 73.0 A wire and the 25.0 A wire is at x = 0.40 m

The magnetic field in a current carrying wire at a distance r from the wire is given by

B = (μ₀I/2πr)

μ₀ = magnetic constant = (4π × 10⁻⁷) H/m

a) If the currents are in the same direction, at what positions is the magnetic field equal to 0.

According to laws describing the direction.of magnetic fields, this position will be at some point between the two wires.

The magnetic field due to the 73.0 A wire points out of the book, at points along the positive x-axis while the magnetic field due to the 25.0 A wire points into the plane of the book, moving in the negative x-direction.

Hence,

For the 73.0 A wire, I₁ = 73.0 A, r₁ = x

For the 25.0 A wire, I₂ = 25.0 A, r₂ = (0.4 - x)

B = B₁ - B₂ = 0

(μ₀/2π) [(I₁/r₁) - (I₂/r₂)] = 0

(I₁/r₁) = (I₂/r₂)

(I₁/x) = [I₂/(0.4-x)]

(73/x) = [25/(0.4-x)]

73(0.4-x) = 25x

29.2 - 73x = 25x

73x + 25x = 29.2

98x = 29.2

x = (29.2/98) = 0.298 m

b) If the currents are in the opposite directions, at what positions is the magnetic field equal to 0?

According to laws describing the direction.of magnetic fields, this position will be at some point beyond the second wire (since we're initially concerned about the positive x-direction).

The magnetic field due to the 73.0 A wire points out of the book, at points along the positive x-axis while the magnetic field due to the 25.0 A wire (whose direction is now in the opposite direction to the current in the first wire) is also along the positive x-direction.

Hence,

For the 73.0 A wire, I₁ = 73.0 A, r₁ = x

For the 25.0 A wire, I₂ = 25.0 A, r₂ = (x - 0.4)

B = B₁ - B₂ = 0

(μ₀/2π) [(I₁/r₁) - (I₂/r₂)] = 0

(I₁/r₁) = (I₂/r₂)

(I₁/x) = [I₂/(x-0.4)]

(73/x) = [25/(x-0.4)]

73(x-0.4) = 25x

73x - 29.2 = 25x

73x - 25x = 29.2

48x = 29.2

x = (29.2/48) = 0.608 m

Hope this Helps!!!

Final answer:

The net magnetic field can be found zero at certain points between or outside two long parallel wires carrying currents, either in the same or opposite directions, by equating the magnetic fields produced by each and solving for the distance.

Explanation:

The problem involves finding the locations where the net magnetic field is zero due to currents carried by two long parallel wires. The Biot-Savart Law or Ampère's Law can be used for such problems; however, in high school problems, we typically use the formula for the magnetic field due to a long straight wire, B = (μ0 * I) / (2π * r), where B is the magnetic field, μ0 is the permeability of free space, I is the current, and r is the distance from the wire.

For wires with currents in the same direction, the magnetic fields due to each wire will point in opposite directions in the region between the wires and in the same direction outside the wires. To find locations where the net magnetic field is zero, we can equalize the magnetic fields coming from each wire and solve for the distance r from one of the wires where the fields cancel out.

For wires with currents in opposite directions, the magnetic fields due to the wires will be in the same direction in the region between the wires and in opposite directions outside of the wires. Again, we can find locations of zero net magnetic field by equalizing the magnitudes of the magnetic fields and solving for r.

Answer:

Q1 = +2.50 x 10^-5C and Q2 = -2.50 x 10^-5C, r = 0.50m, F=?

Using Coulomb's law:

F = 1/(4πE) x Q1 x Q2/ r^2

Where

k= 1/(4πE) = 9 x 10^9Nm2/C2

Therefore,

F = 9x 10^9 x 2.50 x 10^-5 x2.50 x

10^-5/. ( 0.5)^2

F= 5.625/ 0.25

F= 22.5N approximately

F= 23N.

To find the direction of the force: since Q1 is positive and Q2 is negative, the force along Q1 and Q2 is force of attraction.

Hence To = 23N, attractive. C ans.

Thanks.