A square loop of wire consisting of a single turn is perpendicular to a uniform magnetic field. The square loop is then re-formed into a circular loop, which also consists of a single turn and is also perpendicular to the same magnetic field. The magnetic flux that passes through the square loop is 4.5 x 10 -3 Wb. What is the flux that passes through the circular loop?

Answer:

efficiency = 5.4%

Explanation:

Efficiency of heat engine is given as

[tex]\eta = \frac{W}{Q_{in}}[/tex]

now we will have

[tex]W = Q_1 - Q_2[/tex]

so we will have

[tex]\eta = 1 - \frac{Q_2}{Q_1}[/tex]

now we know that

[tex]\frac{Q_2}{Q_1} = \frac{T_2}{T_1}[/tex]

so we have

[tex]\eta = 1 - \frac{T_2}{T_1}[/tex]

[tex]\eta = 1 - \frac{273+6}{273+22}[/tex]

[tex]\eta = 0.054[/tex]

so efficiency is 5.4%

Answer:

750 nm

Explanation:

[tex]d[/tex]  = separation of the slits = 1.8 mm = 0.0018 m

λ = wavelength of monochromatic light

[tex]D[/tex]  = screen distance = 4.8 m

[tex]y[/tex] = position of first bright fringe = [tex]\frac{1cm}{5 fringe} = \frac{0.01}{5} = 0.002 m[/tex]

[tex]n[/tex]  = order = 1

Position of first bright fringe is given as

[tex]y = \frac{nD\lambda }{d}[/tex]

[tex]0.002 = \frac{(1)(4.8)\lambda }{0.0018}[/tex]

λ = 7.5 x 10⁻⁷ m

λ = 750 nm

Answer:

The surface charge density on the plate, [tex]\sigma=7.5\times 10^{-5}\ C/m^2[/tex]

Explanation:

It is given that,

Area of parallel plate capacitor, [tex]A=2\times 10^{-3}\ m^2[/tex]

Separation between the plates, [tex]d=5\times 10^{-2}\ mm[/tex]

Electric field between the plates, [tex]E=8.5\times 10^{6}\ V/m[/tex]

We need to find the surface charge density on the plates. The formula for electric field is given by :

[tex]E=\dfrac{\sigma}{\epsilon}[/tex]

Where

[tex]\sigma[/tex] = surface charge density

[tex]\sigma=E\times \epsilon[/tex]

[tex]\sigma=8.5\times 10^{6}\ V/m\times 8.85\times 10^{-12}\ C^2/Nm^2[/tex]

[tex]\sigma=0.000075\ C/m^2[/tex]

[tex]\sigma=7.5\times 10^{-5}\ C/m^2[/tex]

Hence, this is the required solution.

Final answer:

The surface charge density on the plates of a parallel-plate air-filled capacitor 7.5225 x [tex]10^{-5}[/tex] [tex]C/m^2[/tex].

Explanation:

To determine the surface charge density on the plates of the parallel-plate air-filled capacitor, we can use the relationship between the electric field (E), the permittivity of free space
[tex](\sigma\))[/tex]. The electric field is defined as
[tex](E = \frac{\sigma}{\varepsilon_0}).[/tex]

Given that the electric field (E) is
[tex](8.5 \times 10^6 V/m)[/tex] and the permittivity of free space [tex]((\varepsilon_0))[/tex] is
[tex](8.85 \times 10^{-12} C^2/N \cdot m^2)[/tex], we can rearrange the formula to solve for the surface charge density [tex]((\sigma)):[/tex]
[tex](\sigma = E \cdot \varepsilon_0 = (8.5 \times 10^6 V/m) \times (8.85 \times 10^{-12} C^2/N \cdot m^2) = 7.5225 \times 10^{-5} C/m^2).[/tex]

Thus, the surface charge density on the plates is
[tex](7.5225 \times 10^{-5} C/m^2).[/tex]

Answer:

Part a)

[tex]\Delta U = 4\times 10^5eV[/tex]

Part b)

[tex]\Delta U = -4\times 10^5eV[/tex]

Explanation:

Part a)

Change in potential energy of a charge is given as

[tex]\Delta U = q\Delta V[/tex]

here we know that

[tex]q = e[/tex] for proton

also we have

[tex]\Delta V = 400 kV[/tex]

now we have

[tex]\Delta U = e(400 kV)[/tex]

[tex]\Delta U = 4\times 10^5eV[/tex]

Part b)

Change in potential energy of a charge is given as

[tex]\Delta U = q\Delta V[/tex]

here we know that

[tex]q = -e[/tex] for proton

also we have

[tex]\Delta V = 400 kV[/tex]

now we have

[tex]\Delta U = -e(400 kV)[/tex]

[tex]\Delta U = -4\times 10^5eV[/tex]

Answer:

So recoil speed of the Earth will be

[tex]v = 2.25 \times 10^{-24} m/s[/tex]

Explanation:

Here if we assume that during collision if ball will lose very small amount of energy and rebound with same speed

then the impulse given by the ball is

[tex]Impulse = m(v_f - v_i)[/tex]

[tex]Impulse = (0.155)(43.6 - (-43.6))[/tex]

[tex]Impulse = 13.52 Ns[/tex]

so impulse received by the Earth is same as the impulse given by the ball

so here we will have

[tex]Impulse = mv[/tex]

[tex]13.52 = (6 \times 10^{24})v[/tex]

[tex]v = 2.25 \times 10^{-24} m/s[/tex]

The recoil speed of the Earth is [tex]2.25*10^-24 m/s.[/tex]

Collision can be regarded as the forceful coming together of bodies.

The impulse of the ball can be calculated as;

[tex]Impulse= MV[/tex]

where V=( V1 -Vo)

V1= final velocity=(43m/s)

V0= initial velocity= (-43.6m/s)

m= mass of baseball

Hence,[tex]V= [43.6-(-43.6)]= 87.2m/s[/tex]

Then Impulse given by the ball=[tex]( 0.155*87.2)= 13.53Ns.[/tex]

We can now calculate the recoil speed of the  earth as;

[tex]Impluse= MVV= Impulse/ mass = 13.52/(6*10^24)[/tex]

=2.25*10^-24 m/s

Therefore, the recoil speed of the Earth is 2.25*10^-24 m/s

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Answer:

explained below

Explanation:

solar system consist of sun, planets, asteroids, stars etc.

Milky Way is  a galaxy.

Combination of  many solar systems then forms galaxy.

Our solar system is a part of milky way galaxy.

If you consider marbles as planets and balls as stars.

so, we can say that in milky way is consist of 100-400 billions balls and 100 million marbles.

Answer:

option (b), (c), (g)

Explanation:

When the battery is disconnected, the charge on the plates of a capacitor remains same.

As the capacitance of the capacitor is directly proportional to the dielectric constant.

C = k C0

Now the charge remains same, So

Q = Q0

Q = k C0 x V0 / k

So, potential between the plates is V0 / k.

Energy, E = 1/2 x C X V^2 = 1/2 x k C0 x V0^2 / k^2 = E0 / k

So, energy becomes E0 / k.

Answer:

1.2 x 10¹¹ kgm²/s

Explanation:

m = mass of the airplane = 39043.01

r = altitude of the airplane = 9.2 km = 9.2 x 1000 m = 9200 m

v = speed of airplane = 335 m/s

L = Angular momentum of airplane

Angular momentum of airplane is given as

L = m v r

Inserting the values

L = (39043.01 ) (335) (9200)

L =  (39043.01 ) (3082000)

L = 1.2 x 10¹¹ kgm²/s

Answer:

1. Yes, it can occur adiabatically.

2. The work required is: 86.4kJ

Explanation:

1. The internal energy of a gas is just function of its temperature, and the temperature changes between the states, so, the internal energy must change, but how could it be possible without heat transfer? This process may occur adiabatically due to the energy balance:

[tex]U_{2}-U_{1}=W[/tex]

This balance tell us that the internal energy changes may occur due to work that, in this case, si done over the system.

2. An internal energy change of a gas may be calculated as:

[tex]du=C_{v}dT[/tex]

Assuming [tex]C_{v}[/tex] constant,

[tex]U_{2}-U_{1}=W=m*C_{v}(T_{2}-T_{1})[/tex]

[tex]W=0.72*1*(420-300)=86.4kJ[/tex]

Answer:

False statement = There must be a non-zero net force acting on the object.  

Explanation:

An object is moving at a constant speed along a straight line. If the speed is constant then its velocity must be constant. We know that the rate of change of velocity is called acceleration of the object i.e.

[tex]a=\dfrac{dv}{dt}[/tex]

a = 0

⇒ The acceleration of the object is zero.

The product of force and acceleration gives the magnitude of force acting on the object i.e.

F = m a = 0

⇒  The net force acting on the object must be zero.

So, the option (a) is not true. This is because the force acting on the object is zero. First option contradicts the fact.

After a minute of flight, the plane's altitude changes due to its climb. The speed at which the distance from the plane to the radar station is increasing equals the resultant of its horizontal and vertical speed components. This can be computed as approximately 1.26 km or 1260 meters.

The student's question pertains to both kinematics and trigonometry in Physics. In this scenario, the plane is climbing at an angle, while its horizontal speed is constant. The speed at which the distance from the plane to the radar station increases involves understanding the principle of vector addition and the concept of resultant velocity.

We can construct a right-angled triangle where one side is the horizontal speed component (= 150 km/h), the other side is the vertical speed component (altitude change over time, given by climbing speed = altitude/duration), and the hypotenuse is the resultant velocity, i.e., the speed at which the distance from the plane to the radar station is increasing.

After a minute, the altitude gains due to the climb is 3 km + (150 km/h * sin(30°) * 1/60 hr) = 3.0375 km, where sin(30°) represents the vertical ratio of the velocity. Radar station distance change can be calculated using Pythagoras theorem. In one minute, the plane travels horizontally by 150 km/h * 1/60 hr = 2.5 km. Thus, the change in distance is sqrt{(3.0375 km)^2 + (2.5 km)^2 } - 3 km (original altitude), which approximately equals 1.26 km or 1260 meters when rounded to the nearest whole number.

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The distance from the plane to the radar station is increasing at a rate of approximately 5 km/h one minute later.

Let's define the variables:

v = 150 km/h (speed of the plane)

h = 3 km (altitude of the radar station)

θ = 30° (angle of ascent)

We need to find the rate at which the distance from the plane to the radar station is increasing 1 minute (or 1/60 hours) after the plane passes over the station.

We'll use the following steps:

Determine the horizontal and vertical components of the plane's velocity:

v_x = v × cos(θ) = 150 km/h × cos(30°)

v_x = 150 × (√3 / 2) ≈ 129.9 km/h

v_y = v × sin(θ) = 150 km/h × sin(30°)

v_y = 150 ×0.5 = 75 km/h

d_x = v_x × (1/60) hours

d_x = 129.9 km/h × (1/60) = 2.165 km

New altitude, h_new = h + (v_y × (1/60))

h_new = 3 km + (75 × (1/60)) = 3 + 1.25 km = 4.25 km

Using the Pythagorean Theorem: d = sqrt(d_x² + h_new²)

d = square root of (2.165² + 4.25²)

d = square root of (4.687 + 18.06) = √22.75 = 4.77 km

The rate of distance increase is approximately 4.77 km/h rounded to the nearest whole number, which is 5 km/h

Answer:

The ratio of the pressure at the top to the pressure at the bottom is [tex]\dfrac{701}{1000}[/tex]

Explanation:

Given that,

Number of molecules [tex]n= 10^24[/tex]

Mass [tex]m= 3\times10^{-26}\ kg[/tex]

Temperature = 300 K

Height [tex]h = 5\times10^{3}[/tex]

We need to calculate the  ratio of the pressure at the top to the pressure at the bottom

Using barometric formula

[tex]P_{h}=P_{0}e^{\dfrac{-mgh}{kT}}[/tex]

[tex]\dfrac{P_{h}}{P_{0}}=e^{\dfrac{-mgh}{kT}}[/tex]

Where, m = mass

g = acceleration due to gravity

h = height

k = Boltzmann constant

T = temperature

Put the value in to the formula

[tex]\dfrac{P_{h}}{P_{0}}=e^{\dfrac{-3\times10^{-26}\times9.8\times5\times10^{3}}{1.3807\times10^{-23}\times300}}[/tex]

[tex]\dfrac{P_{h}}{P_{0}}=\dfrac{701}{1000}[/tex]

Hence, The ratio of the pressure at the top to the pressure at the bottom is [tex]\dfrac{701}{1000}[/tex]

Answer:

Top pressure : Bottom pressure = 701 : 1000

Explanation:

Number of molecules = n = 10^24

Height = h = 5 × 10^3 m

Mass = m = 3 × 10^-26 kg  

Boltzman’s Constant = K = 1.38 × 10^-23 J/K

Temperature = T = 300K  

The formula for barometer pressure is given Below:

Ph = P0 e^-(mgh/KT)

Ph/P0 = e^-(3 × 10^-26 × 9.81 × 5 × 10^3)/(1.38 × 10^-23)(300)

Ph/P0 = e^-0.355

Ph/P0 = 1/e^0.355

Ph/p0 =0.7008 = 700.8/1000 = 701/1000

Hence,

Top pressure : Bottom pressure = 701 : 1000

Answer:

Current, I = 2.45 T

Explanation:

It is given that,

Magnetic field, B = 0.92 T

Length of wire, l = 2.6 m

Mass, m = 0.6 kg

We need to find the minimum current needed to levitate the wire. It is given by balancing its weight to the magnetic force i.e.

[tex]Ilb=mg[/tex]

[tex]I=\dfrac{mg}{lB}[/tex]

[tex]I=\dfrac{0.6\ kg\times 9.8\ m/s^2}{2.6\ m\times 0.92\ T}[/tex]

I = 2.45 A

So, the minimum current to levitate the wire is 2.45 T. Hence, this is the required solution.

The minimum current needed to levitate the wire in the given magnetic field is approximately 2.58 Amps, determined by setting the magnetic force acting on the wire equal to the gravitational force and solving for current.

The minimum current necessary to levitate the wire in a uniform magnetic field can be determined by equating the magnetic force acting on the wire to the gravitational force acting on it. The magnetic force exerted on a current-carrying wire in a magnetic field is given by F = IℓBsinθ, where F is the force, I is the current, ℓ is the length of the wire, B is the magnetic field strength, and θ is the angle between the current and the magnetic field. Given the wire is levitating, the angle θ is 90°, meaning sinθ is 1. Additionally, the gravitational force is F = mg, where m is the mass of the wire and g is the acceleration due to gravity. Setting the magnetic force equal to the gravitational force gives IℓB = mg, which we can solve for I to get I = mg/(ℓB). Using the given values, I = (0.60 kg * 9.8 m/s²) / (2.6 m * 0.92 T) = 2.58 A. So, the minimum current needed to levitate the wire is approximately 2.58 Amps.

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Answer:

The number of oscillation is 36.

Explanation:

Given that

Mass = 280 g

Spring constant = 3.3 N/m

Damping constant [tex]b=8.4\times10^{-3}\ Kg/s[/tex]

We need to check the system is under-damped, critical damped and over damped by comparing b with [tex]2m\omega_{0}[/tex]

[tex]2m\omega_{0}=2m\sqrt{\dfrac{k}{m}}[/tex]

[tex]2\sqrt{km}=2\times\sqrt{3.3\times280\times10^{-3}}=1.92kg/s[/tex]

Here, b<<[tex]2m\omega_{0}[/tex]

So, the motion is under-damped and will oscillate

[tex]\omega=\sqrt{\omega_{0}^2-\dfrac{b^2}{4m^2}}[/tex]

The number of oscillation before the amplitude decays to [tex]\dfrac{1}{e}[/tex] of its original value

[tex]A exp(\dfrac{-b}{2m}t)=A exp(-1)[/tex]

[tex]\dfrac{b}{2m}t=1[/tex]

[tex]t = \dfrac{2m}{b}[/tex]

[tex]t=\dfrac{2\times280\times10^{-3}}{8.4\times10^{-3}}[/tex]

[tex]t=66.67\ s[/tex]

We need to calculate the time period of one oscillation

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

[tex]T=\dfrac{2\times3.14}{\sqrt{\omega_{0}^2-\dfrac{b^2}{4m^2}}}[/tex]

[tex]T=\dfrac{2\times3.14}{\sqrt{\dfrac{k}{m}-\dfrac{b^2}{4m^2}}}[/tex]

[tex]T=\dfrac{2\times3.14}{\sqrt{\dfrac{3.3}{280\times10^{-3}}-\dfrac{(8.4\times10^{-3})^2}{4\times(280\times10^{-3})^2}}}[/tex]

[tex]T=1.83\ sec[/tex]

The number of oscillation is

[tex]n=\dfrac{t}{T}[/tex]

[tex]n=\dfrac{66.67}{1.83}[/tex]

[tex]n=36[/tex]

Hence, The number of oscillation is 36.

The radius of the spinning chamber can be calculated using the centripetal force formula, F = m*v^2/r. The values for mass (m), velocity(v) and force(F) are given in the question, which can be substituted into the rearranged version of the formula for radius (r), r = m*v^2/F.

The subject question is based in physics, specifically the 'Circular Motion and Gravitation' topic, and is about finding the radius of a spinning chamber in an amusement park, with the information the force exerted and the speed of the chamber.

To find the radius, we would need to use the formula for the circular force, F = m*v^2/r, where F is the force, m is the mass, v the velocity and r the radius. Rearranging for r, the radius, we get r = m*v^2/F.

Substituting the values from the question, we have m = 84.4 kg (person's weight), v = 3.28 m/s (speed of the outer wall) and F = 488 N. Calculating these, we get r = (84.4 kg * (3.28 m/s)^2)/488 N.

This will give the radius of the chamber.

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The radius of the chamber can be determined using the concept of centripetal force. By substituting the given values into the formula for centripetal force (F = m * v² / r) and rearranging for r, the radius is found to be approximately 2.2 meters.

In order to solve the problem, one must understand the concept of centripetal force, which is described as a force that makes a body follow a curved path, with its direction orthogonal to the velocity of the body, towards the fixed point of the instantaneous center of curvature of the path.

In this scenario, the centripetal force is equal to the force pressing against the rider's back, which is 488N. This force can be calculated using the formula: F = m * v² / r, where F is the centripetal force, m is the mass of the rider, v is the speed of the outer wall, and r is the radius of the chamber.

Given that m = 84.4 kg, v = 3.28 m/s, and F = 488 N, by substituting these values into the formula and rearranging it we find that r = m * v² / F. Consequently, r = (84.4 kg * (3.28 m/s)²) / 488 N, which equals approximately 2.2 meters. Therefore, the radius of the chamber is approximately 2.2 meters.

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Answer:

[tex]8.89\cdot 10^4 V/m[/tex]

Explanation:

The electric field strength between the plates of a parallel plate capacitor is given by

[tex]E=\frac{\sigma}{\epsilon_0}[/tex]

where

[tex]\sigma[/tex] is the surface charge density

[tex]\epsilon_0[/tex] is the vacuum permittivity

Here we have

[tex]A=3.00 \cdot 3.00 = 9.00 cm^2 = 9.0\cdot 10^{-4} m^2[/tex] is the area of the plates

[tex]Q=0.708 nC = 0.708 \cdot 10^{-9} C[/tex] is the charge on each plate

So the surface charge density is

[tex]\sigma=\frac{Q}{A}=\frac{0.708\cdot 10^{-9}}{9.0\cdot 10^{-4} m^2}=7.87\cdot 10^{-7} C/m^2[/tex]

And now we can find the electric field strength

[tex]E=\frac{\sigma}{\epsilon_0}=\frac{7.87\cdot 10^{-7}}{8.85\cdot 10^{-12}}=8.89\cdot 10^4 V/m[/tex]

The electric field strength inside a parallel-plate capacitor with given dimensions and charge is approximately 88.89 kV/m. This is calculated using the capacitance and charge to find the voltage, which then helps determine the electric field strength.

To find the electric field strength inside a parallel-plate capacitor, you can use the relationship between the voltage (V), plate separation (d), and electric field (E). The formula is given by:

E = V / d

However, we are not given the voltage directly, but we have the charge (Q) and the plate area (A). The first step is to determine the capacitance (C) using the formula:

C = ε₀ (A / d)

Where ε₀ (epsilon naught) is the permittivity of free space (8.85 × 10⁻¹² F/m), A is the area of one plate, and d is the separation between the plates.

Given:

Plate area,

A = (3.00 cm × 3.00 cm)

= 9.00 cm²

= 9.00 × 10⁻⁴ m²

Plate separation,

d = 1.30 mm

= 1.30 × 10⁻³ m

Now calculate the capacitance:

C = ε₀ (A / d)

= (8.85 × 10⁻¹² F/m) (9.00 × 10⁻⁴ m² / 1.30 × 10⁻³ m)

≈ 6.13 × 10⁻¹⁵ F (Farads)

Next, use the charge (Q) and capacitance (C) to find the voltage (V):

Q = C × V

V = Q / C

Given charge,

Q = 0.708 nC

= 0.708 × 10⁻⁹ C

V = (0.708 × 10⁻⁹ C) / (6.13 × 10⁻¹⁵ F)

≈ 115.56 V

Finally, calculate the electric field strength:

E = V / d

E = 115.56 V / 1.30 × 10⁻³ m

≈ 88.89 kV/m

Thus, the electric field strength inside the capacitor is approximately 88.89 kV/m.

Answer:

Velocity is 34.42 m/s at an angle of 56.91° below horizontal

Speed is 34.42 m/s

Explanation:

Velocity = 20.0 m/s at an angle of 30° below horizontal

Vertical velocity = 20 sin 20 = 6.84 m/s downward.

Horizontal velocity = 20 cos 20 = 18.79 m/s towards right.

Let us consider the vertical motion of ball we have equation of motion

               v² = u² + 2as

We need to find v,  u = 6.84 m/s, a = 9.81 m/s² and s = 40 m

Substituting

              v² = 6.84² + 2 x 9.81 x 40 = 831.59

               v = 28.84 m/s

So on reaching ground velocity of ball is

        Vertical velocity =  28.84 m/s downward.

        Horizontal velocity = 18.79 m/s towards right.  

Velocity

        [tex]v=\sqrt{28.84^2+18.79^2}=34.42m/s[/tex]

        [tex]tan\theta =\frac{28.84}{18.79}\\\\\theta =56.91^0[/tex]

So velocity is 34.42 m/s at an angle of 56.91° below horizontal

Speed is 34.42 m/s

Answer:3.98cm

Explanation:

given data

mass of object[tex]\left ( m\right )[/tex]=0.5kg

intial extension=5 cm

elevator acceleration=2 m/[tex]s^2[/tex]

From FBD of intial position

Kx=mg

K=[tex]\frac{0.5\times 9.81}{0.05}[/tex]

k=98.1 N/m

From FBD of second situation

mg-k[tex]x_0[/tex]=ma

k[tex]x_0[/tex]=m(g-a)

[tex]x_0[/tex]=[tex]\frac{0.5(9.81-2)}{98.1}[/tex]

[tex]x_0[/tex]=3.98cm

Answer:

376.9911ft²/minute

Explanation:

In the given question the rate of chage of radius in given as

[tex]\frac{\mathrm{d}r }{\mathrm{d} t}[/tex]=5ft per minute

we know ares of circle A=pi r^{2}

differentiating w.r.t. t we get

[tex]\frac{\mathrm{d} A}{\mathrm{d} t}=2\pi r\frac{\mathrm{d}r }{\mathrm{d} t}[/tex]

Now, we have find [tex]\frac{\mathrm{d}A }{\mathrm{d} t} at r=12 feet[/tex]

[tex]\frac{\mathrm{d} A}{\mathrm{d} t}=2\times\pi\times12\times5=120\pi=376.9911ft^{2}/minute[/tex]

The rate at which the area of the ripple is changing when the radius is 12 feet is 120π square feet per minute.

The concept in question is related to mathematical calculus and the principle of rates. The area of a circle is represented by the formula A = πr², with 'A' representing the area and 'r' the radius. To determine how the area changes with changes in the radius, we differentiate this function with respect to time, resulting in dA/dt = 2πr (dr/dt). We know the radius is increasing at a rate of 5 feet per minute (dr/dt = 5 ft/min). At the instant when the radius is 12 feet, we simply substitute into our differentiated equation to find that dA/dt = 2π(12ft)(5ft/min) = 120π ft²/min.

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Answer:

y and length is directly relation

Explanation:

Given data

A single-slit diffraction pattern is formed on a distant scree

angles involved = small

to find out

what factor will the width of the central bright spot on the screen change

solution

we know that  for single slit screen formula is

mass ƛ /area = sin θ and y/L = sinθ

so we can say mass ƛ /area =  y/L

and y = mass length  ƛ / area       .................1

in equation 1 here we can see y and length is directly relation so we can say from equation 1 that  the width of the central bright spot on the screen change if the distance from the slit to the screen is doubled

Answer:

1.16

Explanation:

Let the angle is theta.

tan θ = perpendicular / base

tan θ = 5.9 / 5.1  = 1.16

The player's foot must be in contact with the ball for a distance of 1.48 meters to increase the ball's speed from 2.00 m/s to 6.00 m/s.

To increase the ball's speed from 2.00 m/s to 6.00 m/s, we can use the equation:

v^2 = u^2 + 2as

where v is the final velocity (6.00 m/s), u is the initial velocity (2.00 m/s), a is the acceleration, and s is the distance.

Rearranging the equation, we have:

s = (v^2 - u^2) / (2a)

Plugging in the values, we get:

s = (6.00^2 - 2.00^2) / (2 * 40.0)

s = 1.48 m

Therefore, the player's foot must be in contact with the ball for a distance of 1.48 meters.

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The physics concepts of force, mass, acceleration, and work were used to determine the distance over which the soccer player's foot needs to be in contact with the soccer ball in order to increase its speed from 2.00 m/s to 6.00 m/s. The resulting distance was approximately 0.09 m or 9 cm.

This question involves the relationship between force, mass, and acceleration, as well as the concept of work. Great! For the student to understand this, we need to look at a couple of formulas from physics. The force exerted on an object equals its mass times its acceleration (F = ma). Also, work done on an object is the force applied to it times the distance over which the force is applied (W = Fd).

Firstly, we need to determine the acceleration of the soccer ball when it is kicked. Using the first formula, we rearrange to find a = F/m. Here, F is the force applied (40.0 N) and m is the mass of the soccer ball (0.42 kg). Substituting these values, we get a = 40.0N / 0.42kg = approximately 95.2 m/s².

The initial speed of the ball is 2.00 m/s and the final speed we want is 6.00 m/s. The change in speed (which is also the change in velocity, as the direction doesn't change) is therefore 6.00 m/s - 2.00 m/s = 4.00 m/s. Using the formula for acceleration, which is a change in velocity divided by time (a = Δv / t), we can plug in our known values to solve for the time of contact: t = Δv / a = 4.00 m/s / 95.2 m/s² = approximately 0.042 s.

With the time of contact known, we can now determine the distance over which the soccer player's foot must be in contact with the ball using the formula for distance traveled during uniformly accelerated motion: d = vt + 0.5at². The initial velocity v is 2.00 m/s, the time of contact t is 0.042 s, and the acceleration a is 95.2 m/s². Substituting these values, we find that the distance is approximately 0.09 m or 9 cm.

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Answer:

a) Force must be applied to the input piston = 10N

b) The input piston moved by 7.5 m

Explanation:

a) For a hydraulic lift we have equation

                  [tex]\frac{F_1}{A_1}=\frac{F_2}{A_2}[/tex]

   F₁ = ?

   d₁ = 10 cm = 0.1 m

   F₂ = 250 N

   d₂ = 50 cm = 0.5 m

Substituting

   [tex]\frac{F_1}{\frac{\pi \times 0.1^2}{4}}=\frac{250}{\frac{\pi \times 0.5^2}{4}}\\\\F_1=10N[/tex]

Force must be applied to the input piston = 10N

b) We have volume of air compressed is same in both input and output.    

   That is        A₁x₁ = A₂x₂

   A is area and x is the distance moved

   x₂ = 0.3 m

   Substituting

             [tex]\frac{\pi \times 0.1^2}{4}\times x_1=\frac{\pi \times 0.5^2}{4}\times 0.3\\\\x_1=7.5m[/tex]

The input piston moved by 7.5 m

C. I = 4.2A.

The magnetic field inside a solenoid is given by the equation:

B = μ₀NI/L

Clearing I for the equation above.

I = BL/μ₀N

With B = 1.0T, L = 3 x 10⁻²m, μ₀ = 4π x 10⁻⁷T.m/A and N = 5748turns

I = [(1.0T)(3 x 10⁻²m)]/[(4π x 10⁻⁷T.m/A)(5748turns)]

I = 4.15 ≅ 4.2A

Answer:

18.73 m/s^2

Explanation:

f = 2662 rpm = 2662 / 60 rps

r = 6.725 cm = 0.06725 m

Acceleration, a = r w

a = r x 2 x pi x F

a = 0.06725 × 2 × 3.14 × 2662 / 60

a = 18.73 m/s^2

Answer:

[tex]I_{2}=0.27 W/m^2[/tex]

Explanation:

Intensity is given by the expresion:

[tex]I_{2}=Io (\frac{r1}{r2} )^{2}[/tex]

where:

Io = inicial intensity

r1= initial distance

r= final distance

[tex]I_{2}=10 W/m^2 (\frac{30m}{180m} )^{2}[/tex]

[tex]I_{2}=0.27 W/m^2[/tex]

Answer:

A) Calculate the distance

To find the image distance for a diverging lens with a focal length of -30.0 cm and an object placed 18.0 cm in front, use the lens formula to calculate di = -77.14 cm, indicating a virtual image. The magnification equation yields a magnification of 4.285, indicating an upright image.

To calculate the image distance and magnification for a diverging lens, we can use the lens formula and the magnification equation. Given that a diverging lens has a focal length of -30.0 cm, and an object is placed 18.0 cm in front of this lens, we first need to use the lens formula:

1/f = 1/do + 1/di, where f is the focal length, do is the object distance, and di is the image distance.

Substituting the values, we get:

1/(-30) = 1/18 + 1/di

Following the steps:

Solving the equation for 1/di gives us 1/di = 1/(-30) - 1/18.

Finding a common denominator and subtracting the fractions we get 1/di = (-2 - 5)/(-540), which simplifies to 1/di = -7/540.

Therefore, di = -540/7 = -77.14 cm.

The negative sign indicates that the image is virtual and located on the same side of the lens as the object.

For magnification (m), use the equation m = - di/do:

m = - (-77.14)/18

m = 77.14/18

m = 4.285

The positive magnification value indicates that the image is upright compared to the object.

Answer:

The correct option is : (a) Create a strong magnetic field.

Explanation:

An electromagnet is a substance which produces magnetic field when electric current is passed through it. The magnetic field produced by them disappears when the electric current is turned off.

A Ferromagnetic substance is a substance that gets magnetized when kept in an external magnetic field. Such substances remain magnetized even after the external magnetic field is removed.

When an electromagnet and a ferromagnet is combined, it results in the production of a strong magnetic field.

Answer:

Total solar energy reaches the entire United States per hour is[tex]2.0864\times 10^{16} [/tex] kilo Joules.

Explanation:

Amount of energy reaching 1 square meter of area = 2278 kJ/hour

Total area of the United States = [tex]9,158,960 km^2=9,158,960\times 10^6 m^2[/tex]

[tex](1 km^2=10^6 m^2)[/tex]

Amount of energy reaching [tex]9,158,960\times 10^6 m^2[/tex] of area:

[tex]2278 kJ/hour\times 9,158,960\times 10^6 kJ/hour[/tex]

[tex]=2.0864\times 10^{16} kJ/Hour[/tex]

Total solar energy reaches the entire United States per hour is[tex]2.0864\times 10^{16} [/tex] kilo Joules.

Final answer:

The total solar energy that reaches the entire United States per hour is 20,866,319,680 MJ.

Explanation:

To calculate the total solar energy reaching the entire United States per hour, given that 2278 kJ of energy reaches a square meter in one hour, we would first convert the entire area of the United States into square meters. Since the area is given as 9,158,960 km², we convert this to square meters by multiplying by (1000 m/km)²:

9,158,960 km² * (1000 m/km)² = 9,158,960,000,000 m²

Next, we multiply the area in square meters by the energy received per square meter:

9,158,960,000,000 m² * 2278 kJ/m² = 20,866,319,680,000 kJ

To generate an accurate answer, we can express this in megajoules (MJ) by dividing by 1,000 (since 1 MJ = 1,000 kJ):

20,866,319,680,000 kJ / 1,000 = 20,866,319,680 MJ

Therefore, the total solar energy that reaches the entire United States per hour is 20,866,319,680 MJ.

Answer:

Percentage difference is 4.25 %.

Explanation:

Standard value of the density of wood, [tex]\rho_s=0.47\ g/cc[/tex]

Experimental value of the density of wood, [tex]\rho_e=0.45\ g/cc[/tex]

We need to find the percentage difference on the density of wood. It is given by :

[tex]\%=|\dfrac{\rho_s-\rho_e}{\rho_s}|\times 100[/tex]

[tex]\%=|\dfrac{0.47-0.45}{0.47}|\times 100[/tex]

Percentage difference in the density of the wood is 4.25 %. Hence, this is the required solution.