Given the time of flight for a projectile fired horizontally was o.66 seconds, and it landed 3:3 m from where it was launched, the projectile's muzzle velocity was: (A) 0.20 m/s, (B) 0.22 m/s, (C) 0.50 m/s, (D) 5.0 m/s, (E) none of the above.

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:

[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:

The tension of the rope is T= 172.52 N

Explanation:

m= 21 kg

α= 21º

μ= 0.285

a= 2.09 m/s²

g= 9.81 m/s²

W= m*g

W=206.01 N

Fr= μ*W*cos(α)

Fr= 54.81 N

Fx= W * sin(α)

Fx= 73.82 N

T-Fr-Fx = m*a

T= m*a + Fr + Fx

T= 172.52 N

The magnitude of the force of motion of an object pulled by a rope up an inclined plane is the tension in the rope less the frictional forces as well as the component of the weight of the object acting along the plane

The tension in the rope is approximately 172.53 Newtons

Known:

The mass of the box, m = 21.0 kg

The angle of inclination of the ramp, θ = 21.0°

The coefficient of kinetic friction, μk= 0.285

The acceleration of the box up the ramp, a = 2.09 m/s²

The acceleration due to gravity, g = 9.81 m/s²

Required:

The tension in the rope, T

Solution;

The normal reaction of the box, N = m × g × cos(θ)

∴ N = 21.0 × 9.81 × cos(21.0°) ≈ 192.33

The normal reaction, N ≈ 192.33 N

The frictional force, [tex]F_f[/tex] = μ × N

∴ [tex]F_f[/tex] = 0.285 × 192.33 ≈ 54.81405

The frictional force, [tex]F_f[/tex] ≈ 54.81405 N

The force pulling the box, F = m×a = T - [tex]F_f[/tex] - The component of the weight acting along the plane

The component of the weight acting along the plane = m·g·sin(θ)

∴ m×a = T - [tex]F_f[/tex] - m·g·sin(θ)

T = m×a + [tex]F_f[/tex] + m·g·sin(θ)

Which gives;

T = 21.0 × 2.09 + 54.81405 + 21.0 × 9.81 × sin(21.0°) ≈ 172.53

The tension in the rope, T ≈ 172.53 N

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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:

2.12 rpm

Explanation:

[tex]a_c= Centripetal\ acceleration=3.00\ m/s^2\\r=radius=\frac {120}{2}=60\ m\\a_c=\frac {v^2}{r}\\\Rightarrow v^2=a_c\times r\\\Rightarrow v^2=3\times 60=180\\\Rightarrow v=\sqrt{180}=13.41\ m/s\\[/tex]

[tex]v=\omega r\\\Rightarrow \omega=\frac {v}{r}\\\Rightarrow \omega=\frac {13.41}{60}\\\Rightarrow \omega=0.2236\ rad/s\\1\ rad/s=9.55\ rpm\\\Rightarrow 0.2236\ rad/s=2.12\ rpm\\\therefore Wheel\ rotations=2.12\ rpm[/tex]

Final answer:

To generate an artificial gravity of 3.00 m/s² on a space station with a 120 m diameter, it must rotate at approximately 2.14 revolutions per minute.

Explanation:

To find the rate of the wheel's rotation in revolutions per minute (rpm) that will provide an artificial gravity of 3.00 m/s² for a space station with a diameter of 120 m, we use the formula for centripetal acceleration:

a = ω² × r

where a is the centripetal acceleration, ω is the angular velocity in radians per second, and r is the radius of the circle.

Since we are looking for ω in revolutions per minute and not radians per second, we will need to convert our solution. First, we find ω in radians per second by rearranging the formula:

ω = √(a/r)

Given a = 3.00 m/s² and the radius r = 60 m (half the diameter),

ω = √(3.00 m/s² / 60 m) = √0.05 s²/m = 0.224 rad/s

One revolution is 2π radians and there are 60 seconds in a minute, so we can find the number of revolutions per minute by

rev/min = ω × (60 s/min) / (2π rad/rev)

Plugging in our value of ω:

rev/min = 0.224 rad/s × (60 s/min) / (2π rad/rev) ≈ 2.14 rev/min

Therefore, the space station must rotate at approximately 2.14 revolutions per minute to produce the desired artificial gravity.

Answer:

The combined speed of the woman and the skateboard is 6.67 m/s.

Explanation:

It is given that,

Mass of woman, m₁ = 60 kg

Speed of woman, v = 10 m/s

The woman jumps onto a giant skateboard of mass, m₂ = 30 kg

Let V is the combined speed of the woman and the skateboard. It can be calculated using the conservation of momentum as :

[tex]60\ kg\times 10\ m/s+30(0)=(60\ kg+30\ kg)V[/tex]

On solving the above equation we get V = 6.67 m/s

So, the combined speed of the woman and the skateboard is 6.67 m/s. Hence, this is the required solution.

Answer:

318.5 x 10^4 Pa

Explanation:

weight of woman = m g = 65 x 9.8 = 637 N

Area of both the heels = 1 x 2 = 2 cm^2 = 2 x 10^-4 m^2

Pressure is defined as the thrust acting per unit area.

P = F / A

Where, F is the weight of the woman and A be the area of heels

P = 637 / (2 x 10^-4) = 318.5 x 10^4 Pa

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:

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

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:

Charge, [tex]q=2.98\times 10^{-8}\ C[/tex]

Explanation:

It is given that,

Value of electric field, [tex]E=7.5\times 10^5\ V/m[/tex]

Area of parallel plates, [tex]A=45\ cm^2=0.0045\ m^2[/tex]

Distance between two parallel plates, d = 2.45 mm = 0.00245 m

For a parallel plate capacitor, the capacitance is given by :

[tex]C=\dfrac{\epsilon_oA}{d}[/tex].......(1)

Since, [tex]E=\dfrac{V}{d}[/tex]

V = E . d ............(2)

And [tex]C=\dfrac{q}{V}[/tex].....(3)

From equation (1), (2) and (3) we get :

[tex]\dfrac{q}{V}=\dfrac{\epsilon_oA}{d}[/tex]

[tex]q=\epsilon_o EA[/tex]

[tex]q=8.85\times 10^{-12}\ F/m\times 7.5\times 10^5\ V/m\times 0.0045\ m^2[/tex]

[tex]q=2.98\times 10^{-8}\ C[/tex]

So, the charge on the each plate is [tex]2.98\times 10^{-8}\ C[/tex]. Hence, this is the required solution.

To determine the charge on each plate for a desired electric field in a capacitor with parallel plates, use the formula Q = E x A x ε_0. Convert the area to m^2, insert all values including the permittivity of free space, and solve for Q.

To find the charge required on each of the parallel plates to create a specific electric field, you can use the relationship between the electric field (E), the charge (Q) on the plates, and the permittivity of free space (ε0). The electric field between two parallel plates is given by the equation:

E = Q / (ε0 × A)

where E is the electric field, Q is the charge on each plate, A is the area of the plates, and ε0 is the permittivity of free space (ε0 = 8.854 x 10−12 C2/N·m2). Rearranging this for Q gives:

Q = E × A × ε0

First, convert the area from cm2 to m2 by multiplying by (10−4)2. Then, plug in the values:

Q = (7.50 × 105 V/m) × (0.45 × 10−4 m2) × (8.854 × 10−12 C2/N·m2)

Calculate Q to find the charge required on each plate.

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:

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:

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.

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:

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:

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:

28.87 volt

Explanation:

Let the capacitance of the capacitor is C.

Energy, U = 3 J, V = 50 volt

U = 1/2 C V^2

3 = 0.5 x C x 50 x 50

C = 2.4  x 10^-3 F

Now U' = 1 J

U' = 1/2 C x V'^2

1 = 0.5 x 2.4 x 10^-3 x V'^2

V' = 28.87 volt

Final answer:

Hyperopia, also known as farsightedness, is the condition where close objects are blurry while distant objects are clearly visible. It is corrected with converging lenses. This is the opposite of myopia, which affects distance vision.

Explanation:

The ability to clearly see objects at a distance but not close up is properly called hyperopia, which is a vision problem where close objects are out of focus but distant vision is unaffected; also known as farsightedness. Farsighted individuals have difficulty seeing close objects clearly because their eyes do not converge light rays from a close object enough to make them meet on the retina. To correct this vision problem, converging lenses are used, which help to increase the power of the eyes, allowing for clear vision of nearby objects. This contrasts with myopia (nearsightedness), where distant objects are out of focus while close vision is typically unaffected.

Answer:

The direction of the momentum of the large ball after the collision with respect to east is 146.58°.

Explanation:

Given that,

Mass of large ball = 3.0 kg

Mass of steel ball = 1.0 kg

Velocity = 3.0 kg

After collision,

Velocity = 2.0 m/s

Using conservation of momentum

[tex]m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}[/tex]

[tex]3.0\times0+1.0\times(3.0)(-i)=1.0\times2(-j)+3.0\times v_{2}[/tex]

[tex]-3i+2j=3.0\times v_{2}[/tex]

[tex]v_{2}=-i+0.66j[/tex]

The direction of the momentum

[tex]tan\theta=\dfrac{0.66}{-1}[/tex]

[tex]\theta=tan^{-1}\dfrac{0.66}{-1}[/tex]

[tex]\theta=-33.42^{\circ}[/tex]

The direction of the momentum with respect to east

[tex]\theta=180-33.42=146.58^{\circ}[/tex]

Hence, The direction of the momentum of the large ball after the collision with respect to east is 146.58°.

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:

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: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:

New speed of water is 2.8 m/s.

Explanation:

It is given that,

Speed of water, v₁ = 2.3 m/s

Diameter of pipe, d₁ = 3.2 cm

Radius of pipe, r₁ = 1.6 cm = 0.016 m

Area of pipe, [tex]A_1=\pi(0.016)^2=0.000804\ m^2[/tex]

If the pipe narrows its diameter, d₂ = 2.9 cm

Radius, r₂ = 0.0145 m

Area of pipe, [tex]A_2=\pi(0.0145)^2=0.00066\ m^2[/tex]

We need to find the new speed of the water. It can be calculated using equation of continuity as :

[tex]v_1A_1=v_2A_2[/tex]

[tex]v_2=\dfrac{v_1A_1}{A_2}[/tex]

[tex]v_2=\dfrac{2.3\ m\times 0.000804\ m^2}{0.00066\ m^2}[/tex]

[tex]v_2=2.8\ m/s[/tex]

So, the new speed of the water is 2.8 m/s. Hence, this is the required solution.

The final image position is 4.34 cm to the right of the second lens, and the final image height is 2.4 cm.

First, we will consider the object in front of the first lens (with a focal length of 40 cm). The object is placed 14 cm in front of this lens.

For the first lens, the lens equation is given by:

[tex]\[\frac{1}{f_1} = \frac{1}{d_o} + \frac{1}{d_i}\][/tex]

where [tex]\(f_1\)[/tex] is the focal length of the first lens, [tex]\(d_o\)[/tex] is the object distance from the first lens, and [tex]\(d_i\)[/tex] is the image distance from the first lens.

Rearranging the equation to solve for [tex]\(d_i\)[/tex], we get:

[tex]\[d_i = \frac{f_1 \cdot d_o}{d_o - f_1}\][/tex]

Plugging in the values:

[tex]\[d_i = \frac{40 \cdot 14}{14 - 40}\] \[d_i = \frac{560}{-26}\] \[d_i = -21.54 \text{ cm}\][/tex]

This means the image formed by the first lens is 21.54 cm to the left of the first lens (virtual image).

Next, we calculate the height of this image using the magnification equation:

[tex]\[m_1 = -\frac{d_i}{d_o}\] \[m_1 = -\frac{-21.54}{14}\] \[m_1 = 1.54\][/tex]

The height of the image formed by the first lens is:

[tex]\[h_i = m_1 \cdot h_o\][/tex]

where [tex]\(h_o\)[/tex] is the object height (2.0 cm).

So, [tex]\(h_i = 1.54 \cdot 2.0\)[/tex]

[tex]\[h_i = 3.08 \text{ cm}\][/tex]

Now, this image becomes the object for the second lens (with a focal length of 20 cm). The distance between the two lenses is 16 cm, so the object distance for the second lens is the image distance from the first lens plus 16 cm.

For the second lens, the object distance [tex]\(d_o'\)[/tex] is:

[tex]\[d_o' = -21.54 + 16\] \[d_o' = -5.54 \text{ cm}\][/tex]

Using the lens equation for the second lens:

[tex]\[\frac{1}{f_2} = \frac{1}{d_o'} + \frac{1}{d_i'}\][/tex]

where [tex]\(f_2\)[/tex] is the focal length of the second lens, [tex]\(d_o'\)[/tex] is the object distance from the second lens, and [tex]\(d_i'\)[/tex] is the image distance from the second lens.

Rearranging the equation to solve for [tex]\(d_i'\)[/tex], we get:

[tex]\[d_i' = \frac{f_2 \cdot d_o'}{d_o' - f_2}\][/tex]

Plugging in the values:

[tex]\[d_i' = \frac{20 \cdot (-5.54)}{-5.54 - 20}\] \[d_i' = \frac{-110.8}{-25.54}\] \[d_i' = 4.34 \text{ cm}\][/tex]

This means the final image is 4.34 cm to the right of the second lens (real image).

To find the height of the final image, we use the magnification equation for the second lens:

[tex]\[m_2 = -\frac{d_i'}{d_o'}\] \[m_2 = -\frac{4.34}{-5.54}\] \[m_2 = 0.78\][/tex]

The height of the final image is:

[tex]\[h_i' = m_2 \cdot h_i\] \[h_i' = 0.78 \cdot 3.08\] \[h_i' = 2.4 \text{ cm}\][/tex]

Answer:

1.16

Explanation:

Let the angle is theta.

tan θ = perpendicular / base

tan θ = 5.9 / 5.1  = 1.16

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:

26.67 Volt

Explanation:

number of electrons, n = 5 x 10^21

R = 20 ohm

t = 10 min = 10 x 60 = 600 sec

V = ?

q = n x charge of one electron

q = n x e

q = 5 x 10^21 x 1.6 x 10^-19 = 800 C

i = q / t = 800 / 600 = 4 / 3 A

According to Ohm's law

V = i x R

V = 4 x 20 / 3 = 26.67 Volt

The voltage or potential difference across the resistor is 26.67 Volt.

Given the data in the question;

Ohm’s law states that the potential difference between two points is directly proportional to the current flowing through the resistance.

Hence

[tex]V = IR[/tex]

Where V is the voltage or potential difference, potential difference, I is the current and R is the resistance.

First we determine the total charge.

[tex]Q = n * e[/tex]

Where Q is the charge, n is the number of electrons and e is the charge on one electron. ( [tex]e = 1.6 * 10^{-19}C[/tex] )

Hence

[tex]Q = (5*10^{21}) * (1.6 * 10^{-19C)}\\\\Q = 800C[/tex]

Since current is the electric charge transferred per unit time.

[tex]I = \frac{Q}{t}[/tex]

Hence

[tex]I = \frac{800C}{600s}\\ \\I = 1.33A[/tex]

Now, we input our values into the above expression

[tex]V = I * R\\\\V = 1.33A * 20Ohms\\\\V = 26.67Volt[/tex]

Therefore, the voltage or potential difference across the resistor is 26.67 Volt.

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Answer:

Work done, W = −287500 Joules

Explanation:

It is given that,

Mass of the car, m = 920 kg

The car is travelling at a speed of, u = 90 km/h = 25 m/s

We need to find the amount of work must be done to stop this car. The final velocity of the car, v = 0

Work done is also defined as the change in kinetic energy of an object i.e.

[tex]W=\Delta K[/tex]

[tex]W=\dfrac{1}{2}m(v^2-u^2)[/tex]

[tex]W=\dfrac{1}{2}\times 920\ kg(0-(25\ m/s)^2)[/tex]

W = −287500 Joules

Negative sign shows the work done is done is opposite direction.

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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