A 2.0 m × 4.0 m flat carpet acquires a uniformly distributed charge of −10 μC after you and your friends walk across it several times. A 6.0 μg dust particle is suspended in midair just above the center of the carpet.

What is the charge on the dust particle?

Answer : The volume of the second tank is, [tex]4ft^3[/tex]

Explanation :

Boyle's Law : It is defined as the pressure of the gas is inversely proportional to the volume of the gas at constant temperature and number of moles.

[tex]P\propto \frac{1}{V}[/tex]

or,

[tex]P_1V_1=P_2V_2[/tex]

where,

[tex]P_1[/tex] = initial pressure of gas = 2 atm

[tex]P_2[/tex] = final pressure of gas = 1 atm

[tex]V_1[/tex] = initial volume of gas = [tex]2ft^3[/tex]

[tex]V_2[/tex] = final volume of gas = ?

Now put all the given values in the above equation, we get:

[tex]2atm\times 2ft^3=1atm\times V_2[/tex]

[tex]V_2=4ft^3[/tex]

Therefore, the volume of the second tank is, [tex]4ft^3[/tex]

Answer:

Explanation:

Given

First capacitor magnitude [tex]C=1.75\ \mu F[/tex]

Second capacitor magnitude [tex]C=6.00\ \mu F[/tex]

Voltage of battery [tex]V=3.00\ V[/tex]

Both capacitor are connected in series so net capacitor is given by

[tex]\frac{1}{C_{net}}=\frac{1}{C_{1}}+\frac{1}{C_{2}}[/tex]

[tex]\frac{1}{C_{net}}=\frac{1}{1.75}+\frac{1}{6}[/tex]

[tex]C_{net}=\frac{6\times 1.75}{1.75+6}[/tex]

[tex]C_{net}=1.35\ \mu F[/tex]

So charge Across each capacitor is given by

[tex]Q=C_{net}\times V[/tex]

[tex]Q=1.35\times 10^{-6}\times 3[/tex]

[tex]Q=4.064\ \mu C[/tex]

Normal human body temperature is about 37°C, which is equivalent to 310.15 K. The peak wavelength emitted by a person with this temperature is 9.35 μm, which lies in the infrared part of the spectrum.

Normal human body temperature is about 37°C. To convert this temperature to kelvin, we can use the formula K = °C + 273.15. So, the temperature in kelvins would be:

K = 37 + 273.15 = 310.15 K.

The wavelength emitted by a person with this temperature can be determined using Wien's displacement law: λ = b/T, where b is Wien's constant and T is the absolute temperature. The peak wavelength (λmax) is the wavelength at which the emitted radiation is the greatest. For a temperature of 310.15 K, the peak wavelength can be calculated as: λmax = b/T = 2897.8/T = 9.35 μm.

The peak wavelength of 9.35 μm corresponds to the infrared part of the electromagnetic spectrum.

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Incomplete question.The complete question is here

If the separation between the first and the second minima of a single-slit diffraction pattern is 6.0 mm, what is the distance between the screen and the slit? The light wavelength is 500 nm and the slit width is 0.16 mm.

Answer:

The distance between first and second minimum is 1.92m

Explanation:

Given data

Wavelength of light λ=500nm

Slit width D=0.16 mm

The distance between first and second minima is 6.0 mm

To find

Distance L between the screen and the slit

Solution

As we know that

Yn=(nλL)/D

Where is L is distance between the slit and screen

D is slit width

n is order of minimum

Yn is distance of nth minimum from center maximum

λ is wavelength of light

The Distance between first and second minimum is given as:

Y₂-Y₁=6 mm

(2λL)/D-(1λL)/D=6 mm

(2λL-1λL)/D=6 mm

(λL/D)=6 mm

[tex]L=\frac{6mm(0.16mm)}{500nm}\\ L=1.92m[/tex]

The distance between first and second minimum is 1.92m

Final answer:

The screen distance is 0.19 meters, calculated using the separation between minima, slit width, and light wavelength in single-slit diffraction.

Explanation:

Solve for the distance between the screen and the slit (L).

1. Given Values:

Separation between minima (Δx) = 6.0 mm (convert to meters: 0.0060 m)

Wavelength of light (λ) = 500 nm (convert to meters: 500 x 10⁻⁹ m)

Slit width (a) = 0.16 mm (convert to meters: 0.16 x 10⁻³ m)

2. Relate Separation and Screen Distance:

We can use the relationship between the angular separation (Δθ) and the linear separation (Δx) on the screen:

Δθ ≈ Δx / L

3. Relate Separation and Minimum Position:

We know the separation (Δθ) is related to the difference between the positions of the first (m = 1) and second (m = 2) minima:

Δθ = θ₂ - θ₁ = (λ / a)

4. Combine Equations:

Substitute the equation for Δθ from step 3 into the equation from step 2:

(λ / a) ≈ Δx / L

5. Solve for Screen Distance (L):

Now we can arrange the equation to solve for L:

L ≈ Δx * a / λ

6. Plug in the Values:

L ≈ (0.0060 m) * (0.16 x 10⁻³ m) / (500 x 10⁻⁹ m)

7. Calculate and Round:

L ≈ 0.192 m (round to two significant figures)

Therefore, the distance between the screen and the slit is approximately 0.19 meters.

To solve this problem we will apply the concepts related to electrostatic energy, defined as,

[tex]U_e = \frac{kq_1q_2}{d^2}[/tex]

Here,

k = Coulomb's constant

[tex]q_{1,2}[/tex] = Charge of each object (electron and proton, at this case the same)=

d = Distance

Replacing with our values we have that

[tex]U_e = \frac{(9*10^9Nm^2)(1.6*10^{-19}C)(-1.6*10^{-19}C)}{0.0539*10^{-9}m}[/tex]

[tex]U_e = -4.27*10^{-18}J[/tex]

Therefore for the Part A the answer is [tex]-4.27*10^{-18}J[/tex] and por the Part B the sign indicates that the force between the proton and electron is attractive

Answer:

θ₂ = 35.26°

Explanation:

given,

refractive index of air, n₁ = 1

refractive index of glass, n₂ = 1.5

angle of incidence, θ₁ = 60°

angle of refracted light, θ₂ = ?

using Snell's Law

n₁ sin θ₁ = n₂ sin θ₂

1 x sin 60° = 1.5 sin θ₂

sin θ₂ = 0.577

θ₂ = sin⁻¹(0.577)

θ₂ = 35.26°

Hence, the refracted light is equal to  θ₂ = 35.26°

When a light beam passes from one medium to another, its angle of incidence and angle of refraction are related by Snell's law. The angle of the refracted light beam from the normal in air is 90 degrees.

When a light beam passes from one medium to another, its angle of incidence and angle of refraction are related by Snell's law. The formula for Snell's law is:

n1×sin(θ1) = n2×sin(θ2)

In this case, the light beam is passing from glass (with a refractive index of 1.5) to air. Given that the angle of incidence is 60 degrees, we can use Snell's law to find the angle of refraction from the normal in air.

Using Snell's law, we have:

1.5×sin(60) = 1×sin(θ2)

Simplifying the equation, we find:

sin(θ2) = 1.5×sin(60) = 1.5×0.87 = 1.31

However, the maximum value for the sine function is 1, so the angle of refraction cannot exceed 90 degrees (the maximum angle for a light beam in air). Therefore, the angle of the refracted light beam from the normal in air is 90 degrees.

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Answer:

Because of Moment of inertia.

Explanation:

Larger wheels have larger moment of inertia,when they rotate rotational energy is stored in spinnning of motion so they rotate slowly while lighter wheels move faster comparitively because they use smaller moment of inertia.The thing is to keeping tyre in contact with road,when a vehicla hits a

jolt it is difficult with heavy tyre to be in contact with road. When vehicle loses contact the driver will lose steering control which result in giving a way to sliding friction.

Answer:

854.39 N

Explanation:

The formula for the fundamental frequency of a stretched string is given as,

f = 1/2L√(T/m)..................... Equation 1

Where f = fundamental frequency, L = Length of the wire, T = Tension, m = mass per unit length.

Given: f = 261.6 Hz, L = 0.6 m, m = (5.2×10⁻³/0.6) = 8.67×10⁻³ kg/m.

Substitute into equation 1

261.6 = 1/0.6√(T/8.67×10⁻³)

Making T the subject of the equation,

T = (261.6×0.6×2)²(8.67×10⁻³)

T =854.39 N

Hence the tension of the wire is 854.39 N.

We have that for the Question "To what tension must this wire be stretched so that the fundamental vibration corresponds to middle C "

Answer:

From the question we are told

a piano has a length of 0.6m and a mass of [tex]5.2 * 10^{−3} kg[/tex]

Generally the equation for frequency is mathematically given as

[tex]F = \frac{V}{2L}[/tex]

where,

[tex]V = \sqrt\frac{T}{U}[/tex]

Therefore,

[tex]261.6 = \frac{V}{2*0.76}\\\\V = 261.6*2*0.6\\\\V = 313.92m/s[/tex]

so,

[tex]v^2 = \frac{T}{U}\\\\T = V^2 * U\\\\T = 512.43N[/tex]

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Explanation:

The visible light coming from celestial bodies from space are affected most by the atmosphere. The visible light suffer blurring and absorption.  So, to avoid such situation optical telescopes are kept in mountains where atmosphere is thin.

Whereas radio signal are not affected by the atmosphere so, they need not be placed in mountains.However, it is affected by various noises from the man made devices. So, to avoid these noises radio telescopes are kept in deep valleys.

Final answer:

Optical telescopes are placed on mountains to minimize atmospheric interference, reduce light pollution, and capture clearer images, while radio telescopes are located in valleys to shield them from man-made radio interference for clearer radio signal observations.

Explanation:

Optical astronomers often put their telescopes at the tops of mountains because these locations offer several advantages that are critical for observing the cosmos. Sir Isaac Newton mentioned that a serene and quiet air, often found on the tops of high mountains, is beneficial for reducing the confusion of rays caused by the atmosphere's tremors. This is echoed by modern practices where observatories are located in high, dry, and dark sites to minimize atmospheric interference, reduce light pollution, and avoid water vapor which absorbs infrared light.

The Andes Mountains in Chile, the desert peaks of Arizona, and the summit of Maunakea in Hawaii are examples of such ideal locations. On the other hand, radio astronomers sometimes place their telescopes in deep valleys to shield them from radio interference from human-made sources such as cell phones and electrical circuits. The natural geography of valleys can act as a barrier against these disturbances, providing a clearer signal for radio observations.

Answer:

616.223684211 N

Explanation:

[tex]F_r[/tex] = Resistive force on the wheel = 115 N

F = Force acting on sprocket

[tex]r_2[/tex] = Radius of sprocket = 4.75 cm

[tex]r_1[/tex] = Radius of wheel = 25 cm

Moment of inertia is given by

[tex]I=mr_1^2\\\Rightarrow I=1.7\times 0.25^2\\\Rightarrow I=0.10625\ kgm^2[/tex]

Torque

[tex]\tau=I\alpha\\\Rightarrow \tau=0.10625\times 4.9\\\Rightarrow \tau=0.520625\ Nm[/tex]

Torque is given by

[tex]\tau=Fr_2-F_rr_1\\\Rightarrow F=\dfrac{\tau+F_rr_1}{r_2}\\\Rightarrow F=\dfrac{0.520625+115\times 0.25}{0.0475}\\\Rightarrow F=616.223684211\ N[/tex]

The force on the chain is 616.223684211 N

1)

first you find the maxium force that the car can produce.

f=ma

Fmax=(1100kg)(6m/s^2)

then use f = ma again to find the accel with the passengers

Fmax=(1100kg +1650kg)(a)

=> a = (1100kg)(6m/s^2)/( 1100kg +1650kg)

= 2.4 m/s^2

The maximum magnitude of the acceleration of the two‑car system is  [tex]2.44 \;\rm m/s^{2}[/tex].

Given data:

The mass of car is, m = 1200 kg.

The maximum acceleration is, [tex]a =6.00 \;\rm m/s^{2}[/tex].

The mass of stall is, m' = 1750 kg.

As per the Newton's Second law of motion, the applied force on the car is equal to the product of mass and acceleration of car. So, the maximum force to push the car is given as,

F = ma

[tex]F = 1200 \times 6.00\\\\F= 7200\;\rm N[/tex]

Now for two-car system, the total mass is,

M = m + m'

M = 1200 +1750 = 2950 kg.

So, the acceleration for the two-car system is,

[tex]F = M \times a'\\\\7200 =2950 \times a'\\\\a'=2.44 \;\rm m/s^{2}[/tex]

Thus, the maximum magnitude of the acceleration a of the two‑car system is  [tex]2.44 \;\rm m/s^{2}[/tex].

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Answer:

(A) When you hold the ball still in your hands after catching it

(E) When you hold the ball still in your hands before it is thrown.

Explanation:

According to Newton's 1st law, objects subjected to 0 net force would maintain a constant velocity or staying at rest. This is not the case for the ball when it leaves your hands. This ball would always be subjected to gravitational acceleration g so its velocity changes. So only (A) and (E) are correct when the ball stays still in your hands.

Final answer:

A zero net force acts on the basketball when it is held still after catching it (A) and before throwing it (E) because these are the points where there is no acceleration due to balanced forces. Points B, C, and D have non-zero net forces because the ball is accelerating due to the force of gravity.

Explanation:

When considering a basketball being thrown straight up and ignoring air resistance, a zero net force acts on the basketball:

(A) When holding the ball still after catching it because there is no acceleration and gravity is balanced by the upward force from your hands.

(E) When holding the ball still before throwing it for the same reason as in (A).

The points at which the net force is not zero are:

(B) Just after the ball leaves the hand as the only force acting on the ball is gravity, resulting in acceleration and therefore a net force downwards.

(C) At the highest point because gravity is still acting downwards, although the ball is temporarily at rest.

(D) As the ball hits the hands because the hands apply an upward force to decelerate the ball, which is different from the force of gravity, hence not zero net force.

Overall, the net force on an object is related to its acceleration due to Newton's second law of motion, and since acceleration occurs whenever the ball is in motion and not at rest, there is a net force acting on the ball during these periods.

Answer:

[tex]q=2.08*10^{-8}C\\ q=20.8nC[/tex]

Explanation:

Given data

Electric potential V=1.50 kV

Diameter d=25.0 cm

radius=diameter/2=25/2=12.5 cm=0.125 m

to find

We are asked to find excess charge

Solution

As inside the sphere the electric field is zero everywhere and potential is same at every point inside the sphere and on its surface

So we can use the value of the potential to get the charge q on the sphere

[tex]V=\frac{1}{4\pi E}\frac{q}{R}\\ q=4\pi ERV\\Where\\4\pi E=\frac{1}{9.0*10^{9}N.m^{2}/C^{2} }\\ q=(\frac{1}{9.0*10^{9}N.m^{2}/C^{2} })(0.125m)(1.50*10^{3}V )\\q=2.08*10^{-8}C\\ q=20.8nC[/tex]

To make the potential of a copper sphere's center 1.50 kV relative to infinity, an excess charge of approximately 2.09 x 10^-8 C must be placed on it.

The potential V of a charged sphere is given by the equation V = kQ/R, where k is Coulomb's constant (8.99 x 109 Nm2/C2), Q is the charge on the sphere, and R is the radius. We can rearrange this to calculate the sphere's charge: Q = VR/k. The sphere's radius is half its diameter, hence 25.0 cm / 2 = 12.5 cm = 0.125 m. Substituting the given values in, we have Q = 1.50 x 103 V x 0.125 m / 8.99 x 109 Nm2/C2, recalling that 1 kV = 103 V.  Doing the calculation, we find the excess charge Q to be approximately 2.09 x 10-8 C.

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Answer:

(a). The mass of the block of ice is 62.8 kg.

(b). The distance is 24.192 m.

Explanation:

Given that,

Horizontal force = 80.5 N

Distance = 13.0 m

Time = 4.50 s

(a). We need to calculate the acceleration of the block of ice

Using equation of motion

[tex]s=ut+\dfrac{1}{2}at^2[/tex]

Put the value into the formula

[tex]13=0+\dfrac{1}{2}\times a\times(4.50)^2[/tex]

[tex]a=\dfrac{13\times2}{(4.50)^2}[/tex]

[tex]a=1.28\ m/s^2[/tex]

We need to calculate the mass of the block of ice

Using formula of force

[tex]F = ma[/tex]

[tex]m=\dfrac{F}{a}[/tex]

Put the value into the formula

[tex]m=\dfrac{80.5}{1.28}[/tex]

[tex]m=62.8\ kg[/tex]

(b). If the worker stops pushing at the end of 4.50 s,

We need to calculate the velocity

Using equation of motion

[tex]v =u+at[/tex]

Put the value into the formula

[tex]v=0+1.28\times4.50[/tex]

[tex]v=5.76\ m/s[/tex]

We need to calculate the distance

Using formula of distance

[tex]v = \dfrac{d}{t}[/tex]

[tex]d=v\times t[/tex]

Put the value into the formula

[tex]d=5.76\times4.20[/tex]

[tex]d=24.192\ m[/tex]

Hence, (a). The mass of the block of ice is 62.8 kg.

(b). The distance is 24.192 m.

Answer:

v = 0.2035 m/s

Explanation:

given,

Amplitude,A = 0.0127 m

Frequency, f = 2.55 Hz

maximum speed of the needle = ?

we know,

x = A cos ω t

velocity of the motion

v = - A ω sin ω t

for maximum speed

sin ω t = 1

v = - A ω

ω = 2 π f

now,

v =  A 2 π f

v = 0.0127 x 2 π x 2.55

v = 0.2035 m/s

Hence, the maximum speed of the needle is equal to 0.2035 m/s

Answer:

Torque=6.261×10^-43Nm

Explanation:

Torque is given by÷

Torque=dqEsin(phi)

Where E is a constant=1.6022×10^-19

d= distance = 6.186×10^-30cm

Changing to metres=6.186×10^-28m

q= the charge=8500NC

Phi =42°

Torque= (6.186×10^-28)×8500×(1.6022×10^-19)sin42°

Torque= 8.425×10^-43 × 0.7431

Torque= 6.261 ×10^-43

The torque exerted by an 8500 N/C electric field exerts on the dipole is 6.261 ×10⁻⁴³ Nm.

The force which causes the object to rotate about any axis is called perpendicular distance.

Torque is a twisting or turning force that frequently results in rotation around an axis, which may be a fixed point or the center of mass. The ability of something rotating, such as a gear or a shaft, to overcome turning resistance is another way to think of torque.

Given:

A single water molecule was oriented such that its dipole moment of magnitude 6.186 × 10 − 30 Cm is along a z-axis,

The electric field  = 8500 N/C,

∅ = 42°

Calculate the torque as shown below,

Torque = d × q ×  E × sin(∅)

Where E is a constant = 1.6022×10⁻¹⁹

distance = 6.186×10⁻³⁰ cm =  6.186×10⁻²⁸ m

Torque = 6.186×10⁻²⁸ × 8500 × (1.6022 ×  10⁻¹⁹) sin42°

Torque = 8.425×10⁻⁴³ × 0.7431

Torque = 6.261 ×10⁻⁴³

Thus, the torque is 6.261 ×10⁻⁴³ Nm.

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Answer:

103.3°C

Explanation:

k = Thermal conductivity of stainless steel = 14 W/m² K

P = Power = 800 W

d = Diameter = 22 cm

t = Thickness of kettle = 2.2 mm

[tex]\Delta T[/tex] = Change in temperature = [tex](T_2-100)[/tex]

100°C boiling point of water

Power is given by

[tex]P=\dfrac{kA\Delta T}{t}\\\Rightarrow P=\dfrac{kA(T_2-T_1)}{t}\\\Rightarrow T_2=\dfrac{Pt}{kA}+T_1\\\Rightarrow T_2=\dfrac{800\times 2.2\times 10^{-3}}{14\times \dfrac{\pi}{4} 0.22^2}+100\\\Rightarrow T_2=103.3\ ^{\circ}C[/tex]

The temperature of the bottom of the surface 103.3°C

The temperature of the bottom surface of the kettle is approximately 1.593 Kelvin higher than the boiling water.

To calculate the temperature of the bottom surface of the kettle, we can use the formula for heat conduction: q = (k * A * ΔT) / t. Here, q is the heat transfer rate, k is the thermal conductivity of stainless steel, A is the surface area of the bottom of the kettle, ΔT is the temperature difference between the bottom surface of the kettle and the boiling water, and t is the thickness of the bottom of the kettle. Rearranging the formula, we can solve for ΔT as:

ΔT = (q * t) / (k * A)

Plugging in the given values, we find:

ΔT = (800 W * 0.0022 m) / (14 W/(m·K) * π * (0.22 m/2)²)

Calculating this, we find ΔT ≈ 1.593 K. Therefore, the temperature of the bottom surface of the kettle would be approximately 1.593 Kelvin higher than the boiling water.

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Answer:

[tex]6.945326087\times 10^{-11}\ C[/tex]

5434.78260873 N/C

2963369.48874 m/s

Explanation:

[tex]\epsilon_0[/tex] = Permittivity of free space = [tex]8.85\times 10^{-12}\ F/m[/tex]

q = Charge of electron = [tex]1.6\times 10^{-19}\ C[/tex]

V = Voltage = 25 V

Side = 3.8 cm

d = Separation = 4.6 mm

m = Mass of electron = [tex]9.11\times 10^{-31}\ kg[/tex]

Charge is given by

[tex]Q=\dfrac{VA\epsilon_0}{d}\\\Rightarrow Q=\dfrac{25\times (3.8\times 10^{-2})^2\times 8.85\times 10^{-12}}{4.6\times 10^{-3}}\\\Rightarrow Q=6.945326087\times 10^{-11}\ C[/tex]

Charge on each plate is [tex]6.945326087\times 10^{-11}\ C[/tex]

Electric field is given by

[tex]E=\dfrac{Q}{A\epsilon_0}\\\Rightarrow E=\dfrac{6.945326087\times 10^{-11}}{(3.8\times 10^{-2})^2\times 8.85\times 10^{-12}}\\\Rightarrow E=5434.78260873\ N/C[/tex]

The electric field strength is 5434.78260873 N/C

Acceleration is given by

[tex]a=\dfrac{qE}{m}\\\Rightarrow a=\dfrac{1.6\times 10^{-19}\times 5434.78260873}{9.11\times 10^{-31}}\\\Rightarrow a=9.5451725291\times 10^{14}\ m/s^2[/tex]

[tex]v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 9.5451725291\times 10^{14}\times 4.6\times 10^{-3}+0^2}\\\Rightarrow v=2963369.48874\ m/s[/tex]

The velocity is 2963369.48874 m/s

Answer:

Angular acceleration = 5 rad /s ^2

Kinetic energy = 0.391 J

Work done = 0.391 J

P =6.25 W

Explanation:

The torque is given as moment of inertia × angular acceleration

angular acceleration = torque/ moment of inertia

= 10/2= 5 rad/ s^2

The kinetic energy is = 1/2 Iw^2

w = angular acceleration/time

=5/8= 0.625 rad /s

1/2 × 2× 0.625^2

=0.391 J

The work done is equal to the kinetic energy of the motor at this time

W= 0.391 J

The average power is = torque × angular speed

= 10× 0.625

P = 6.25 W

A. The angular acceleration is 5 rad/s². B. The kinetic energy of the shaft after 8 seconds is 16000 J. C. The work done by the motor in this time is 1600 J. D. The average power delivered by the motor over this time interval is 200 W.

A. Angular acceleration can be found using the rotational analog to Newton's second law a = net τ/I. The moment of inertia I is given and the torque τ can be found from the given torque value. So, τ = 10 Nm and I = 2.0 kgm2. Substituting these values, we get a = 10 Nm / 2.0 kgm2 = 5 rad/s².

B. Using the equation K = (1/2) I ω², where I is the moment of inertia and ω is the angular velocity, we can calculate the initial angular velocity ω₁ as 0 rad/s. Then, the final angular velocity ω can be found using the relationship ω = ω₁ + αt. So, ω = 0 rad/s + (5 rad/s²)(8 s) = 40 rad/s. Finally, substituting these values into the kinetic energy equation, we have K = (1/2)(2.0 kgm2)(40 rad/s)² = 16000 J.

C. The work done by the motor can be found using the equation W = τθ, where τ is the torque and θ is the angular displacement. In this case, θ = ω₁t + (1/2)αt² = 0 rad/s(8 s) + (1/2)(5 rad/s²)(8 s)² = 160 rad. Therefore, W = (10 Nm)(160 rad) = 1600 J.

D. The average power can be calculated using the equation P = W/t, where W is the work done and t is the time interval. Substituting the values we found in the previous calculations, we have P = 1600 J / 8 s = 200 W.

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Answer:

a) 75.5 degree relative to the North in north-west direction

b) 309.84 km/h

Explanation:

a)If the pilot wants to fly due west while there's wind of 80km/h due south. The north-component of the airplane velocity relative to the air must be equal to the wind speed to the south, 80km/h in order to counter balance it

So the pilot should head to the West-North direction at an angle of

[tex]cos(\alpha) = 80/320 = 0.25[/tex]

[tex]\alpha = cos^{-1}(0.25) = 1.32 rad = 180\frac{1.32}{\pi} = 75.5^0[/tex] relative to the North-bound.

b) As the North component of the airplane velocity cancel out the wind south-bound speed. The speed of the plane over the ground would be the West component of the airplane velocity, which is

[tex]320sin(\alpha) = 320sin(75.5^0) = 309.84 km/h[/tex]

Answer:

a) Ф=14°, north of west

b) 310 km/h

Explanation:

we have,

[tex]v_{p/G}[/tex], velocity of plane relative to the ground(west)

[tex]v_{p/A}[/tex],velocity of plane relative to the air(320 km/h)

[tex]v_{A/G}[/tex],velocity of air relative to the ground(80 km/h, due south).

                 [tex]v_{p/G}[/tex]=[tex]v_{p/A}[/tex]+[tex]v_{A/G}[/tex].........(1)

a)     sin(Ф)=[tex]\frac{v_{A/G}}{v_{p/A}}[/tex]

                 =[tex]\frac{80km/h}{320km/h}[/tex]              

Ф=14°, north of west

b)    using Pythagorean theorem

[tex]v_{p/G}=\sqrt{v^2_{p/A}+v^2_{A/G}}[/tex]

[tex]v_{p/G}[/tex]=310 km/h

note:

diagram is attached

Answer:

0.08 N/C

Explanation:

Electric Field: This is defined as the force per unit charge exerted at a point. The expression for electric field is given as,

E = Kq/r².............................. Equation 1

Where E = Electric Field, q = Charge, k = proportionality constant, r = distance.

making q the subject of the equation,

q = Er²/k............................... Equation 2

Given: E = 2 N/C, r = 4 m,

Substitute into equation 2

q = 2(4)²/k

q = 32/k C.

When r is increased to 20 m,

E = k(32/k)/20²

E = 32/400

E = 0.08 N/C.

Hence the electric Field = 0.08 N/C

The initial electric field from a charge is 2 N/C at 4m distance. When the distance increases fivefold, the strength of the electric field decreases by the square of this factor, resulting in a new electric field magnitude of 0.08 N/C.

This question is about the relationship between an electric field and its distance from a point charge. According to the formula for the magnitude of the electric field generated by a point charge, which is E = kQ / r^2, where E is the electric field strength, k is Coulomb's constant, Q is the charge, and r is the distance from the charge, the electric field is inversely proportional to the square of the distance.

At a distance of 4m, the electric field's magnitude is 2 N/C. If you increase the distance to 20m (which is 5 times more than the initial distance), the new electric field magnitude will be the initial magnitude divided by the square of this factor, i.e., 2 N/C divided by 5^2 = 2 N/C / 25, which equals to 0.08 N/C.

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Answer:

Explanation:

The electric field outside the sphere is given as,

E = k Q /r²

here Q = n x 1.6 x 10⁻¹⁹ C

where n is the number of electons

if the dimeter of sphere d= 25 cm= 0.25 m

then the radius r = 0.125 m

we get

n= E r²/ k x 1.6 x 10⁻¹⁹ C

n = 1350N/C x (0.125m)² / (8.99 x 10⁹ N m²/C² x 1.6 x 10⁻¹⁹ C)

n = 14664731646

Answer:

[tex]\frac{4}{3}\rho\pi(r_2^3 - r_1^3)[/tex]

Explanation:

The volume of the shell is the difference between the volume of the outer sphere and the volume of the inner sphere

[tex]V = V_o - V_i = \frac{4}{3}\pi r_2^3 - \frac{4}{3}\pi r_1^3[/tex]

[tex]V = \frac{4}{3}\pi(r_2^3 - r_1^3)[/tex]

The mass of the shell is the product of the volume and its density

[tex]m = V\rho = \frac{4}{3}\rho\pi(r_2^3 - r_1^3)[/tex]

Answer:

0.00000103149529075 m²/s

0.0103149529075 stokes

Explanation:

A = [tex]2.4\times 10^-5[/tex]

B = 250 K

C = 140 K

T = 20°C

[tex]\rho[/tex] = Density of water = 998 kg/m³

Viscosity is given by

[tex]\mu=A\times 10^{\dfrac{B}{T-C}}\\\Rightarrow \mu=2.4\times 10^{-5}\times 10^{\dfrac{250}{273.15+20-140}}\\\Rightarrow \mu=0.00102943230017\ Pas[/tex]

Kinematic viscosity is given by

[tex]\nu=\dfrac{\mu}{\rho}\\\Rightarrow \nu=\dfrac{0.00102943230017}{998}\\\Rightarrow \nu=0.00000103149529075\ m^2/s[/tex]

The kinematic viscosity is 0.00000103149529075 m²/s

In BG units [tex]0.00000103149529075\times 10^4=0.0103149529075\ stokes[/tex]

In the SI system of units, the dynamic viscosity of water can be computed using the equation μ=A10B/(T-C). The dimensions of A are (m⋅s²)/kg. To find the kinematic viscosity of water at 20°C, we can use the formula ν=μ/ρ, where μ is the dynamic viscosity and ρ is the density of water. The kinematic viscosity of water at 20°C is approximately 0.0001 m²/s in SI units and 1 cm²/s in BG units.

(a) Dimensions of A:

To determine the dimensions of A, we need to rearrange the equation and analyze the units on both sides. We know that viscosity has units of kilograms per meter per second (kg/m/s). Plugging in the given values for B and C, we have μ=A10B/(T-C). The dimensions of 10B/(T-C) are (dimensionless). Therefore, the dimensions of A must be kg/m/s multiplied by the inverse of the dimensions of 10B/(T-C), which is 1/(kg/m/s) or (m⋅s²)/kg.

(b) Kinematic viscosity at 20°C:

To find the kinematic viscosity of water at 20°C, we can use the formula ν=μ/ρ, where μ is the dynamic viscosity and ρ is the density of water. The density of water at 20°C is approximately 998 kg/m³. Plugging in the given values for A and C, we can calculate μ using the equation μ=A10B/(T-C). Substituting the values into the formula ν=μ/ρ, we get ν≈μ/ρ≈(A10B/(T-C))/ρ.

In the SI unit, ν≈(2.4×10-5×10250 K/(293 K-140 K))/998 kg/m³≈0.0001 m²/s.

In the BG unit, ν≈0.0001 m²/s×104 cm²/1 m²≈1 cm²/s.

Answer:

a. -7.11+5.7

Explanation:

Given:

Now the acceleration:

[tex]a=\frac{v_w}{t} +\frac{v_n}{t}[/tex]

[tex]a=\frac{-25\hat i}{3.5} +\frac{20\hat j}{3.5}[/tex]

[tex]a=\frac{-50}{7} \hat i+\frac{40}{7} \hat j[/tex]

[tex]a=-7.1428\hat i+5.7142\hat j[/tex]

[tex]a=\sqrt{(\frac{50}{7} )^2+(\frac{40}{7} )^2}[/tex]

[tex]a=9.1473\ m.s^{-2}[/tex]

Answer:

(a) E = 3.065 * 10^21 N/C

(b) The direction is radially outward.

Explanation:

Parameters given:

Number of protons = 94

Charge of proton = 1.6023 * 10^-19 C

Radius of nucleus = 6.65 fm = 6.65 * 10^-15

At the surface of the nucleus, Electric field, E is given as:

E = kq/r^2

Where k = Coulombs constant

r = radius of nucleus

=> E = (9 * 10^9 * 94* 1.6023 * 10^-19)/(6.65 * 10^21)

E = 3.065 * 10^21 N/C

The direction of the field is radially outward.

The magnitude of the electric field at the surface of a plutonium-239 nucleus is roughly 7.94 x 10^21 NC^-1. The direction of the electric field, created by positively charged protons, is radially outward.

The question pertains to the electric field created by the protons of a plutonium-239 atom. The electric field resulting from a spherical charge distribution can be determined by using Coulomb's law transformed to a volumetric charge distribution. The formula is E = kQ/r2, where k is the Coulomb constant (8.99 x 109 Nm2/C2), Q is the total charge, and r is the distance from the center of the field (which in this case is the radius of the nucleus).

(a) Magnitude of the electric field: Here, the total charge Q is the charge of a single proton (1.60 x 10-19 C) multiplied by the total number of protons which is 94. Therefore, upon inserting these values into the formula, the calculated electric field at the surface of the nucleus is around 7.94 x 1021 NC-1.

(b) Direction of the electric field: The direction of the electric field is always from positive to negative. Since protons are positively charged and they are creating the field, the electric field direction will be radially outward.

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To solve this problem we will apply the concepts related to the Force of gravity given by Newton's second law (which defines the weight of an object) and at the same time we will apply the Hooke relation that talks about the strength of a body in a system with spring.

The extension of the spring due to the weight of the object on Earth is 0.3m, then

[tex]F_k = F_{W,E}[/tex]

[tex]kx_1 = mg[/tex]

The extension of the spring due to the weight of the object on Moon is a value of [tex]x_2[/tex], then

[tex]kx_2 = mg_m[/tex]

Recall that gravity on the moon is a sixth of Earth's gravity.

[tex]kx_2 = m\frac{g}{6}[/tex]

[tex]kx_2 = \frac{1}{6} mg[/tex]

[tex]kx_2 = \frac{1}{6} kx_1[/tex]

[tex]x_2 = \frac{1}{6} x_1[/tex]

We have that the displacement at the earth was [tex]x_1 = 0.3m[/tex], then

[tex]x_2 = \frac{1}{6} 0.3[/tex]

[tex]x_2 = 0.05m[/tex]

Therefore the displacement of the mass on the spring on Moon is 0.05m

The same spring and mass system will have a displacement of 0.05 m on the Moon compared to 0.3 m on Earth due to the Moon's lower gravitational acceleration.

The subject of the question relates to how a spring and mass system will behave on Earth versus the Moon. The solution requires understanding of Hooke's Law and gravitation. Hooke's Law states that the displacement of a spring is directly proportional to the force applied. On Earth, when a 0.500 kg mass is suspended and displaced by 0.3 m, this sets up a certain relationship of force (F = mg, where g is Earth's gravity, 9.8 m/s²): F = 0.500 kg * 9.8 m/s² = 4.9 Newtons. Displacement, d, is proportional to the force:F = kd, so d = F/k where k is the spring constant.

On the Moon, g is not equal to Earth's, it's 1.67 m/s². The same hanging mass on the Moon would exert a force of Fm=0.500 kg * 1.67 m/s² = 0.835 Newtons. Since the spring constant doesn't change, and F = kd still holds, the new displacement on the Moon will be greater because the force is smaller. Displacement on the Moon (dM) will be dM = Fm/k, which is Fm divided by the same k we had on Earth. Not knowing k, we do know dM = (Fm/F) * d, and (Fm/F) is equal to the ratio of the Moon's gravity to Earth's gravity, 1/6, so dM = (1/6)*0.3 m = 0.05 m.

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Answer: Water on Earth was transported here by comets. The correct option is A.

Explanation:

Comets are made up of water with ice, rock and minerals.

Alot of research and hypotheses has been made to prove the origin of water on planet earth. Extraplanetary source such as comets, trans-Neptunian objects, and water-rich meteoroids (protoplanets) are believed to have delivered water to Earth.

Final answer:

Water on Earth primarily originated from the (b) solar nebula during planetary accretion, with minor contributions from comets and volcanic activity. Hence, (b) is the correct option.

Explanation:

As Earth coalesced from the protoplanetary disk around the young Sun, volatile compounds, including water, were incorporated into the developing planet. Comets may have contributed some water, but the predominant source was the solar nebula.

While volcanic activity released water vapor, it played a minor role compared to the water acquired during accretion. Chemical reactions involving hydrogen and oxygen likely occurred after Earth's formation, but the bulk of the water was established during the early stages of planetary accretion from the solar nebula.

Answer:

Explanation:

Each square meter of earth surface intercepts 1110 protons per second.

Total surface of the earth will intercept  no of proton equal to

= 1110 x area of surface

= 1110 x 5.1 x 10¹⁴ m²

= 5661 x 10¹⁴

Total charge

= 1.6 x 10⁻¹⁹ x  5661 x 10¹⁴ coulomb per second

= .09 A .

electric current intercepted by the total surface area of the planet = .09 A.

Final answer:

To find the horizontal distance, we use the equation d = horizontal velocity * T, where T is the time of flight. Using the given values, the horizontal distance is 25.71 m.

Explanation:

To determine the horizontal distance the helicopter should be from the launch site, we need to find the time it takes for the projectile to reach the helicopter. Since the projectile only travels in the vertical direction, we can use the equation:

T = (2 * v0) / g

where T is the time of flight, v0 is the initial vertical velocity, and g is the acceleration due to gravity. Plugging in the values, we get:

T = (2 * 28 m/s) / 9.8 m/s^2 = 5.71 s

Now, we can find the horizontal distance using the equation:

d = horizontal velocity * T

The horizontal velocity of the helicopter is its constant speed, which is 4.5 m/s. Plugging in the values, we get:

d = 4.5 m/s * 5.71 s = 25.71 m