A student performs this experiment and measures the bar to have a mass of 150g and length of 36cm. What is the moment of inertia of the bar?

To find the work done by the wind on the boat, we first find the southward component of the wind's force, since the boat is sailing south. We find this to be about 243 lb. Multiplying this force by the distance the boat travels, we find that the wind does approximately 31590 ft-lb of work on the boat.

In order to find the work done by the wind on the boat, we need to find the component of the wind force that acts in the same direction as the displacement of the boat. The wind is blowing in the direction of S36°E, meaning it has a southward component and an eastward component. However, since the boat is moving south, only the southward component of the wind's force will do work on the boat.

We use the equation F_s = F * cos(θ) to find the southward component of the force, where F is the magnitude of the total wind force and θ is the angle between the force of the wind and the direction of displacement. Plugging in the given values, we get F_s = 300 lb * cos(36°) = 243 lb.

To find work, we use the equation W = F * d * cos(θ), where F is the force, d is the distance traveled, and θ is the angle between the force and the displacement. Since the force and the displacement are in the same direction, the angle θ is 0, so cos(θ) is 1. Plugging in the appropriate values, we get W = 243 lb * 130 ft * 1 = 31590 ft-lb.

Rounded to the nearest whole number, the wind does approximately 31590 ft-lb of work on the boat.

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Answer:

Wavelength of microwaves is 3 m.

Explanation:

In this question, we need to find the wavelength of 100 MHz microwaves in muscle and in fat.

Frequency of the microwaves, [tex]\nu=100\ MHz=100\times 10^6\ Hz=10^8\ Hz[/tex]

The relation between the frequency and the wavelength is given by :

[tex]c=\nu\times \lambda[/tex]

[tex]\lambda=\dfrac{c}{\nu}[/tex]

Where

c is the sped of light

[tex]\lambda=\dfrac{3\times 10^8\ m/s}{10^8\ Hz}[/tex]

[tex]\lambda=3\ m[/tex]

So, the wavelength of microwaves is 3 m. Hence, this is the required solution.

Explanation:

We need to find the speed of an electron that has fallen through a potential difference of  125 volts. It can be calculated using the De-broglie hypothesis as :

(a) V = 125 volts

[tex]\dfrac{1}{2}mv^2=qV[/tex]

Where

v = speed of electron

V is potential difference

[tex]v=\sqrt{\dfrac{2qV}{m}}[/tex]

[tex]v=\sqrt{\dfrac{2\times 1.6\times 10^{-19}\times 125\ V}{9.1\times 10^{-31}}}[/tex]

v = 6629935.44 m/s

[tex]v=6.62\times 10^6\ m/s[/tex]

(b) V = 125 megavolts

[tex]V=1.25\times 10^8\ V[/tex]

[tex]v=\sqrt{\dfrac{2qV}{m}}[/tex]

[tex]v=\sqrt{\dfrac{2\times 1.6\times 10^{-19}\times 1.25\times 10^8\ V}{9.1\times 10^{-31}}}[/tex]

[tex]v=6.62\times 10^9\ m/s[/tex]

Hence, this is the required solution.

Answer:

14 m/s

Explanation:

u = 0, h = 10 m, g = 9.8 m/s^2

Use third equation of motion

v^2 = u^2 + 2 g h

Here, v be the velocity of ball as it just strikes with the ground

v^2 = 0 + 2 x 9.8 x 10

v^2 = 196

v = 14 m/s

Answer:

Gravitational force, [tex]F=6.67\times 10^{-7}\ N[/tex]    

Explanation:

Masses of two iron balls, m₁ = m₂ = 10 kg

Distance between balls, d = 0.1 m

We need to find the gravitational force between two balls. It is given by :

[tex]F=G\dfrac{m_1m_2}{d^2}[/tex]

[tex]F=6.67\times 10^{-11}\times \dfrac{(10\ kg)^2}{(0.1\ m)^2}[/tex]

[tex]F=6.67\times 10^{-7}\ N[/tex]

Hence, this is the required solution.

Final answer:

The gravitational force between two 10 kg Iron balls separated by a distance of 0.1 m is 6.67 x 10^-11 N, calculated using Newton's law of universal gravitation.

Explanation:

The gravitational force between two 10 kg Iron balls separated by a distance of 0.1 m can be calculated using Newton's law of universal gravitation.

The formula for gravitational force is F = G(m1 * m2) / d^2, where G is the gravitational constant, m1 and m2 are the masses of the objects, and d is the distance between their centers.

Plugging in the values, the force would be 6.67 x 10^-11 N.

Answer:

847.45 N

Explanation:

F₁=force of exerted by smaller piston

F₂=force of exerted by larger piston=1.44×10⁴ N

A₁=Area of smaller piston= 2.62 cm =0.0265 m

A₂=Area of larger piston= 10.8 cm =0.108 m

Pressure exerted by both the pistons will be equal

[tex]P_1=P_2\\\Rightarrow \frac{F_1}{A_1}=\frac{F_2}{A_2}\\\Rightarrow F_1=\frac{F_2}{A_2} A_1\\\Rightarrow F_1=\frac{14400}{\pi\times  0.108^2}\pi\times  0.0262^2\\\Rightarrow F_1=847.45\ N[/tex]

Hence, force exerted to lift a 14400 N car is 847.45 N

Answer:

0.4375 cm

Explanation:

f = frequency of the chirp emitted by the bats = 7.84 x 10⁴ Hz

v = speed of sound in air = 343 m/s

λ = smallest wavelength = size of the smallest insect a bat can detect

Using the equation

v = f λ

inserting the values

343 = (7.84 x 10⁴) λ

λ = [tex]\frac{343}{7.84\times 10^{4}}[/tex]

λ = 43.75 x 10⁻⁴ m

λ = 0.4375 cm

Explanation:

The speed of light in vacuum is, c = 3 × 10⁸ m/s

We have to find the speed of light :

(a) In air :

[tex]n_1_{air}=1.0003[/tex]

The equation of refractive index is given as :

[tex]n_1=\dfrac{c}{v_1}[/tex]

[tex]v_1=\dfrac{c}{n_1}[/tex]

[tex]v_1=\dfrac{3\times 10^8\ m/s}{1.0003}[/tex]

[tex]v_1=299910026.9\ m/s[/tex]

[tex]v_1=2.99\times 10^8\ m/s[/tex]

(b) In diamond :

[tex]n_2_{diamond}=2.42[/tex]

The equation of refractive index is given as :

[tex]n_2=\dfrac{c}{v_2}[/tex]

[tex]v_2=\dfrac{c}{n_2}[/tex]

[tex]v_2=\dfrac{3\times 10^8\ m/s}{2.42}[/tex]

[tex]v_2=123966942.1\ m/s[/tex]

[tex]v_2=1.23\times 10^8\ m/s[/tex]

(c) In crown glass :

[tex]n_3_{glass}=1.5[/tex]

The equation of refractive index is given as :

[tex]n_3=\dfrac{c}{v_3}[/tex]

[tex]v_3=\dfrac{c}{n_3}[/tex]

[tex]v_3=\dfrac{3\times 10^8\ m/s}{1.5}[/tex]

[tex]v_3=200000000\ m/s[/tex]

[tex]v_3=2\times 10^8\ m/s[/tex]

(4) In water :

[tex]n_4_{glass}=1.34[/tex]

The equation of refractive index is given as :

[tex]n_4=\dfrac{c}{v_4}[/tex]

[tex]v_4=\dfrac{c}{n_4}[/tex]

[tex]v_4=\dfrac{3\times 10^8\ m/s}{1.34}[/tex]

[tex]v_4=223880597.01\ m/s[/tex]

[tex]v_4=2.23\times 10^8\ m/s[/tex]

Hence, this is the required solution.

Answer:

(a) 5.43 x 10⁵ J

(b) 3.07 x 10⁵ J

(c) 45 °C

Explanation:

(a)

[tex]L_{f}[/tex] = Latent heat of fusion of ice to water = 3.33 x 10⁵ J/kg

m = mass of ice = 1.63 kg

[tex]Q_{f}[/tex] = Energy required to melt the ice

Energy required to melt the ice is given as

[tex]Q_{f}[/tex] = m [tex]L_{f}[/tex]

[tex]Q_{f}[/tex] = (1.63) (3.33 x 10⁵)

[tex]Q_{f}[/tex] = 5.43 x 10⁵ J

(b)

E = Total energy transferred = 8.50 x 10⁵ J

Q  = Amount of energy remaining to raise the temperature

Using conservation of energy

E = [tex]Q_{f}[/tex] + Q

8.50 x 10⁵ = 5.43 x 10⁵ + Q

Q = 3.07 x 10⁵ J

(c)

T₀ = initial temperature = 0°C

T = Final temperature

m = mass of water = 1.63 kg

c = specific heat of water = 4186 J/(kg °C)

Q = Amount of energy to raise the temperature of water = 3.07 x 10⁵ J

Using the equation

Q = m c (T - T₀)

3.07 x 10⁵ = (1.63) (4186) (T - 0)

T = 45 °C

(a) The energy to melt the ice is 5.43 × 10^5 J.

(b) The remaining energy is 3.07 × 10^5 J used to raise the temperature of the water.

(c) The final temperature of the water is approximately 45 °C.

(a) Energy to Melt the Ice:

To find the energy required to melt the ice, we use the formula Q_melt = mass * latent heat of fusion.

Given:

Mass of the ice = 1.63 kg

Latent heat of fusion (Lf) = 3.33 × 10^5 J/kg

Q_melt = 1.63 kg * (3.33 × 10^5 J/kg) = 5.43 × 10^5 J

(b) Energy to Raise the Temperature of Water:

The remaining energy after melting is used to raise the temperature of the water. Subtracting Q_melt from the total energy transferred gives Q_raise.

Q_raise = 8.50 × 10^5 J - 5.43 × 10^5 J = 3.07 × 10^5 J

(c) Final Temperature:

To find the temperature increase, we use the formula ΔT = Q_raise / (mass * specific heat).

Given:

Mass of the water = 1.63 kg

Specific heat of water (c) = 4186 J/(kg · °C)

ΔT = (3.07 × 10^5 J) / (1.63 kg * 4186 J/(kg · °C)) ≈ 45 °C

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The question probable may be:

Suppose 8.50 ✕ 10^5 J of energy are transferred to 1.63 kg of ice at 0°C. The latent heat of fusion and specific heat of water are Lf = 3.33 ✕ 1065 J/kg and c = 4186 J (kg · °C) . HINT

(a) Calculate the energy (in J) required to melt all the ice into liquid water. (Enter your answer to at least three significant figures.)

(b) How much energy (in J) remains to raise the temperature of the liquid water? (Enter your answer to at least three significant figures.)

c) Determine the final temperature of the liquid water in Celsius.

Answer:

The ratio of stopping distances is 4 i.e by a factor 4 the stopping distances differ

Explanation:

Using 3rd equation of motion we have

For car 1

[tex]v_{1}^{^{2}}=u_{1}^{2}+2a_{1}s_{1}[/tex]

For car 2 [tex]v_{2}^{^{2}}=u_{2}^{2}+2a_{2}s_{2}[/tex]

Since the initial speed of both the cars are equal and when the cars stop the final velocities of both the cars become zero thus the above equations reduce to

[tex]u^{2}=-2a_{1}s_{1}\\\\s_{1}=\frac{-u^{2}}{2a_{1}}[/tex].............(i)

Similarly for car 2 we have

[tex]s_{2}=\frac{-u^{2}}{2a_{2}}[/tex]..................(ii)\

Taking ratio of i and ii we get

[tex]\frac{s_{1}}{s_{2}}=\frac{a_{2}}{a_{1}}[/tex]

Let

[tex]\frac{a_{2}}{a_{1}}=4[/tex]

Thus

[tex]\frac{s_{1}}{s_{2}}=4[/tex]

The ratio of stopping distances is 4

Final answer:

The stopping distances of two cars with different deceleration rates will differ by the inverse ratio of their decelerations. If one car's deceleration is a quarter of the other's, the car with the lower deceleration will require four times the stopping distance of the other car.

Explanation:

The question at hand is about how the stopping distances of two cars differ when their deceleration rates are different. If one car's deceleration is a quarter of the other's, then to find the factor by which the stopping distances differ, we can use the equation of motion [tex]v^2 = u^2 + 2as[/tex] v is the final velocity, u is the initial velocity, a is the deceleration, and s is the stopping distance.

Since the final velocity (v) for both cars will be zero (they come to a stop), and assuming the initial velocities (u) are the same for both cars, we can set the equation for both as follows: 0 = u^2 + 2a1s1 and 0 = u^2 + 2a2s2, with a1 being the higher deceleration of the first car and a2 being a quarter of that, s1 and s2 being the stopping distances respectively.

When we solve the equations for s1 and s2, we find that s1 is proportional to 1/a1 and s2 is proportional to 1/a2, so if a2 = 1/4 a1, then the ratio of the stopping distances s2/s1 is equal to the inverse ratio of the decelerations, which is 4. Hence, the car with the lower deceleration requires four times the distance to stop compared to the car with the higher deceleration.

Answer:

I) It becomes two times larger

II) It becomes four times larger

Explanation:

I) Electric field is directly proportional to Magnetic field and as such, if one is increased, the other is also increased by the same proportion.

The formula is given as

[tex]E_{1}  = cB_{1}[/tex]

c = speed of light, therefore

[tex]2E_{1}  = c2B_{1}[/tex]

II) Intensity of is proportional to the square of amplitude, as such

If amplitude is doubled ([tex]X2^{2}[/tex]), intensity is [tex]X4^{}[/tex]

This means the intensity becomes four times bigger.

Answer:

μ = 0.375

Explanation:

F = Applied force on the trash can = 75 N

W = weight of the trash can = 200 N

f = frictional force acting on trash can

Since the trash can moves at constant speed, force equation for the motion of can is given as

F - f = 0

75 - f = 0

f = 75 N

μ = Coefficient of friction

frictional force is given as

f = μ W

75 = μ (200)

μ = 0.375

Answer: The mass of nitrogen gas required to fill the balloon is 1.12 grams.

Explanation:

We are given:

Mass of helium gas = 0.16 g

We need to calculate the mass of nitrogen gas that can fill the balloon at same pressure, volume and temperature. This means that the moles of both the gases filling up balloon will be same.

So, to calculate the number of moles, we use the equation:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]  

Given mass of Helium = 0.16 g

Molar mass of helium = 4.00 g/mol

Putting values in above equation, we get:

[tex]\text{Moles of helium}=\frac{0.16g}{4g/mol}=0.04mol[/tex]

Now, calculating the mass of nitrogen gas using same above equation, we get:

Moles of nitrogen gas = 0.04 moles

Molar mass of nitrogen gas = 28.01 g/mol

Putting values in above equation, we get:

[tex]0.04mol=\frac{\text{Mass of }N_2}{28.01g/mol}\\\\\text{Mass of }N_2=1.12g[/tex]

Hence, the mass of nitrogen gas required to fill the balloon is 1.12 grams.

To fill the balloon with nitrogen (N2) under the same conditions as helium (He), you would need 1.12 grams, based on the molar masses of each gas and Avogadro's law.

The subject matter of your question lies in the field of chemistry, specifically the concepts of Avogadro's law and the ideal gas law. Avogadro's law states that equal volumes of gases, at the same temperature and pressure, contain the same number of molecules. Therefore, given that N2 and He are both gases, the same volume of each gas will contain the same number of molecules.

The ideal gas law can also be applied here. The molar mass of nitrogen (N₂) is 28.01 g/mol, and the molar mass of helium (He) is 4 g/mol. Given that we know the mass of helium required to fill the balloon (0.16 g), we can use the relationship between molar mass and the actual mass to determine the equivalent mass for nitrogen.

This relationship is represented by the ratio of the molar mass of nitrogen to the molar mass of helium, which is 28.01 / 4 = 7. Thus, to fill the balloon with nitrogen under the same conditions of pressure, volume, and temperature, the required mass would be 7 times that of the mass of the helium, or 0.16 * 7 = 1.12 g.

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Answer:

Its focal length is positive

Explanation:

A concave mirror is shown in attached figure. The distance from the pole to the focus of the mirror is called its focal length. Spherical mirrors are a part of a sphere.

As per conventions, we know that the axis opposite to x axis is taken as negative.

So, it is clear that the focal length of spherical concave mirror is negative.

Hence, the incorrect option is (c) " its focal length is positive".

Answer:

Rhodium has FCC structure.

Explanation:

Formula used :  

[tex]\rho=\frac{Z\times M}{N_{A}\times a^{3}}[/tex]

where,

[tex]\rho[/tex] = density

Z = number of atom in unit cell

M = atomic mass

[tex](N_{A})[/tex] = Avogadro's number  

a = edge length of unit cell

1) If it FCC cubic lattice

Number of atom in unit cell of FCC (Z) = 4

Atomic radius of Rh= 0.1345 nm = [tex]1.345\times 10^{-8} cm[/tex]

Edge length = a

For FCC,  a = 2.828 × r :

a = [tex]2.828\times 1.345\times 10^{-8} cm=3.80366\times 10^{-8}cm[/tex]

Density of Rh= [tex]12.41 g/cm^3[/tex]

Atomic mass of Rh(M) = 102.91 g/mol

On substituting all the given values , we will get the value of 'a'.

[tex]12.41 g/cm^3=\frac{4\times 102.91 g/mol}{6.022\times 10^{23} mol^{-1}\times (3.80366\times 10^{-8}cm)^{3}}[/tex]

[tex]12.41 g/cm^3\approx 12.4214 g/cm^3[/tex]

2) If it BCC cubic lattice

Number of atom in unit cell of BCC (Z) = 2

Atomic radius of Rh= 0.1345 nm = [tex]1.345\times 10^{-8} cm[/tex]

Edge length = a

For BCC,  a = 2.309 × r :

a = [tex]2.828\times 1.345\times 10^{-8} cm=3.105605\times 10^{-8}cm[/tex]

Density of Rh= [tex]12.41 g/cm^3[/tex]

Atomic mass of Rh(M) = 102.91 g/mol

On substituting all the given values , we will get the value of 'a'.

[tex]12.41 g/cm^3=\frac{2\times 102.91 g/mol}{6.022\times 10^{23} mol^{-1}\times (3.105605\times 10^{-8}cm)^{3}}[/tex]

[tex]12.41 g/cm^3[/tex] ≠ [tex]11.41 g/cm^3[/tex]

Rhodium has FCC structure.

Rhodium has a FCC structure because the edge length of the unit cell is smaller than 4 times the atomic radius divided by the square root of 3.

BCC stands for Body-Centered Cubic structure and FCC stands for Face-Centered Cubic structure. To determine which structure rhodium has, we need to compare the atomic radius of rhodium with the edge length of the unit cell for both structures.

For BCC, the diagonal distance of the body-centered cube is equal to 4 times the atomic radius. Therefore, the edge length of the unit cell for BCC is given by 4 times the atomic radius divided by √3.

For FCC, the face diagonal distance of the face-centered cube is equal to 2 times the atomic radius. Therefore, the edge length of the unit cell for FCC is given by 2 times the atomic radius.

By comparing the given atomic radius of 0.1345 nm to the calculated edge lengths, we can determine that rhodium has a FCC structure because 2 times the atomic radius (2 x 0.1345 nm) is smaller than 4 times the atomic radius divided by √3 (4 x 0.1345 nm / √3).

Answer: 259.2 KJ

Explanation:

The formula calculate work don in a circuit is given by :-

[tex]W=QV[/tex], where Q is charge and V is the potential difference.

The formula to calculate charge in circuit :-

[tex]Q=It[/tex], where I is current and t is time.

Given : Current : [tex]I=3A[/tex]

Potential difference : [tex]V=12\ V[/tex]

Time : [tex]t=2\ hr=2(3600)\text{ seconds}=7200\text{ seconds}[/tex]

Now, [tex]Q=3(7200)=21,600\ C[/tex]

Then,  [tex]W=(21600)(12)=259,200\text{ Joules}=259.2\text{ KJ}[/tex]

Hence, the work done = 259.2 KJ

The work done by the battery charger is 259.2 kilojoules.

To calculate the work done by the battery charger, we can use the formula:

Work (W) = Power (P) x Time (t)

In this case, the power is given as 3A (current) and 12 Volts (voltage), and the time is given as 2 hours. We need to convert the time to seconds:

2 hours = 2 x 60 x 60 = 7200 seconds

Now we can substitute the values into the formula:

W = 3A x 12V x 7200s = 259,200 Joules

To convert the work into kilojoules, we divide by 1000:

W = 259,200 J = 259.2 kJ

Therefore, the work done by the battery charger is 259.2 kilojoules.

Answer:

21.6 A

Explanation:

n = number density of free electrons in copper = 8.5 x 10²² cm⁻³ = 8.5 x 10²⁸ m⁻³

e = magnitude of charge on electron = 1.6 x 10⁻¹⁹ C

d = diameter of copper wire = 0.790 mm = 0.790 x 10⁻³ m  

Area of cross-section of copper wire is given as  

A = (0.25) πd²

A = (0.25) (3.14) (0.790 x 10⁻³)²

A = 4.89 x 10⁻⁷ m²

v = drift speed = 3.25 mm/s = 3.25 x 10⁻³ m /s

the electric current is given as

i = n e A v

i = (8.5 x 10²⁸) (1.6 x 10⁻¹⁹) (4.89 x 10⁻⁷ ) (3.25 x 10⁻³)

i = 21.6 A

Answer:

a)

346.67 N/C, downward

b)

1.3 m

Explanation:

(a)

q = magnitude of charge on the particle = 1.95 x 10⁻⁹ C

r = distance of location from the charged particle = 0.225 m

E = magnitude of electric field at the location

Magnitude of electric field at the location is given as

[tex]E = \frac{kq}{r^{2}}[/tex]

Inserting the values

[tex]E = \frac{(9\times 10^{9})(1.95 \times 10^{-9})}{(0.225)^{2}}[/tex]

E = 346.67 N/C

a negative charge produce electric field towards itself.

Direction :  downward

(b)

E = magnitude of electric field at the location = 10.5 N/C

r = distance of location from the charged particle = ?

q = magnitude of charge on the particle = 1.95 x 10⁻⁹ C

Magnitude of electric field at the location is given as

[tex]E = \frac{kq}{r^{2}}[/tex]

Inserting the values

[tex]10.5 = \frac{(9\times 10^{9})(1.95 \times 10^{-9})}{r^{2}}[/tex]

r = 1.3 m

The magnitude and direction of the electric field due to a particle of charge -1.95 nC at a point 0.225 m directly above it is approximately -3.57 x 10^5 N/C towards the particle. The distance from this particle at which its electric field has a magnitude of 10.5 N/C is approximately 0.036 m or 36 mm.

The magnitude of the electric field (E) due to a charged particle can be calculated using Coulomb's Law, which states E = k*Q/r^2, where k is Coulomb's constant (k ≈ 8.99 × 10^9 N m^2/C^2), Q is the charge of the particle, and r is the distance from the particle.

(a) Incorporating the given values: Q = -1.95 nC = -1.95 × 10^-9 C, and r = 0.225 m, we find E = k*Q/r^2 = (8.99 × 10^9 N m^2/C^2) * (-1.95 × 10^-9 C) / (0.225 m)^2 ≈ -3.57 x 10^5 N/C. The negative sign indicates the direction of the electric field is towards the particle.

(b) To find the distance at which the electric field has a magnitude of 10.5 N/C, we rearrange the equation to solve for r. That is r = sqrt(k*Q/E). By plugging in the given values, we find r = sqrt((8.99 × 10^9 N m^2/C^2 * -1.95 × 10^-9 C) / 10.5 N/C) ≈ 0.036 m or 36 mm.

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Answer:

Increased by 16 times

Explanation:

F = Gravitational force between two bodies

G = Gravitational constant = 6.67408 × 10⁻¹¹ m³/kg s²

m₁ = Mass of one body

m₂ = Mass of other body

d = distance between the two bodies

[tex]F=\frac{Gm_1m_2}{d^2}\\ F=\frac{1}{d^2}\quad \text {(as G and masses are constant)}[/tex]

[tex]F_{new}=\frac{1}{\left (\frac{d}{4}\right )^2}\\\Rightarrow F_{new}=\frac{1}{\frac{d^2}{16}}\\\Rightarrow F_{new}={16}\times \frac{1}{d^2}\\\Rightarrow F_{new}=16\times F[/tex]

∴Force will increase 16 times

Vb - Va = -366.7 V.

Vab = Va - Vb, the potential of a with respect to b, is equal to the work done by the electric force when a unit of charge moves from a to b, it is given by:

Vab = Va - Vb = Wab/q,

So, in order to determinate the potential difference Vb - Va we have to multiply by -1 both side of the equation above:

- (Va - Vb) = - (Wab/q)

Resulting

Vb - Va = -(Wab/q)

Given a positive charge q = 6.0μC = 6.0x10⁻⁶C, Wab = 2.2x10⁻³J. Determine Vb - Va.

Vb - Va = - (2.2x10⁻³J/6.0x10⁻⁶C)

Vb - Va = -366.7 J/C = -366.7 V

The potential difference between point A and point B is 366.67 V; point B is at a lower potential than point A as derived from the work done WAB and the charge q.

The potential difference (also known as voltage) between two points in an electric field, such as point A to point B (VB - VA), can be calculated using the work done by or against the electric field to move a charge q from point A to point B. The formula for work done by the electric force is W = q(VB - VA), and given that work WAB = +2.2 × 10-3 J and the charge q = +6.0 µC (or +6.0 × 10-6 C), we can rearrange the formula to solve for the potential difference: VB - VA = WAB / q.

The calculation yields VB - VA = +2.2 × 10-3 J / (+6.0 × 10-6 C) which equals +366.67 V. Therefore, the electric potential difference between point A and point B is 366.67 V, with point B being at a lower potential than point A since the charge is positive and the work done is positive, indicating that it has moved in the direction of the electric field, from higher to lower potential.

Answer:

Minimum uncertainty in position is [tex]\Delta x= 1157808.48\ m[/tex]

Explanation:

It is given that,

Uncertainty in the position of an electron, [tex]\Delta x=0.1\ nm=0.1\times 10^{-9}\ m[/tex]

According to uncertainty principle,

[tex]\Delta x.\Delta p\geq \dfrac{h}{4\pi}[/tex]

[tex]\Delta x.m\Delta v\geq \dfrac{h}{4\pi}[/tex]

[tex]\Delta v\geq \dfrac{h}{4\pi \times \Delta x\times m}[/tex]

[tex]\Delta v\geq \dfrac{6.62\times 10^{-34}\ J-s}{4\pi \times 0.1\times 10^{-9}\ m\times 9.1\times 10^{-31}\ kg}[/tex]

[tex]\Delta v\geq 578904.24\ m/s[/tex]

Let [tex]\Delta x[/tex] is the uncertainty in position after 2 seconds such that,

[tex]\Delta x=\Delta v\times t[/tex]

[tex]\Delta x=578904.24\ m/s\times 2\ s[/tex]

[tex]\Delta x= 1157808.48\ m[/tex]

or

[tex]\Delta x= 1.15\times 10^6\ m[/tex]

Hence, this is the required solution.

Answer:

a)

0.245 m/s

b)

0.904 m

Explanation:

a)

[tex]v_{d}[/tex] = speed of duck ahead of wave

[tex]v_{s}[/tex] = speed of surface wave = 0.32 m/s

T = time for paddling = 1.6 s

d = spacing between the waves = 0.12 m

speed of duck ahead of wave is given as

[tex]v_{d}[/tex] = [tex]v_{s}[/tex] - [tex]\frac{d}{T}[/tex]

[tex]v_{d}[/tex] = 0.32 - [tex]\frac{0.12}{1.6}[/tex]

[tex]v_{d}[/tex] = 0.245 m/s

b)

[tex]v_{w}[/tex] = speed of wave behind the duck

speed of wave behind the duck is given as

[tex]v_{w}[/tex] = [tex]v_{s}[/tex] + [tex]v_{d}[/tex]

[tex]v_{w}[/tex] = 0.32 + 0.245

[tex]v_{w}[/tex] = 0.565 m/s

D = spacing between the crests

spacing between the crests is given as

D = [tex]v_{w}[/tex] T

D = (0.565) (1.6)

D = 0.904 m

Answer:

F = 318.1 N

Explanation:

As we know that torque to open the nut is given by formula

[tex]\tau = \vec r \times \vec F[/tex]

so we can write it as

[tex]\tau = rFsin\theta[/tex]

now we know that

[tex]\tau = 48 Nm[/tex]

r = 0.23 m

angle between force and the wrench is 41 degree

so we have

[tex]48 = (0.23)Fsin41[/tex]

[tex]F = \frac{48}{(0.23)sin41}[/tex]

[tex]F = 318.1 N[/tex]

Answers: (1) a. weight, (2)b. Force changes by 2/9, (3)b. movement, (4)a. 40,000 Joules, (5)c. the soil will be 5°C.

Mass and weight are very different concepts.  

Mass is the amount of matter that exists in a body, which only depends on the quantity and type of particles within it. This means mass is an intrinsic property of each body and remains the same regardless of where the body is located.  

On the other hand, weight is a measure of the gravitational force acting on an object and is directly proportional to the product of the mass [tex]m[/tex] of the body by the acceleration of gravity [tex]g[/tex]:  

[tex]W=m.g[/tex]  

Then, since the Earth and the Moon have different values ​​of gravity, the weight of an object in each place will vary, but its mass will not.

According to the law of universal gravitation, which is a classical physical law that describes the gravitational interaction between different bodies with mass:  

[tex]F=G\frac{m_{1}m_{2}}{r^2}[/tex] (1)

Where:  

[tex]F[/tex] is the module of the force exerted between both bodies  

[tex]G[/tex] is the universal gravitation constant

[tex]m_{1}[/tex] and [tex]m_{2}[/tex] are the masses of both bodies.

[tex]r[/tex] is the distance between both bodies

If we double the mass of one object (for example [tex]2m_{1}[/tex]) and triple the distance between both (for example [tex]3r[/tex]). The equation (1) will be rewritten as:

[tex]F=G\frac{2m_{1}m_{2}}{(3r)^2}[/tex] (2)

[tex]F=\frac{2}{9}G\frac{m_{1}m_{2}}{r^2}[/tex] (3)

If we compare (1) and (2) we will be able to see the force changes by 2/9.

The Work [tex]W[/tex] done by a Force [tex]F[/tex] refers to the release of potential energy from a body that is moved by the application of that force to overcome a resistance along a path.  

When the applied force is constant and the direction of the force and the direction of the movement are parallel, the equation to calculate it is:  

[tex]W=(F)(d)[/tex]

Now, when they are not parallel, both directions form an angle, let's call it [tex]\alpha[/tex]. In that case the expression to calculate the Work is:  

[tex]W=Fdcos{\alpha}[/tex]

Therefore, pushing on a rock accomplishes no work unless there is movement (independently of the fact that movement is parallel to the applied force or not).

The Kinetic Energy is given by:

[tex]K=\frac{1}{2}mV^{2}[/tex]   (4)

Where [tex]m[/tex] is the mass of the body and [tex]V[/tex] its velocity

For the first case (kinetic energy [tex]K_{1}=10000J[/tex]  for a car at [tex]V_{1}=30 mph=13.4112m/s[/tex]):

[tex]K_{1}=\frac{1}{2}mV_{1}^{2}[/tex]   (5)

Finding [tex]m[/tex]:

[tex]m=\frac{2K_{1}}{V_{1}^{2}}[/tex]   (6)

[tex]m=\frac{2(10000J)}{(13.4112m/s)^{2}}[/tex]   (7)

[tex]m=111.197kg[/tex]   (8)

For the second case (unknown kinetic energy [tex]K_{2}[/tex]  for a car with the same mass at [tex]V_{2}=60 mph=26.8224m/s[/tex]):

[tex]K_{2}=\frac{1}{2}mV_{2}^{2}[/tex]   (9)

[tex]K_{2}=\frac{1}{2}(111.197kg)(26.8224m/s)^{2}[/tex]   (10)

[tex]K_{2}=40000J[/tex]   (11)

The formula to calculate the amount of calories [tex]Q[/tex] is:

[tex]Q=m. c. \Delta T[/tex]   (12)

Where:

[tex]m[/tex]  is the mass

[tex]c[/tex]  is the specific heat of the element. For water is [tex]c_{w}=1 kcal/g\°C[/tex]  and for soil is [tex]c_{s}=0.20 kcal/g\°C[/tex]  

[tex]\Delta T[/tex]  is the variation in temperature (the amount we want to find for both elements)

This means we have to clear [tex]\Delta T[/tex] from (12) :

[tex]\Delta T=\frac{Q}{m.c} [/tex]   (13)

For Water:

[tex]\Delta T_{w}=\frac{Q_{w}}{m_{w}.c_{w}} [/tex]   (14)

[tex]\Delta T_{w}=\frac{1kcal}{(1kg)(1 kcal/g\°C)}[/tex]   (15)

[tex]\Delta T_{w}=1\°C)}[/tex]   (16)

For Soil:

[tex]\Delta T_{s}=\frac{Q_{s}}{m_{s.c_{s}} [/tex]   (17)

[tex]\Delta T_{s}=\frac{1kcal}{(1kg)(0.20 kcal/g\°C)}[/tex]   (18)

[tex]\Delta T_{s}=5\°C)}[/tex]   (19)

Hence the correct option is c.

Answer:

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Answer:

The 99% confidence interval for the weights = [71.36lb, 78.64lb]

Explanation:

Mean weight

        [tex]\bar{x} =75 lb[/tex]

Variance of weights

        [tex]\sigma^2 =100lb[/tex]

Standard deviation,

       [tex]\sigma =\sqrt{100}=10lb[/tex]

Confidence interval  is given by

       [tex]\bar{x}-Z\times \frac{\sigma}{\sqrt{n}}\leq \mu\leq \bar{x}+Z\times \frac{\sigma}{\sqrt{n}}[/tex]

For 99% confidence interval Z = 2.576,

Number of weights, n = 50

Substituting

       [tex]75-2.576\times \frac{10}{\sqrt{50}}\leq \mu\leq 75+2.576\times \frac{10}{\sqrt{50}}\\\\71.36\leq \mu\leq 78.64[/tex]

The 99% confidence interval for the weights = [71.36lb, 78.64lb]

The 99 percent confidence interval for the population mean (μ) can be calculated using the formula X ± Z*(σ/√n) where X is the sample mean, Z is the Z-score for a 99% confidence interval (approximately 2.57), σ is the standard deviation of the population and n is the sample size. Substituting the given values results in 75 ± 2.57*(10/√50).

To calculate a 99 percent confidence interval for the population mean (μ), you can use the formula:

X± Z*(σ/√n)

Where:

Substituting these values into the formula gives the following 99 percent confidence interval:

75 ± 2.57*(10/√50).

Calculate the range and you will find your answer for the 99 percent confidence interval for μ.

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Answer:

2.5 x 10^-8 C

Explanation:

Diameter = 6 cm, radius = 3 cm = 0.03 m, d = 2 mm = 2 x 10^-3 m

E = 1 x 10^6 N/C, q = ?

q = C V

As we know that, V = E x d and C = ∈0 A / d

q = ∈0 x A x E x d / d

q = ∈0 x A x E

q = 8.854 x 10^-12 x 3.14 x 0.03 x 0.03 x 1 x 10^6

q = 2.5 x 10^-8 C

Electric field exerts a force on all charged particles. The charge on the electrode is 2.505 x 10⁻⁸ C.

An electric field can be thought to be a physical field that surrounds all the charged particles and exerts a force on all of them.

Given to us

Plate dimensions diameter, d = 6 cm

Area of the plate, A = πr² = π(0.03)² = 0.00283 m²

Distance between the two plates, d = 2 mm = 0.002 m

Electric field strength, E  = 1.0 x 10⁶ N/C

We know that electric field inside a parallel plate capacitor is given as,

[tex]E = \dfrac{Q}{A\epsilon_0}[/tex]

Substitute the value,

[tex]1 \times 10^6 = \dfrac{Q}{0.00283 \times 8.854 \times 10^{-12}}[/tex]

Q = 2.505 x 10⁻⁸ C

Hence, the charge on the electrode is 2.505 x 10⁻⁸ C.

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Answer:

61.85 ohm

Explanation:

L = 12 m H = 12 x 10^-3 H, C = 15 x 10^-6 F, Vrms = 110 V, R = 45 ohm

Let ω0 be the resonant frequency.

[tex]\omega _{0}=\frac{1}{\sqrt{LC}}[/tex]

[tex]\omega _{0}=\frac{1}{\sqrt{12\times 10^{-3}\times 15\times 10^{-6}}}[/tex]

ω0 = 2357 rad/s

ω = 2 x 2357 = 4714 rad/s

XL = ω L = 4714 x 12 x 10^-3 = 56.57 ohm

Xc = 1 / ω C = 1 / (4714 x 15 x 10^-6) = 14.14 ohm

Impedance, Z = [tex]\sqrt{R^{2}+\left ( XL - Xc \right )^{2}}[/tex]

Z = \sqrt{45^{2}+\left ( 56.57-14.14 )^{2}} = 61.85 ohm

Thus, the impedance at double the resonant frequency is 61.85 ohm.

In an LRC circuit, impedance at resonance is equal to resistance. When frequency is doubled, the impedance will increase as the inductive reactance becomes greater than the capacitive reactance. Calculation of the new impedance would involve re-computing inductive and capacitive reactances and substituting these in the impedance formula.

In an LRC series circuit, the impedance (Z) at resonance is equal to the resistance (R), because the reactance of the inductor (L) and the capacitor (C) cancel each other. The resonant frequency is determined by the values of L and C. If we increase the frequency to twice the resonance frequency, the reactance of the inductor becomes higher than the reactance of the capacitor, leading to an increase in total impedance.

Impedance (Z) is given by the square root formula Z = sqrt( R^2 + (X_L - X_C)^2 ). X_L and X_C are the inductive reactance (2πfL) and capacitive reactance (1/2πfC) respectively. If the frequency (f) is doubled, then X_L will be doubled while X_C will be halved.

We can calculate X_L and X_C under these conditions and substitute these values into the impedance formula to find the new impedance. Without the values of R, L, and C, it's not possible to give a numeric answer to this question, but the overall concept of how frequency affects impedance in an LRC circuit can be understood from the explanation.

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Answer:

90 miles

Explanation:

speed = 30 miles per hour, time = 3 hour

The formula for the speed is given by

speed = distance / time

Distance = speed x time

distance = 30 x 3 = 90 miles

Answer:

90 Miles

Explanation:

This is more simple than you thought...

The equation is 30 miles per hour,in 3 hours you would travel  (30*3)=90 miles.

Formula used Distance = Speed x Time

Speed = 30 mph

Time = 3 hours

So, Distance = Speed * Time = 30 * 3 = 90 miles.

So all in all, the answer is 90 miles

Answer:

The coefficient of x³y⁴ in the expansion of ( x+2y)⁷ is 560.

Explanation:

The given expression is

[tex](x+2y)^7[/tex]

According to binomial expansion,

[tex](a+b)^n=^nC_0a^nb^0+^nC_1a^{n-1}b^1+...+^nC_{n-1}a^1b^{n-1}+^nC_0a^0b^n[/tex]

The r+1th term of the expansion is

[tex]^nC_rx^{n-r}(2y)^r=^nC_r(2^r)x^{n-r}(y)^r[/tex]       ... (1)

In the term  x³y⁴ the power of x is 3 and the power of y is 4. It means the value of r is 4 and the value n-r is 3.

[tex]n-r=3[/tex]

[tex]n-4=3\Rightarrow n=7[/tex]

Put n=7 and r=4 in equation (1)

[tex]^7C_4(2^4)x^{7-4}(y)^4[/tex]

[tex]\frac{7!}{4!(7-4)!}(16)x^3y^4[/tex]

[tex]\frac{7\times 6\times 5\times 4!}{4!(3)!}(16)x^3y^4[/tex]

[tex]560x^3y^4[/tex]

Therefore the coefficient of x³y⁴ in the expansion of ( x+2y)⁷ is 560.

The coefficient of x^3 y^4 in the expansion of (x + 2y)^7 is calculated using the binomial theorem, resulting in 560.

To find the coefficient of x^3 y^4 in the expansion of (x + 2y)^7, we use the binomial theorem. The general term in the expansion of (a + b)^n is given by T(r+1) = C(n, r) * a^(n-r) * b^r, where C(n, r) is the binomial coefficient n choose r. In this question, we need to find the term where x is raised to the power of 3 and y to the power of 4.

Plug in the values: a = x, b = 2y, n = 7, and r = 4 (since y^4). Thus, T(5) = C(7, 4) * x^(7-4) * (2y)^4 = C(7, 4) * x^3 * 16y^4. The binomial coefficient C(7, 4) is 35.

Now multiply 35 by 16 to get the coefficient: 35 * 16 = 560. Therefore, the coefficient of x^3 y^4 is 560.