What is the kinetic energy of a 1450 kg sports car traveling down the road with a speed of 30 m/s?

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

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: 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:

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:

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:

The diameter of the bull-wheel is 3.82

Explanation:

Given that,

Velocity = 2.0 m/s

Angular velocity = 10 rev/m

[tex]\omega=10\times\dfrac{2\pi}{60}[/tex]

[tex]\omega=1.0472\ rad/s[/tex]

We need to calculate the diameter of bull-wheel

Using formula of angular velocity

[tex]v= r\omega[/tex]

[tex]r=\dfrac{v}{\omega}[/tex]

Put the value into the formula

[tex]r=\dfrac{2.0}{1.0472}[/tex]

[tex]r=1.91\ m[/tex]

The diameter of the bull-wheel

[tex]D=2r[/tex]

[tex]D=2\times1.91[/tex]

[tex]D=3.82\ m[/tex]

Hence, The diameter of the bull-wheel is 3.82 m.

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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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:

option (b)

Explanation:

Th weight of the body is equal to the product of mass of the body and acceleration due to gravity. So, it depends on the acceleration due to gravity.

Inertia is the inherent property of the body which always resists any change in the body. It does not depend on any factor. mass is the measure of inertia.

Momentum is the product of mass of body and the velocity of the body.

Force is the product of mass and the acceleration of the body.

Answer:

The distance is 19.58 m.

Explanation:

Given that,

Mass = 3.0 kg

Radius = 1.3 m

Angular speed = 6.8 rad/s

Angle = 24°

Acceleration of gravity = 9.81 m/s²

We need to calculate the distance

Using formula of kinetic energy

Total Initial kinetic energy,

[tex]K.E = \dfrac{1}{2}(mv^2+I\omega^2)[/tex]

[tex]K.E = \dfrac{1}{2}(mv^2+mr^2\omega^2)[/tex]

[tex]K.E =\dfrac{(mr^2+mr^2)\omega^2}{2}[/tex]

[tex]K.E =mr^2\omega^2[/tex]....(I)

Now, Total potential energy

[tex]P.E=mgs\sin\theta[/tex]

[tex]P.E=mgs\sin24^{\circ}[/tex]....(II)

Equating equation (I) and (II)

[tex]mr^2\omega^2==mgs\sin24^{\circ}[/tex]

[tex](1.3)^2\times6.8^2=9.81\times s\times\sin24^{\circ}[/tex]

[tex]s = \dfrac{1.3^2\times6.8^2}{9.81\times\sin24^{\circ}}[/tex]

[tex]s=19.58\ m[/tex]

Hence, The distance is 19.58 m.

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.

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]

Answer:

The temperature of hot reservoir is 669.24 K.

Explanation:

It is given that,

Efficiency of gasoline engine, [tex]\eta=35.3\%=0.353[/tex]

Temperature of cold reservoir, [tex]T_C=160^{\circ}C=433\ K[/tex]

We need to find the temperature of hot reservoir. The efficiency of Carnot engine is given by :

[tex]\eta=1-\dfrac{T_C}{T_H}[/tex]

[tex]T_H=\dfrac{T_C}{1-\eta}[/tex]

[tex]T_H=\dfrac{433}{1-0.353}[/tex]

[tex]T_H=669.24\ K[/tex]

So, the temperature of hot reservoir is 669.24 K. Hence, this is the required solution.

Answer:

Option (4)

Explanation:

There are two types of collision.

Perfectly elastic collision: the collision in which the momentum and kinetic energy is conserved. There is no loss of energy in other forms of energy.

Perfectly plastic collision: The collision in which the momentum is conserved and kinetic energy is not conserved. The two bodies stick after the collision.

Here, the bullet hits the block and then embedded in the block, it is the example of plastic collision.

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:

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:

The the enthalpy of the reaction is 241.36 KJ.

Explanation:

Given that,

Weight of sodium = 0.511 g

We need to calculate the number of moles

Using formula of moles

[tex]moles=\dfrac{given\ mass}{molar\ mass}[/tex]

[tex]moles=\dfrac{0.511}{23}[/tex]

[tex]moles= 0.022[/tex]

We need to calculate the energy in 1 mole

In 0.022 moles of sodium metal = 5310 J

In 1 moles of sodium metal =[tex]\dfrac{5310}{0.022}[/tex]

The Energy in 1 moles of sodium is 241363.63 J.

Hence, The the enthalpy of the reaction is 241.36 KJ.

Final answer:

To find the reaction enthalpy per mole of sodium reacting with hydrochloric acid, divide the heat produced (5310 J) by the number of moles of sodium (0.0222 mol). This calculation results in an enthalpy change of -239 kJ/mol for sodium.

Explanation:

To calculate the enthalpy of the reaction per mole of sodium metal reacting with hydrochloric acid, we can use the information provided about the heat produced during the reaction. Given that 0.511 g of sodium metal yields 5310 J of heat, we first need to determine the amount in moles of sodium. We know that the molar mass of sodium (Na) is approximately 23.0 g/mol. Thus, 0.511 g Na amounts to 0.511 g / 23.0 g/mol = 0.0222 mol Na.

Given the heat released, we can now calculate the enthalpy change per mole of sodium reacted. The enthalpy change (H) for the reaction can be calculated by dividing the heat by the number of moles of sodium:
H = Heat Produced / Moles of Sodium
H = 5310 J / 0.0222 mol Na
H = -239,189 J/mol Na (note the negative sign because the reaction is exothermic)

Converting this result to kilojoules, the enthalpy change per mole of sodium is approximately -239.189 kJ/mol Na, which we often express as -239 kJ/mol to match conventional significant figures for such values.

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: 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:

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:

0.003181 radians

0.003005 radians

Explanation:

Number of slits = 140 /cm

λ = Wavelength = 434 nm = 434×10⁻⁹ m

m = 3 Third order spectrum

Space between slits

[tex]d=\frac{0.01}{140} =7.14\times 10^{-5}\ m[/tex]

Now,

[tex]dsin\theta = m\lambda\\\Rightarrow \theta=sin^{-1}\left(\frac{m\lambda}{d}\right)\\\Rightarrow \theta=sin^{-1}\left(\frac{3\times 434\times 10^{-9}}{7.14\times 10^{-5}}\right)\\\Rightarrow \theta=sin^{-1}0.018228\\\Rightarrow \theta=0.01823^{\circ}=0.01823\times \frac{\pi}{180}=0.003181 radians[/tex]

0.003181 radians

When λ = 410 nm = 410×10⁻⁹ m

[tex]dsin\theta = m\lambda\\\Rightarrow \theta=sin^{-1}\left(\frac{m\lambda}{d}\right)\\\Rightarrow \theta=sin^{-1}\left(\frac{3\times 410\times 10^{-9}}{7.14\times 10^{-5}}\right)\\\Rightarrow \theta=sin^{-1}0.01722\\\Rightarrow \theta=0.01722^{\circ}=0.01722\times \frac{\pi}{180}=0.003005 radians[/tex]

0.003005 radians

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:

Part a)

[tex]f_w = f_g = 4.57 \times 10^{14} Hz[/tex]

Part b)

[tex]\lambda_w = 492 nm[/tex]

[tex]\lambda_g = 437.3 nm[/tex]

Part c)

[tex]v_w = 2.25 \times 10^8 m/s[/tex]

[tex]v_g = 2.0 \times 10^8 m/s[/tex]

Explanation:

Part a)

frequency of light will not change with change in medium but it will depend on the source only

so here frequency of light will remain same in both water and glass and it will be same as that in air

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

[tex]f = \frac{3 \times 10^8}{656 \times 10^{-9}}[/tex]

[tex]f = 4.57 \times 10^{14} Hz[/tex]

Part b)

As we know that the refractive index of water is given as

[tex]\mu_w = 4/3[/tex]

so the wavelength in the water medium is given as

[tex]\lambda_w = \frac{\lambda}{\mu_w}[/tex]

[tex]\lambda_w = \frac{656 nm}{4/3}[/tex]

[tex]\lambda_w = 492 nm[/tex]

Similarly the refractive index of glass is given as

[tex]\mu_w = 3/2[/tex]

so the wavelength in the glass medium is given as

[tex]\lambda_g = \frac{\lambda}{\mu_g}[/tex]

[tex]\lambda_g = \frac{656 nm}{3/2}[/tex]

[tex]\lambda_g = 437.3 nm[/tex]

Part c)

Speed of the wave in water is given as

[tex]v_w = \frac{c}{\mu_w}[/tex]

[tex]v_w = \frac{3 \times 10^8}{4/3}[/tex]

[tex]v_w = 2.25 \times 10^8 m/s[/tex]

Speed of the wave in glass is given as

[tex]v_g = \frac{c}{\mu_g}[/tex]

[tex]v_g = \frac{3 \times 10^8}{3/2}[/tex]

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

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:

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:

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:

Explanation:

Let east be towards X-axis ant north be Y-axis. Let initial position of Dog be at

A . O be the police station (centre). Vector OA can be written as follows

OA = 1.2 Cos 33 + 1.2 Sin 33

O B =2 Cos 75 + 2 Sin 75.

Displacement A   → B = OB - OA  =2Cos75i +2 Sin 75j -1.2 Cos 33i -1.2 Sin 33j

= 2 x .2588 i+ 2x .966j - 1.2 x .8387i - 1.2 x .5446j = .5176i + 1.932 j- 1.0064i - .65356j

= -.4888 i + 1.27844j

Magnitude of displacement =√( .4888)² + ( 1.27844)²

= 1.405 km

Average velocity =1.405 / 57-23 km / min  = 1.405 /34 x 60 =2.48 km/h

angle with x-axis ( east towards north ) ∅

 Tan∅ =- 1.27844/.4888 = - 2.615

∅ = -69° or 111° towards north from east or 21° towards west from north.

The meteorite's speed just before striking the sand is approximately 3836.82 m/s, calculated using conservation of energy principles from its initial potential energy at 750 km above Earth.

Let's break down the solution step by step:

Step 1: Calculate the initial potential energy of the meteorite when it is 750 km above the Earth's surface.

[tex]\[PE_{initial} = mgh\][/tex]

Where:

- [tex]\(m\)[/tex] is the mass of the meteorite,

- [tex]\(g\)[/tex] is the acceleration due to gravity, and

- [tex]\(h\)[/tex] is the height above the Earth's surface.

Given:

- [tex]\(h = 750 \, \text{km} = 750 \times 10^3 \, \text{m}\)[/tex] (converted to meters)

- [tex]\(g = 9.8 \, \text{m/s}^2\)[/tex] (acceleration due to gravity)

[tex]\[PE_{initial} = mg \times h\][/tex]

[tex]\[PE_{initial} = (m)(9.8 \, \text{m/s}^2)(750 \times 10^3 \, \text{m})\][/tex]

Step 2: Calculate the final kinetic energy of the meteorite just before it strikes the sand.

[tex]\[KE_{final} = \frac{1}{2}mv^2\][/tex]

Where:

- [tex]\(v\)[/tex] is the final velocity of the meteorite just before it strikes the sand.

Given:

- [tex]\(d = 3.35 \, \text{m}\)[/tex] (distance traveled while coming to rest)

Step 3: Apply the principle of conservation of energy.

Since energy is conserved, the initial potential energy of the meteorite is converted into kinetic energy just before it strikes the sand.

[tex]\[PE_{initial} = KE_{final}\][/tex]

[tex]\[mg \times h = \frac{1}{2}mv^2\][/tex]

[tex]\[9.8 \times 750 \times 10^3 = \frac{1}{2} \times v^2\][/tex]

[tex]\[v^2 = \frac{2 \times 9.8 \times 750 \times 10^3}{1}\][/tex]

[tex]\[v^2 = 2 \times 9.8 \times 750 \times 10^3\][/tex]

[tex]\[v^2 = 2 \times 7.35 \times 10^6\][/tex]

[tex]\[v^2 = 14.7 \times 10^6\][/tex]

[tex]\[v^2 = 1.47 \times 10^7\][/tex]

[tex]\[v = \sqrt{1.47 \times 10^7}\][/tex]

[tex]\[v \approx 3836.82 \, \text{m/s}\][/tex]

So, the speed of the meteorite just before striking the sand is approximately [tex]\(3836.82 \, \text{m/s}\)[/tex].

Answer:

[tex]C_2=\frac{C_1}{4}[/tex]

Explanation:

Given:

Initial capacitance, C = C₁

Initial Frequency, F₁ = 600kHz

Final Frequency, F₂ = 1200kHz

let the adjusted capacitance be, C₂

Now

the  frequency (f) is given as:

[tex]f={\frac{1}{2\pi \sqrt{LC}}}[/tex]

where, L = inductance (same for the same material)

thus substituting the values we have

[tex]600={\frac{1}{2\pi \sqrt{LC_1}}}[/tex]     ...............(1)

and

[tex]1200={\frac{1}{2\pi \sqrt{LC_2}}}[/tex]        ..............(2)

dividing the equation 1 with 2, we get

[tex]\frac{600}{1200}=\frac{{\frac{1}{2\pi \sqrt{LC_1}}}}{{\frac{1}{2\pi \sqrt{LC_2}}}}[/tex]

or

[tex]\frac{1}{2}=\sqrt{\frac{C_2}{C_1}}[/tex]

or

[tex]\frac{C_2}{C_1}=\frac{1}{4}[/tex]

or

[tex]C_2=\frac{C_1}{4}[/tex]

hence, the new capacitance C, must be one-fourth times the initial capacitance

The capacitance for the frequency of 1200 kHz must be one-fourth times the capacitance for the frequency of 800 kHz.

Capacitance is the ability of a circuit to store energy in the form of an electrical charge.it is an energy-storing device. generally, it is defined by the ratio of electric charge stored to the potential difference.

Given:

Initial capacitance is C

Capacitance for the frequency of 1200 kHz is [tex]\rm{C=C_1}[/tex]

The capacitance for the frequency of 800 kHz is [tex]\rm{C=C_2}[/tex]

The  frequency (f) is given by

[tex]f=\frac{1}{2\pi \sqrt{LC} }[/tex]

The capacitance for the frequency of 1200 kHz finds by

[tex]f_1=\frac{1}{2\pi \sqrt{LC_1} }[/tex]

[tex]1200=\frac{1}{2\pi \sqrt{LC_1} }[/tex]        ------------------1

The capacitance for the frequency of 800 kHz

[tex]f_2=\frac{1}{2\pi \sqrt{LC_2} }[/tex]

[tex]600=\frac{1}{2\pi \sqrt{LC_2} }[/tex]         ------------------2

On dividing the equation 2 by 1

[tex]\frac{1}{2} =\sqrt{\frac{C_2}{C_1} }[/tex]

[tex]{\frac{C_2}{C_1} }=\frac{1}{4}[/tex]

[tex]C_2=\frac{C_1}{4}[/tex]

Hence the capacitance for the frequency of 1200 kHz must be one-fourth times the capacitance for the frequency of 800 kHz.

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