A 2170 kg space station orbits Earth at an altitude of 5.27 x 10^5 m. Find the magnitude of the force with which the space station attracts Earth. The mass and mean radius of Earth are 5.98 x 10^24 kg and 6.37 x 10^6 m, respectively.

Answer:

the visible wavelength is 480 nm

Explanation:

Given data

thick film = 4.0 × 10² nm

n = 1.2

wavelength range = 380 nm to 750 nm

to find out

the visible wavelength in air

solution

we know that index of water is 1 and kerosene is 1.2

we can say that when light travel reflected path difference is = 2 n t

and for maximum intensity it will be k × wavelength

so it will be  2 n t = k × wavelength

2 × 1.2 × 4.0 × 10² = k × wavelength

wavelength = 2 × 1.2 × 4.0 × 10² / k

here k is 2 for visible

so wavelength = 2 × 1.2 × 4.0 × 10² / 2

wavelength  = 480 nm

the visible wavelength is 480 nm

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:

The Magnetic field is 59.13 mT.

Explanation:

Given that,

Number of turns = 400

Radius = 1.0 cm

Frequency = 90 Hz

Emf = 4.2 V

We need to calculate the angular velocity

Using formula of angular velocity

[tex]\omega=2\pi f[/tex]

[tex]\omega=2\times3.14\times90[/tex]

[tex]\omega=565.2\ rad/s[/tex]

We need to calculate the magnetic flux

Relation between magnetic flux and induced emf

[tex]\epsilon=NA\omega B[/tex]

[tex]B=\dfrac{\epsilon}{NA\omega}[/tex]

Put the value into the formula

[tex]B=\dfrac{4.2}{400\times\pi\times(1.0\times10^{-2})^2\times 565.2}[/tex]

[tex]B=0.05913\ T[/tex]

[tex]B=59.13\ mT[/tex]

Hence, The Magnetic field is 59.13 mT.

Answer:

c. Throw the items away from the spaceship.

Explanation:

By the Principle of action and reaction yu can get back to your spacecraft throwing the items away from the spaceship.

Answer:

c. Throw the items away from the spaceship

Explanation:

The propulsion that throwing items away from the spaceship will propel you to the opposite direction in which you are throwing the objects, this means that if you throw the items away from the spaceship you will be getting force of rpopulsion towards the spaceship.

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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a. The resistance of the bulb is equal to 4.58 Ohms.

b. The fraction of the chemical energy transformed which appears as internal energy in the batteries is 8.4%.

Given the following data:

a. To find the resistance of the bulb:

First of all, we would determine the total electromotive force (E) of the electric circuit:

[tex]E = 2 \times 1.50[/tex]

E = 3.0 Volts

Total internal resistance = [tex]0.240 + 0.180 = 0.42 \;Ohms[/tex]

Mathematically, electromotive force (E) is given by the formula:

[tex]E = V + Ir[/tex]    ....equation 1.

Where:

According to Ohm's law, voltage is given by:

[tex]V = IR[/tex]    ....equation 2

Substituting eqn 2 into eqn 1, we have:

[tex]E = IR + Ir\\\\E = I(R + r)\\\\R + r = \frac{E}{I} \\\\R = \frac{E}{I} - r\\\\R = \frac{3}{0.6} - 0.42\\\\R = 5 - 0.42[/tex]

Resistance, R = 4.58 Ohms

b. To determine what fraction of the chemical energy transformed appears as internal energy in the batteries:

First of all, we would determine the electromotive force (E) in the batteries.

[tex]E_B = IR\\\\E_B = 0.6 \times 0.42[/tex]

[tex]E_B[/tex] = 0.252 Volt

[tex]Percent = \frac{E_B}{E} \times 100\\\\Percent = \frac{0.252}{3} \times 100\\\\Percent = \frac{25.2}{3}[/tex]

Percent = 8.4 %

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The resistance of the bulb and the fraction of the chemical energy transformed into internal energy in the batteries can be determined using Ohm's law and the power dissipation formula. The total resistance in the series circuit is the sum of the individual resistances, and the total voltage is the sum of the individual voltages.

The total voltage supplied by the batteries is the sum of their voltages, so Vtot = 1.5V + 1.5V = 3.0V. The total internal resistance of the batteries is the sum of their internal resistances, Rt = 0.240Ω + 0.180Ω = 0.420Ω. For the bulb's resistance, we can rearrange Ohm's law to R = V/I. Knowing the total voltage (3.0V) and the current (600 mA or 0.600 A), we can find the total resistance in the circuit.

Subtracting the total internal resistance gives us the resistance of the bulb. As for the second part of the question, the power dissipated in the internal resistances can be found using P = I²r for each battery, and then add these together. The total power supplied by the batteries is P = IV, using the total current and total voltage. The fraction of the chemical energy that appears as internal energy in the batteries is then Pinternal / Ptotal.

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Answer:

Explanation:

KE_s: Kinetic Energy Son

KE_f: Kinetic Energy Father.

Relationship

KE_f: =  (1/4) KE_s

m_s: = (1/3) m_f

v_f: = velocity of father

v_s: = velocity of the son

Relationship

1/2 mf (v_f + 1.2)^2 = 1/2 m_s (v_s)^2      Multiply both sides by 2.

mf (v_f + 1.2)^2 = m_s * (v_s)^2               Substitute for the mass of the m_s

mf (v_f + 1.2)^2 = (m_f/3) * (v_s)^2         Divide both sides by father's mass

(v_f + 1.2)^2 = 1/3 * (v_s)^2                      multiply both sides by 3

3*(v_f + 1.2)^2 = (v_s) ^2                         Take the square root both sides

√3 * (v_f + 1.2) = v_s

Note

KE_f = (1/4) KE_s                                                  Multiply by 4

4* KE_f = KE_s                                                     Substitute (again)

4*(1/2) m_f (v_f + 1.2)^2 = 1/2* (1/3)m_f *v_s^2   Divide by m_f

2* (v_f + 1.2)^2 = 1/6 * (v_s)^2                              multiply by 6

12*(vf + 1.2)^2 = (v_s)^2                                        Take the square root

2*√(3* (v_f + 1.2)^2) = √(v_s^2)

2*√3 * (vf + 1.2) = v_s

Use the second relationship to substitute for v_s so you can solve for v_f

2*√3 * ( v_f + 1.2) = √3 * (v_f + 1.2)                     Divide by sqrt(3)

2(v_f + 1.2) = vf + 1.2

Edit

2vf + 2.4 = vf + 1.2

2vf - vf + 2.4 = 1.2

vf = 1.2 - 2.4

vf = - 1.2

This answer is not possible, but 2 of us are getting the same answer. The other person is someone whose math I would never question. She rarely makes an error. And I do mean rarely. Could you check to see that you have copied this correctly?

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:

Energy, [tex]E=9\times 10^{13}\ J[/tex]

Explanation:

It is given that,

Mass of matter, m = 1 g = 0.001 Kg

If this matter is completely converted into energy, we need to find the yield. The mass of an object can be converted to energy as :

[tex]E=mc^2[/tex]

c = speed of light

[tex]E=0.001\ kg\times (3\times 10^8\ m/s)^2[/tex]

[tex]E=9\times 10^{13}\ J[/tex]

So, the energy yield will be [tex]9\times 10^{13}\ J[/tex]. Hence, this is the required solution.

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:

0.032 m

Explanation:

Consider the forces acting on the block

m = mass of the block = 1.50 kg

[tex]f_{s}[/tex] = Static frictional force

[tex]F_{n}[/tex] = Normal force on the block from the wall

[tex]F_{s}[/tex] = Spring force due to compression of spring

[tex]F_{g}[/tex] = Force of gravity on the block = mg = 1.50 x 9.8 = 14.7 N

k = spring constant = 860 N/m

μ = Coefficient of static friction between the block and wall = 0.54

x = compression of the spring

Spring force is given as

[tex]F_{s}[/tex] = kx

From the force diagram of the block, Using equilibrium of force along the horizontal direction, we get the force equation as  

[tex]F_{n}[/tex] = [tex]F_{s}[/tex]

[tex]F_{n}[/tex] = kx                                             eq-1

Static frictional force is given as

[tex]f_{s}[/tex] = μ [tex]F_{n}[/tex]

Using eq-1

[tex]f_{s}[/tex] = μ k x                                                eq-2

From the force diagram of the block, Using equilibrium of force along the vertical direction, we get the force equation as

[tex]f_{s}[/tex] = [tex]F_{g}[/tex]

Using eq-2

μ k x = 14.7

(0.54) (860) x = 14.7

x = 0.032 m

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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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:

Distance=538.51m

Explanation:

The echo is heard after covering double distance .

Therefore 2d=(t1+t2)×speed.

2d={(1.76+1.38)×343}

2d=743.02

d=1077.02÷2

=538.51m

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:

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.

Answer:

374 N

Explanation:

N = normal force acting on the skier

m = mass of the skier = 82.5

From the force diagram, force equation perpendicular to the slope is given as

N = mg Cos18.7

μ = Coefficient of friction = 0.150

frictional force is given as

f = μN

f =  μmg Cos18.7

F = force applied by the rope

Force equation parallel to the slope is given as

F - f - mg Sin18.7 = 0

F - μmg Cos18.7 - mg Sin18.7 = 0

F = μmg Cos18.7 + mg Sin18.7

F = (0.150 x 82.5 x 9.8) Cos18.7 + (82.5 x 9.8) Sin18.7

F = 374 N

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:

f = 632 Hz

Explanation:

As we know that for destructive interference the path difference from two loud speakers must be equal to the odd multiple of half of the wavelength

here we know that

[tex]\Delta x = (2n + 1)\frac{\lambda}{2}[/tex]

given that path difference from two loud speakers is given as

[tex]\Delta x = 5.80 m - 3.90 m[/tex]

[tex]\Delta x = 1.90 m[/tex]

now we know that it will have fourth lowest frequency at which destructive interference will occurs

so here we have

[tex]\Delta x = 1.90 = \frac{7\lambda}{2}[/tex]

[tex]\lambda = \frac{2 \times 1.90}{7}[/tex]

[tex]\lambda = 0.54 m[/tex]

now for frequency we know that

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

[tex]f = \frac{343}{0.54} = 632 Hz[/tex]

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:

The second wavelength is 482 nm.

Explanation:

Given that,

Wavelength = 630 nm

Distance from central maxim = 0.51 m

Distance from central maxim of another wavelength = 0.39 m

We need to calculate the second wavelength

Using formula of width of fringe

[tex]\beta=\dfrac{\lambda d}{D}[/tex]

Here, d and D will be same for both wavelengths

[tex]\lambda[/tex] = wavelength

[tex]\beta [/tex] = width of fringe

The width of fringe for first wavelength

[tex]\beta_{1}=\dfrac{\lambda_{1} d}{D}[/tex]....(I)

The width of fringe for second wavelength

[tex]\beta_{2}=\dfrac{\lambda_{2} d}{D}[/tex]....(II)

Divided equation (I) by equation (II)

[tex]\dfrac{\beta_{1}}{\beta_{2}}=\dfrac{\lambda_{1}}{\lambda_{2}}[/tex]

[tex]\lambda_{2}=\dfrac{630\times10^{-9}\times0.39}{0.51}[/tex]

[tex]\lambda_{2}=4.82\times10^{-7}[/tex]

[tex]\lambda=482\ nm[/tex]

Hence, The second wavelength is 482 nm.

To find the second wavelength, we can use the formula for the wavelength of light from a diffraction grating. In this case, we know the first wavelength is 630 nm, the first bright spot is separated from the central maximum by 0.51 m, and we need to find the second wavelength.

To find the second wavelength, we can use the formula for the wavelength of light from a diffraction grating: wavelength = (m * d * sin(theta)) / n. In this case, we know the first wavelength is 630 nm, the first bright spot is separated from the central maximum by 0.51 m, and we need to find the second wavelength. The m and n values are the same for both cases, so we can set up the equation:

630 nm = (m * 0.51 m * sin(theta)) / n
wavelength = (m * 0.51 m * sin(theta)) / n

Next, we can solve for the second wavelength by rearranging the equation:

wavelength = (630 nm * n) / (m * 0.51 m * sin(theta))

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Answer:

No we cannot perform any experiment that tells us weather we are moving or not at a constant speed.

Explanation:

Motion of any body is relative to other reference bodies. We can perceive motion only if our position with respect to a fixed object changes. This is the fundamental concept of reference frame in classical physics or Newtonian Physics  that motion is always from a reference.

We choose a fixed body as reference and ,measure the distance we cover from this fixed point and also our speed with respect to this fixed point. The choice of frame of reference is completely dependent on the observer.

Since in the given case a reference cannot be established outside the plane thus we cannot detect our motion.

Answer:

a= 2 m/s^2

Explanation:

take to the right as positive

let Ftot be the total forces acting on the box , m be the mass of the box and a be the acceleration of the box.

Ftot = 10 - 2 = 8 N

and,

Ftot = ma

   a = Ftot/m

      = 8/4

      = 2 m/s^2

therefore, the acceleration of the box is of magnitude of 2 m/s^2.

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.

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:

Length of pendulum, l = 1.74 meters

Explanation:

The time period of simple pendulum is, [tex]T=2\pi\sqrt{\dfrac{l}{g}}[/tex]

Where

l is the length of simple pendulum

The time period of  uniform thin rod is hung vertically from one end is, [tex]T=2\pi\sqrt{\dfrac{2l'}{3g}}[/tex]

l' is the length of uniform rod, l' = 2.6 m

It is given that the rod and pendulum have same time period. So,

[tex]2\pi\sqrt{\dfrac{l}{g}}=2\pi\sqrt{\dfrac{2l'}{3g}}[/tex]

After solving above expression, the value of length of the pendulum is, l = 1.74 meters. Hence, this is the required solution.

Answer:

Length of the chain, l = 0.99 m

Explanation:

Given that,

A playground tire swing has a period of 2.0 s on Earth i.e.

T = 2 s

We need to find the length of this chain. The relationship between the length and the time period is given by :

[tex]T=2\pi \sqrt{\dfrac{l}{g}}[/tex]

Where

l = length of chain

g = acceleration due to gravity

[tex]l=\dfrac{T^2g}{4\pi^2}[/tex]

[tex]l=\dfrac{(2)^2\times 9.8}{4\pi^2}[/tex]

l = 0.99 meters

So, the length of the chain is 0.99 meters. Hence, this is the required solution.

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