A 155 kg satellite is orbiting on a circular orbit 5995 km above the Earth's surface. Determine the speed of the satellite. (The mass of the Earth is 5.97×1024 kg, and the radius of the Earth is 6370 km.)

Answer:

563712.04903 Pa

Explanation:

m = Mass of material = 3.3 kg

r = Radius of sphere = 1.25 m

v = Volume of balloon = [tex]\frac{4}{3}\pi r^3[/tex]

M = Molar mass of helium = [tex]4.0026\times 10^{-3}\ kg/mol[/tex]

[tex]\rho[/tex] = Density of surrounding air = [tex]1.19\ kg/m^3[/tex]

R = Gas constant = 8.314 J/mol K

T = Temperature = 345 K

Weight of balloon + Weight of helium = Weight of air displaced

[tex]mg+m_{He}g=\rho vg\\\Rightarrow m_{He}=\rho vg-m\\\Rightarrow m_{He}=1.19\times \frac{4}{3}\pi 1.25^3-3.3\\\Rightarrow m_{He}=6.4356\ kg[/tex]

Mass of helium is 6.4356 kg

Moles of helium

[tex]n=\frac{m}{M}\\\Rightarrow n=\frac{6.4356}{4.0026\times 10^{-3}}\\\Rightarrow n=1607.85489[/tex]

Ideal gas law

[tex]P=\frac{nRT}{v}\\\Rightarrow P=\frac{1607.85489\times 8.314\times 345}{\frac{4}{3}\pi 1.25^3}\\\Rightarrow P=563712.04903\ Pa[/tex]

The absolute pressure of the Helium gas is 563712.04903 Pa

Final answer:

To find the absolute pressure of helium gas in the balloon, we can use the ideal gas law. Since the thickness of the material is negligible compared to the radius of the balloon, we can consider the balloon as a sphere. Once we have the number of moles, we can substitute the values into the ideal gas law and solve for P, the absolute pressure.

Explanation:

To find the absolute pressure of helium gas in the balloon, we can use the ideal gas law: PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature. In this case, we need to solve for P. Since the thickness of the material is negligible compared to the radius of the balloon, we can consider the balloon as a sphere and use the formula for the volume of a sphere: V = (4/3)πr^3, where r is the radius. Given that the radius is 1.25 m and the volume is known, we can calculate the number of moles of helium using the ideal gas law and the molar mass of helium.

Once we have the number of moles, we can substitute the values into the ideal gas law and solve for P, the absolute pressure. Remember to convert the temperature from Celsius to Kelvin by adding 273.15 to the given temperature.

Answer:0.548 gallon

Explanation:

Given

Average coefficient of volume expansion for carbon Tetrachloride [tex]\beta =5.81\times 10^{-4} /^{\circ}C[/tex]

Volume of steel container [tex]V=50 gallon[/tex]

Initial temperature [tex]T_i=10^{\circ}C[/tex]

Final temperature [tex]T_f=30^{\circ}C[/tex]

[tex]\Delta T=20^{\circ}C[/tex]

Coefficient of expansion for steel is [tex]\alpha =11\times 10^{-6}[/tex]

[tex]\beta =3\alpha =3\times 11\times 10^{-6}[/tex]

[tex]\beta =33\times 10^{-6}/^{\circ}C[/tex]

[tex]\Delta V_{spill}=\Delta V_{liquid}-\Delta V_{steel}[/tex]

[tex]\Delta V_{spill}=(\beta _{carbon}-\beta _{steel})V_0(\Delta T)[/tex]

[tex]\Delta V_{spill}=(5.81\times 10^{-4}-33\times 10^{-6})50\times 20[/tex]

[tex]\Delta V_{spill}=0.548\ gallon[/tex]

Approximately 0.04927 gal of carbon tetrachloride will spill over when the temperature rises to 30.0°C inside a 50.0-gal steel container.

To calculate how much carbon tetrachloride will spill over when the temperature rises, we need to find the change in volume for the steel container and the carbon tetrachloride. The change in volume can be calculated using the formula:

ΔV = V * β * ΔT

where ΔV is the change in volume, V is the initial volume, β is the coefficient of volume expansion, and ΔT is the change in temperature.

Using the given values, the change in volume for the steel container is 0.000605 gal and for the carbon tetrachloride is 0.04927 gal. Therefore, approximately 0.04927 gal of carbon tetrachloride will spill over when the temperature rises to 30.0°C.

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Final answer:

Decreasing the distance between successive crests of a wave results in a shorter wavelength and an increased frequency without affecting the speed of the wave. The correct answer to the student's question is E. Both the wavelength and the frequency.

Explanation:

If you decrease the distance between successive crests of a wave, this changes the wavelength and the frequency of the wave. Decreasing the distance between the crests means the wavelength is shorter, and since wavelength and frequency are inversely related, the frequency increases. However, this does not necessarily change the speed of the wave, which is determined by the medium through which the wave is traveling. Therefore, the correct answer is E. Both the wavelength and the frequency.

Wavelength is defined as the distance between any two consecutive identical points on the waveform, typically measured from crest to crest or trough to trough. When the period of a wave increases, its frequency decreases. This is because the frequency is the number of wave cycles that pass a point in one second, and if the period (the time for one cycle) increases, fewer cycles can pass in the same amount of time.

Final answer:

The specific gravity of a liquid can be calculated using the formula: Specific Gravity = mL / mA. Specific gravity is a dimensionless number used to compare the density of a substance with the density of water.

Explanation:

The specific gravity of a liquid can be calculated using the formula:

Specific Gravity = mL / mA

Here, mL is the apparent mass of the object in the unknown liquid, and mA is the apparent mass of the object in water. Specific gravity is a dimensionless number, so it doesn't have any units. It is used to compare the density of a substance with the density of water, which is 1.0 g/mL.

Answer:

0.84 cm

Explanation:

u = Object distance =  0.35 cm

v = Image distance = -0.6 cm (near point is considered as image distance and negative due to sign convention)

f = Focal length

From lens equation

[tex]\frac{1}{f_2}=\frac{1}{u}+\frac{1}{v}\\\Rightarrow \frac{1}{f_2}=\frac{1}{0.35}+\frac{1}{-0.6}\\\Rightarrow \frac{1}{f_2}=\frac{25}{21}\\\Rightarrow f_2=\frac{21}{25}=0.84\ cm[/tex]

Focal length of the lens is 0.84 cm

Answer:

Velocity of the bullet was 800 m/s.

Explanation:

It is given that,

Mass of the block, m₁ = 10 kg

Mass of the bullet, m₂ = 3 g

Height reached by the block, h = 3 mm

Let v is the velocity of the wooden block. Using the conservation of energy to find it as :

[tex]v=\sqrt{2gh}[/tex]

[tex]v=\sqrt{2\times 9.8\times 3\times 10^{-3}}[/tex]

v = 0.24 m/s

Let v' is the velocity of the bullet. The momentum of the system remains conserved. It can be calculated as :

[tex]m_1v=m_2v'[/tex]

[tex]v'=\dfrac{m_1v}{m_2}[/tex]

[tex]v'=\dfrac{10\times 0.24}{3\times 10^{-3}}[/tex]

v' = 800 m/s

So, the velocity of the bullet was 800 m/s. Hence, this is the required solution.

The velocity of the bullet which is fired horizontally into a 10-kg block of wood suspended by a rope from the ceiling is 800 m/s.

When the two objects collides, then the initial collision of the two body is equal to the final collision of two bodies by the law of conservation of momentum.

A 3-g bullet is fired horizontally into a 10-kg block of wood suspended by a rope from the ceiling.

The kinetic energy will be equal to the gravitation potential energy for this case. Thus,

[tex]\dfrac{1}{2}mv^2=mgh\\v^2=2gh[/tex]

As the maximum height is 3mm or 0.003 m. Thus, the velocity of the wooden block is,

[tex]v^2=2(9.8)(0.003)\\v=0.24\rm\; m/s[/tex]

Suppose the velocity of the bullet is v'. Thus by the law of conservation of momentum,

[tex]m_1v=m_2v'\\10(0.24)=0.003(v')\\v'=800\rm\; m/s[/tex]

Hence, the velocity of the bullet which is fired horizontally into a 10-kg block of wood suspended by a rope from the ceiling is 800 m/s.

Learn more about the conservation of momentum here;

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Answer:

The maximum power is 23.89 k watt.

Explanation:

Given that,

Diameter = 0.015 m

Long = 0.3 m

Initial temperature = 20°C

Final temperature = 150°C

Suppose the material is copper and here no maintain at time so we assuming heat supplied per unit time

We need to calculate the energy

Using formula of energy

[tex]\Delta Q=mc_{p}\Delta T[/tex]

[tex]\Delta Q=\rho\times V\times c_{p}\times\Delta T[/tex]

[tex]\Delta Q=8960\times\dfrac{\pi}{4}\times(0.015)^2\times0.3\times385\times(423-293)[/tex]

[tex]\Delta Q=23897.6\ J[/tex]

[tex]\Delta Q=23.89\ kJ[/tex]

We need to calculate the power

Using formula of power

[tex]P=\dfrac{\Delta Q}{dt}[/tex]

[tex]P=23.89\ k watt[/tex]

Hence, The maximum power is 23.89 k watt.

One of the scientific characteristics that can make the difference between the Earth and the moon is the so-called Rayleigh dispersion effect. This concept is identified as the dispersion of visible light or any other electromagnetic radiation by particles whose size is much smaller than the wavelength of the dispersed photons. Our atmosphere allows even our 'shadows' to be clear.

On the moon under the absence of the atmosphere or any other mechanism that allows absorbing or failing to re-irradiate sunlight towards the area in its shadows, which makes the shadows on the moon look darker.

1. Spring constant k is 233.17 N/m.

2. Oscillation frequency f is 0.924 Hz.

3. Speed of the block after 0.42 s is 2.62 m/s.

4. Magnitude of the maximum acceleration is 23.8 m/s².

5. Magnitude of the net force at 0.42 s is 105.02 N.

[tex]\\\( \omega_i \)[/tex]is the initial angular velocity (given as 90.0 rpm), Let's solve the problem step-by-step.

1. What is the spring constant of the spring?

The spring constant  k  can be determined using Hooke's Law  F = kx , where  F  is the force exerted by the spring and  x  is the displacement.

At equilibrium, the force exerted by the spring equals the weight of the block:

F = mg

Given:

m = 6.9 kg

g = 9.8 m/s²

x = 0.29 m

We calculate the force:

F = 6.9 kg * 9.8 m/s² = 67.62 N

Now, using  F = kx :

k = F/x = 67.62 N/0.29 m = 233.17 N/m

Spring constant  k  is 233.17 N/m

2. What is the oscillation frequency?

The frequency  [tex]\( f \)[/tex] can be found using the formula for the angular frequency [tex]\( \omega \):[/tex]

[tex]\[ \omega = \sqrt{\frac{k}{m}} \][/tex]

Given:

[tex]\[ k = 233.17 \text{ N/m} \]\[ m = 6.9 \text{ kg} \][/tex]

Calculate [tex]\( \omega \):[/tex]

[tex]\[ \omega = \sqrt{\frac{233.17 \text{ N/m}}{6.9 \text{ kg}}} = \sqrt{33.8 \text{ s⁻²}} = 5.81 \text{ rad/s} \][/tex]

The frequency[tex]\( f \)[/tex]is:

[tex]\[ f = \frac{\omega}{2\pi} = \frac{5.81 \text{ rad/s}}{2\pi} \approx 0.924 \text{ Hz} \][/tex]

Oscillation frequency f is 0.924 Hz

3. After t = 0.42 s, what is the speed of the block?

The speed of the block in simple harmonic motion can be described by:

[tex]\[ v(t) = \omega A \cos(\omega t + \phi) \][/tex]

The amplitude A and phase [tex]\( \phi \)[/tex]are determined from initial conditions.

The initial conditions:

- Initial position [tex]\( x(0) = 0 \)[/tex](at equilibrium)

- Initial velocity [tex]\( v(0) = -4.1 \text{ m/s} \)[/tex]

Since [tex]\( x(0) = A \cos(\phi) = 0 \), \(\phi = \pi/2\) (or 90°).[/tex]

Then, [tex]\( v(0) = -\omega A \sin(\phi) = -\omega A \):[/tex]

[tex]\[ -4.1 = -5.81 A \]\[ A = \frac{4.1}{5.81} \approx 0.705 \text{ m} \][/tex]

Now, calculate [tex]\( v(t) \) at \( t = 0.42 \text{ s} \):[/tex]

[tex]\[ v(0.42) = 5.81 \times 0.705 \times \cos(5.81 \times 0.42 + \frac{\pi}{2}) \]\[ v(0.42) = 5.81 \times 0.705 \times \cos(2.44 + 1.57) \]\[ v(0.42) = 4.096 \times \cos(4.01) \]\[ v(0.42) = 4.096 \times -0.64 = -2.62 \text{ m/s} \][/tex]

The speed of the block after 0.42 s is 2.62 m/s

4. What is the magnitude of the maximum acceleration of the block?

The maximum acceleration [tex]\( a_{max} \)[/tex] occurs at the maximum displacement [tex]\( A \):[/tex]

[tex]a_max} = \omega[/tex]² A

Given:

[tex]\[ \omega = 5.81 \text{ rad/s} \]\[ A = 0.705 \text{ m} \][/tex]

Calculate:

[tex]\[ a_{max} = (5.81)^2 \times 0.705 = 33.76 \times 0.705 = 23.8 \text{ m/s^2} \][/tex]

The magnitude of the maximum acceleration is 23.8 m/s²

5. At t = 0.42 s, what is the magnitude of the net force on the block?

The net force ( F(t) ) at any time ( t ) can be found using:

[tex]\[ F(t) = ma(t) \][/tex]

Where [tex]\( a(t) = -\omega^2 x(t) \). From \( x(t) = A \sin(\omega t + \phi) \):[/tex]

[tex]\[ x(0.42) = 0.705 \sin(5.81 \times 0.42 + \frac{\pi}{2}) \]\[ x(0.42) = 0.705 \sin(2.44 + 1.57) \]\[ x(0.42) = 0.705 \sin(4.01) \]\[ x(0.42) = 0.705 \times -0.64 = -0.451 \text{ m} \][/tex]

Now, [tex]\( a(0.42) = -\omega^2 x(0.42) \):[/tex]

[tex]\[ a(0.42) = - (5.81)^2 \times (-0.451) \]\[ a(0.42) = -33.76 \times -0.451 = 15.22 { m/s^2}[/tex]

Then, the net force:

[tex]\[ F(0.42) = 6.9 \text{ kg} \times 15.22 \text{ m/s^2} = 105.02 \text{ N} \][/tex]

The magnitude of the net force at 0.42 s is 105.02 N

Answer:

b. 25 N to the left.

Explanation:

Hi there!

The static friction force is calculated as follows:

Fr = N · μ

Where:

Fr = friction force.

N = normal force.

μ = coefficient of static friction.

If the object is not being accelerated in the vertical direction, the normal force is equal to the weight of crate (with opposite sign). Then:

Fr = 50 N · 0.50

Fr = 25 N

Since the 20-N force is applied to the right of the box, the friction force will be directed to the left because the friction force opposes the displacement. Then, the right answer is "b": 25 N to the left.

The resulting static friction force on the crate will be 20 N to the left, matching the applied force but acting in the opposite direction to prevent the crate from moving.

The question is related to static and kinetic friction in physics. A 50-N crate on a horizontal floor has a coefficient of static friction (μs) of 0.50. When a 20-N force is applied, the force of static friction (fs) that acts on the crate in response can be calculated. Since the applied force is less than the maximum possible static friction force (μs × normal force), the static friction will equal the applied force but act in the opposite direction, preventing the crate from moving. Here, the normal force (N) is equal to the weight of the crate because the floor is horizontal and there is no vertical acceleration.

The maximum static friction force would be fs(max) = μs × N = 0.50 × 50 N = 25 N. However, the applied force is only 20 N. Since this is less than the maximum, the actual static frictional force will match the applied force of 20 N, opposing the direction of the applied force, which is to the left. Thus, the correct answer is a. 20 N to the left.

Answer:

411087.52089 J

[tex]\frac{K_r}{K_b}=154.31213[/tex]

Explanation:

R = Radius of Earth = 6370000 m

h = Altitude of satellite = 810 km

r = R+h = 63700000+810000 m

m = Mass of bullet = 3.7 g

Velocity of bullet = 1200 m/s

The relative velocity between the pellets and satellite is 2v

Now, the square of velocity is proportional to the kinetic energy

[tex]K\propto v^2[/tex]

[tex]\\\Rightarrow 4K\propto (2v)^2\\\Rightarrow 4K\propto 4v^2[/tex]

Kinetic energy in terms of orbital mechanics is

[tex]K=\frac{GMm}{2r}[/tex]

In this case relative kinetic energy is

[tex]K_r=4\frac{GMm}{2r}\\\Rightarrow K_r=2\frac{6.67\times 10^{-11}\times 5.98\times 10^{24}\times 3.7\times 10^{-3}}{(6370+810)\times 10^3}\\\Rightarrow K_r=411087.52089\ J[/tex]

The relative kinetic energy is 411087.52089 J

The ratio of kinetic energies is given by

[tex]\frac{K_r}{K_b}=\frac{411087.52089}{\frac{1}{2}\times 3.7\times 10^{-3}\times 1200^2}\\\Rightarrow \frac{K_r}{K_b}=154.31213[/tex]

The ratio is [tex]\frac{K_r}{K_b}=154.31213[/tex]

Answer:

20J

Explanation:

Using conservation law of momentum;

since the bodies were at rest, their initial momentum is zero

0 =  M1Vx + M2Vy

- M1Vx = M2Vy where Vx is the final velocity of x after the spring has been release and Vy is final velocity of y and M1 and M2 are the masses of x and y

also M1 = 2/5 M2

substitute M1 into the the equation above

-2/5 M2Vx =  M2Vy

cancel M2 on both side

-2/5Vx =  Vy

comparing the kinetic energy of both x and y

for x K.E = 1/2 M1 Vx²

and y K.E = 1/2M2 Vy²

substitute for M1 = 2/5 M2

K.Ex = 1/2 × 2/5 M2 Vx²

divide  K.Ex / K.Ey = (1/2 × 2/5 M2 Vx²) / 1/2 M2 Vy²

cancel the common terms

K.Ex / K.Ey = (2/5 Vx²) / Vy²

substitute -2/5Vx for Vy

(2/5 Vx²) / ( -2/5 Vx)² = (2/5 Vx²) / ( 4/25 Vx²)

cancel Vx²

(2/5) / (4/25) = 2/5 ÷ 4/25 = 2/5 × 25/4 = 5/2

the ratio of x and y kinetic energy is 5:2

since the kinetic energy of x is 50

50 : 20 = 5 : 2 if 10 is used to divide both sides

the kinetic energy of y = 20 J

Final answer:

The kinetic energy of object Y is 312.5 J.

Explanation:

The kinetic energy of an object is given by the equation KE = 1/2 mv^2, where m is the mass of the object and v is its velocity. Since object X has a kinetic energy of 50 J, we can use this equation to find the velocity of X. Rearranging the equation, we have v = sqrt(2KE/m). Plugging in the values, we get v = sqrt(2*50 / (2/5)m) = sqrt(500/m).

Since object Y has a mass that is 2/5 times the mass of X, its mass is (2/5)m. Therefore, its velocity can be calculated as v = sqrt(500 / (2/5)m) = sqrt(1250/m).

To find the kinetic energy of Y, we use the formula KE = 1/2 mv^2. Plugging in the mass of Y and its velocity, we get KE = 1/2 ((2/5)m) (sqrt(1250/m))^2 = 1/2 (5/10) m (1250/m) = 1/2 * (5/10) * 1250 = 312.5 J.

Para resolver este problema es necesario aplicar los conceptos de Fuerza, dados en la segunda Ley de Newton y el concepto de Trabajo, como expresión de la fuerza necesaria para realizar una actividad en una distancia determinada.

El trabajo se define como

W = F*d

Where,

F = Force

d = Distance

At the same time we have that the Force by second's Newton law is equal to

F = mg

Where,

m = mass

g = Gravitational acceleration

PART A) Using our values and replacing we have that

[tex]W = F*d\\W = mg*d\\W=281.5*9.8(17.1*10^{-2}\\W = 471.738 J\approx 472J[/tex]

PART B) Using Newton's Second law we have that,

[tex]F = mg \\F= 281.5*9.8\\F= 2758.7 N \approx 2.76kN[/tex]

Answer:

The approximate value of Planck's constant is [tex]6.377\times10^{-34}\ J[/tex]

Explanation:

Given that,

Frequency [tex]f_{1}= 547.5\ THz[/tex]

Kinetic energy [tex]K.E=1.260\times10^{-19}\ J[/tex]

Frequency [tex]f_{2}=738.8\ THz[/tex]

Kinetic energy [tex]K.E=2.480\times10^{-19}\ J[/tex]

We need to calculate the approximate value of Planck's constant

Using formula of change in energy

[tex]E = hf[/tex]

[tex]K.E_{2}-K.E_{1}=h(f_{2}-f_{1})[/tex]

[tex]h=\dfrac{K.E_{2}-K.E_{1}}{(f_{2}-f_{1})}[/tex]

[tex]h=\dfrac{2.480\times10^{-19}-1.260\times10^{-19}}{738.8\times10^{12}-547.5\times10^{12}}[/tex]

[tex]h=6.377\times10^{-34}\ J[/tex]

Hence, The approximate value of Planck's constant is [tex]6.377\times10^{-34}\ J[/tex]

To solve this problem it is necessary to resort to the concepts expressed in the Buoyancy Force.

The buoyancy force is given by the equation

[tex]F = \rho Vg[/tex]

Where,

[tex]\rho =[/tex] Density

V =Volume

g = Gravitational Acceleration

PART A) From the given data we can find the volume, so

[tex]9800 = 1007*V*9.8[/tex]

[tex]V = 0.99m^3[/tex]

PART B) The mass can be expressed from the Newton equation in which

[tex]F = mg[/tex]

Where,

m = mass

g = Gravitational acceleration

Replacing with our values we have that

[tex]29000 = m*9.8[/tex]

[tex]m = 2959.18Kg[/tex]

Therefore the Density can be calculated with the ratio between the Volume and Mass

[tex]\rho = \frac{m}{V}[/tex]

[tex]\rho = \frac{2959.18Kg}{0.99m^3}[/tex]

[tex]\rho = 2989.074kg/m^3[/tex]

Therefore the Density of the log is [tex]2989.074kg/m^3[/tex]

The depth at which the amplitude of the incident wave is down to 1 μV/m in water can be calculated using the equations for wave propagation. At 1 GHz, the skin depth of water is approximately 5 mm. Using the exponential attenuation equation, we can find that the depth is approximately 35.3 mm.

To find the depth at which the amplitude of the incident wave is down to 1 μV/m, we can use the equations for wave propagation in a medium.

First, we need to calculate the skin depth (δ) of the water at 1 GHz using the equation:

δ = √(2/πfμσ)

Substituting the given values, we get:

δ = √(2/π * 1 * 10^9 * 4π * 10^-7 * 1) = 0.005 m = 5 mm

Next, we can use the equation for exponential attenuation of the wave:

A = A0 * e^(-x/δ)

Here, A0 is the initial amplitude, A is the amplitude at depth x, and δ is the skin depth. We can rearrange the equation to solve for x:

x = -δ * ln(A/A0)

Substituting the given values, we get:

x = -5 mm * ln(1 * 10^-6/20) ≈ 35.3 mm

Therefore, the depth at which the amplitude of the incident wave is down to 1 μV/m is approximately 35.3 mm.

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Answer:

ω = 26.35 rad/s

Explanation:

given,                            

speed of discus thrower = 25.3 m/s

whirling through  = 1.29 rev          

radius of circular arc = 0.96 m          

final angular speed of the discus = ?

using formula                      

v =  ω r                  

v  is the velocity of disk

ω is the angular speed of the discus

r is the radius of arc                      

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

[tex]\omega = \dfrac{25.3}{0.96}[/tex]                      

      ω = 26.35 rad/s

Final angular speed of the discus is equal to ω = 26.35 rad/s

The final angular speed is approximately 26.354 rad/s. To find the final angular speed, we use the relation [tex]\omega = v / r[/tex]. With v = 25.3 m/s and r = 0.96 m.

Calculate the final angular speed (ω) using the linear speed provided. Rotational motion relies on the relationship between linear speed (v) and angular speed (ω), which is expressed as [tex]v = r\cdot \omega[/tex], where r is the circular path radius. Rearranging this equation yields [tex]\omega = v / r[/tex], demonstrating the angular speed's direct and inverse relationship.

Here, v = 25.3 m/s (final linear speed) and r = 0.96 m (radius of the circle):

Conclusion:

Thus, the final angular speed of the discus is approximately 26.354 rad/s.

Answer:

option B

Explanation:

Current unit is ampere (A)

Ampere will be equivalent to = ?

we know,

Current can be define as the charge per unit time

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

unit of charge(Q) is coulomb which is equal to C.

unit of time(t) is equal to 's'.

now,

[tex]I = \dfrac{C}{s}[/tex]

unit of I = C/s

ampere(A) is equivalent to C/s

The correct answer is option B

Final answer:

The unit equivalent to the ampere (A) is C/s (Coulombs per second), making the correct option B) C/s. This illustrates the fundamental relationship between electrical charge, time, and current in physics.

Explanation:

The unit of electrical current is the ampere (A), which is defined as the amount of electrical charge in coulombs that passes through a conductor in one second. The correct unit combination that is equivalent to the ampere is C/s (Coulombs per second). Therefore, the correct option from the given combinations is B) C/s. This unit measures the flow of electrical charge and is a fundamental concept in understanding electric currents and how they are quantified in the International System of Units (SI)

Therefore, as per the above  explaination,  the correct answer is option C

To solve the problem it is necessary to apply the concepts related to torque, as well as the concepts where the Force is defined as a function of mass and acceleration, which in this case is gravity.

Considering the system in equilibrium, we perform sum of moments in the rear wheel (R2)

[tex]\sum M = 0[/tex]

[tex]F_g*1.68-R_1*d = 0[/tex]

[tex]mg*1.68-R_1*d = 0[/tex]

Another of the parameters given in the problem is that the front wheel supports 43% of the weight, that is

[tex]R1=0.43F_g[/tex]

[tex]R1=0.43mg[/tex]

Replacing, we have to

[tex]mg*1.68 -R_1*d = 0[/tex]

[tex]mg*1.68 -0.43mg*d = 0[/tex]

[tex]mg*1.68 =0.43mg*d[/tex]

[tex]1.68 = 0.43*d[/tex]

[tex]d =3.9m[/tex]

Therefore the wheelbase of the truck is 3.9m between the front and the rear.

With the use of physics principles, the truck's wheelbase, given that the rear wheels support 57.0% of the weight, and this weight acts at the center of gravity 1.68 meters in front of the rear wheels, is found to be 3.90 meters.

The question asks for the wheelbase of the truck, which can be solved using physics principles such as torque and equilibrium. The center of gravity is the point where all the weight can be considered to be concentrated for the purpose of calculations. Here, we know that the rear wheels support 57.0% of the weight of the truck, and this weight acts at the center of gravity, which is 1.68 meters in front of the rear wheels. Given that the truck is in equilibrium (i.e., not tipping over), the torques about any point caused by the weight must cancel out. Hence, the total distance from the rear wheels (where the 57% weight acts) to the front wheels (where the 43% weight acts) is (1.68 m * 57.0 / 43.0) m = 2.22 meters. Therefore, the wheelbase of the truck is 1.68 m + 2.22 m = 3.90 meters.

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Answer:

v= 0.0316 m/s

Explanation:

We need to use the Doppler Effect defined as the change in frequency of a wave in relation to an observer who is moving relative to the wave source.

Notation

Let v= magnitude of the heart wall speed

V= speed  of sound

fh= the frequency the heart receives (and  reflects)

fi= original frequency

ff= reflected frequency

fb= frequency for the beats

Apply the Doppler Effect formula

Since the heart is moving observer then the device is a stationary source, and we have this formula

fh = [(V+ v)/(v)] fi  (1)

We can consider the heart as moving source and the device as a stationary observer, and we have this formula

ff = [(V)/(V-v)] fh  (2)

The frequency for the beats would be the difference from the original and the reflected frequency

fb = ff -fi (3)

Replacing equations (1) and (2) into equation (3) we have:

[tex] f_b = \frac{V}{V-v} \frac{V+v}{V}f_i - f_i [/tex]

[tex]f_b = f_i ( \frac{V+v}{V-v} -1) [/tex]

fb = fi(V+v -V+v)/(V-v)

[tex] f_b = \frac{2v}{V-v}[/tex]

Solving for v we have:

[tex] v = V (\frac{f_b}{2f_o - f_b}) [/tex]

[tex] v = 1510 m/s (\frac{90 Hz}{2∗2150000Hz - 90 Hz})= 0.0316 m/s [/tex]

The speed of the fetal heart wall at the instant this measurement is made can be calculated using the Doppler shift in frequency. Given the transmitted sound frequency of 2.15 MHz and beat frequency detected as 90 Hz, the speed of the heart wall can be calculated using the provided speed of sound in body tissue. The calculated speed is approximately 0.063 m/s.

The question is based on the application of the principle of the Doppler effect in medical physics, specifically radiology. The Doppler effect involves a change in frequency and wavelength of a wave in relation to an observer who is moving relative to the wave source. In the case of ultrasound waves being used to monitor a fetus in the womb, the opening and closing of the heart valves reflect the waves back, and by calculating the shift in frequency (the Doppler shift), we can find out the speed at which the heart wall is moving.

In this particular scenario, two major factors are at play, namely the frequency of the transmitted sound (2.15 MHz), and the beat frequency detected (90 Hz). To calculate the speed of the fetal heart wall at the instant this measurement is made, we use the formula for Doppler shift in frequency: Δf/f=V/Vw, where Δf is the change in frequency (i.e., the beat frequency), V is the velocity of the moving observer (i.e., the fetal heart wall), and Vw is the speed of sound in the medium (body tissue in this case, which is given as 1510 m/s).

So our calculation will be as follows: V=(Δf/f) * Vw. Plugging in our values, V = (90 s-1 / 2.15 x 106 s-1) * 1510 m/s, which yields a value for V, the speed of the fetal heart wall, as approximately 0.063 m/s.

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Answer:

True A and B

Explanation:

Let's propose the solution of the exercise before seeing the affirmations.

We use the law of refraction

      n₁ sin θ₁ = n₂. Sin θ₂

Where n₁ and n₂ are the refractive indices of the two means, θ₁ and θ₂ are the angles of incidence and refraction, respectively

      sin θ₂ = (n1 / n2) sin θ₁

Let's apply this equation to the case presented. The index of refraction and airs is 1 (n1 = 1)

     Sin θ₂ = (1 / n2) sin θ₁

 the angle  θ₂ which is the  refracted angle is less than the incident angle

Let's analyze the statements time

A. False. We saw that it deviates

B. True Approaches normal (vertical axis)

C. False It deviates, but it is not parallel to normal

D. False It deviates, but approaching the normal not moving away

E. True. Because its refractive index is higher than air,

When a beam of light with an angle of incidence less than 90° enters a denser medium like a glass slab, it bends closer to the normal, bends away from the normal, and slows down, dependent on the refractive indices of the two media. Here options B, D, and E are correct.

When a beam of light travels from air into a denser medium, such as a glass slab, it undergoes refraction. Refraction is the bending of light as it passes from one medium to another with a different optical density.

The angle of incidence, the angle formed between the incident ray and the normal (a line perpendicular to the surface at the point of incidence), plays a crucial role in determining the behavior of the refracted light.

These statements are correct. The degree to which the light bends depends on the refractive indices of the two media. In this case, as light enters the glass slab, it slows down due to the higher refractive index of glass compared to air.

The bending of light towards the normal and slowing down are characteristic behaviors of light when it travels from a less dense to a denser medium. Here options B, D, and E are correct.

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Answer:

The thickness of the oil film is 198 nm.

Explanation:

Given that,

Refractive index of glass plate = 1.60

Refractive index of oil = 1.29

Wavelength = 511 nm

We need to calculate the thickness of the oil film

Using formula of path difference

[tex]2nt=k\lambda[/tex]

[tex]t=\dfrac{k\times\lambda}{2n}[/tex]

Where, n = refractive index

t = thickness

[tex]\lambda[/tex] = wavelength

Put the value into the formula

[tex]t=\dfrac{1\times511\times10^{-9}}{2\times1.29}[/tex]

[tex]t=198\times10^{-9}\ m[/tex]

[tex]t= 198\ nm[/tex]

Hence, The thickness of the oil film is 198 nm.

Answer:

(a) [tex]2.26\times 10^{13}\ N/C.s[/tex]

(b) 5 A

Solution:

As per the question:

Radius of the circular plate, R = 9 cm = 0.09 m

Distance, d = 2.0 mm = [tex]2.0\times 10^{- 3}\ m[/tex]

At [tex]t_{1}[/tex], current, I = 5 A

Now,

Area, A = [tex]\pi R^{2} = \pi 0.09^{2} = 0.025[/tex]

We know that the capacitance of the parallel plate capacitor, C = [tex]\frac{\epsilon_{o} A}{d}[/tex]

Also,

[tex]q = CV[/tex]

[tex]q = \frac{\epsilon A}{d}V[/tex]

Also,

[tex]V = \frac{E}{d}[/tex]

Now,

(a) The rate of change of electric field:

[tex]\frac{dE}{dt} = \frac{dq}{dt}(\frac{1}{A\epsilon_{o}})[/tex]

where

[tex]I = \frac{dq}{dt} = 5\ A[/tex]

[tex]\frac{dE}{dt} = 5\times (\frac{1}{0.025\times 8.85\times 10^{- 12}}) = 2.26\times 10^{13}\ N/C.s[/tex]

(b) To calculate the displacement current:

[tex]I_{D} = epsilon_{o}\times \frac{d\phi}{dt}[/tex]

where

[tex]\frac{d\phi}{dt}[/tex] = Rate of change of flux

[tex]I_{D} = Aepsilon_{o}\times \frac{dE}{dt}[/tex]

[tex]I_{D} = 0.025\times 8.85\times 10^{- 12}\times 2.26\times 10^{13} = 5\ A[/tex]

Answer:

The sound level will be 1.870 dB louder.

Explanation:

Given that,

Power = 130 W

Power = 200 W

We need to calculate the sound level

Using formula of sound level

[tex]I_{dB}=10\log(\dfrac{I}{I_{0}})[/tex]

For one amplifier,

[tex]I_{1}=10\log(\dfrac{130}{I_{0}})[/tex]...(I)

For other amplifier,

[tex]I_{2}=10\log(\dfrac{200}{I_{0}})[/tex]...(II)

For difference in dB levels

[tex]I_{2}-I_{1}=10\log(\dfrac{200}{I_{0}})-10\log(\dfrac{130}{I_{0}})[/tex]

[tex]I_{2}-I_{1}=10\log(\dfrac{200}{I_{0}}\times\dfrac{I_{0}}{130})[/tex]

[tex]I_{2}-I_{1}=10\log(\dfrac{200}{130})[/tex]

[tex]I_{2}-I_{1}=1.870\ dB[/tex]

Hence, The sound level will be 1.870 dB louder.

Answer:

a)[tex]v=\dfrac{2.KE}{qBR}[/tex]

b)[tex]m=\dfrac{(qBR)^2}{2.KE}[/tex]

Explanation:

Given that

Charge = q

Magnetic filed = B

Radius = R

We know that kinetic energy KE

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

m v² = 2 .KE            

The magnetic force F = q v B

Radial force

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

For uniform force these two forces should be equal

[tex]q v B=\dfrac{1}{r}mv^2[/tex]

q v B R =m v²

q v B R =  2 .KE

[tex]v=\dfrac{2.KE}{qBR}[/tex]

Now put the velocity v in the equation

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

[tex]m=\dfrac{2 .KE}{v^2}[/tex]

[tex]m=\dfrac{2.KE}{\left(\dfrac{2.KE}{qBR}\right)^2}[/tex]

[tex]m=\dfrac{(qBR)^2}{2.KE}[/tex]

Final answer:

To find the speed and mass of a charged particle moving in a magnetic field, we use its uniform circular motion to derive formulas for speed (v = qBR/m) and mass (m = q^2B^2R^2/(2KE)). These expressions help us understand the relationship between the particle's physical properties and its motion within the magnetic field.

Explanation:

To find the expressions for the speed v and mass m of a charged particle moving in a circular path of radius R inside a uniform magnetic field, we utilize the motion equations for a particle undergoing uniform circular motion due to a magnetic force. The magnetic force provides the centripetal force required for this motion and is given by qvB, where q is the charge, v is the velocity, and B is the magnetic field strength. This force is equal to the centripetal force needed for circular motion, which is given by mv2/R.

(a) Rearranging the formula qvB = mv2/R to solve for v, we get:
 v = qBR/m

(b) To find the mass m, we use the particle’s kinetic energy KE and the velocity from part (a). Since KE = 1/2 mv2 and the value of v is qBR/m, we can substitute it to get:
 m = q2B2R2/(2KE)

The units of these expressions can also be derived to ensure they're correct. Additionally, if we consider changes to the magnetic field's strength or the particle's charge, we can infer how these changes affect the radius R using the derived formulas.

a The kinetic energy of an object always has a positive value.

Explanation:

The kinetic energy of an object is the energy possessed by an object due to its motion, and it can be calculated as

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

where

m is the mass of the object

v is its speed

Let's now analyze each statement:

a The kinetic energy of an object always has a positive value. --> TRUE. In fact, the mass of the object is always positive, and the term [tex]v^2[/tex] is always positive as well, so the kinetic energy is always positive.

b The kinetic energy of an object is directly proportional to its speed. --> FALSE. Looking at the formula, we see that the kinetic energy is proportional to the square of the speed, [tex]K\propto v^2[/tex].

c The kinetic energy of an object is expressed in watts. --> FALSE. Watts is the units for measuring power, while the kinetic energy is measured in Joules, the units for the energy.

d The kinetic energy of an object is a quantitative measure of its inertia. --> FALSE. The inertia of an object depends only on its mass, not on its speed.

e The kinetic energy of an object is always equal to the object’s potential energy. --> FALSE. The potential energy depends on the altitude from the ground, not from the speed, so the two energies can be different.

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Kinetic energy, a concept in physics, refers to the energy that an object possesses due to its motion and is always positive. It is proportional to the square of speed, expressed in Joules, and distinct from inertia and potential energy.

In physics, kinetic energy pertains to the energy possessed by an object due to its motion. The kinetic energy (K) of a moving object can be calculated using the formula K = (1/2)mv², where 'm' stands for mass and 'v' for velocity.

Concerning the statements included in the question, option a) 'The kinetic energy of an object always has a positive value.' is correct. This is due to the fact that kinetic energy is product of the mass and the square of the speed, which are always positive or zero, thereby making kinetic energy always positive.

Regarding option b) 'The kinetic energy of an object is directly proportional to its speed', it's essential to note that kinetic energy is proportional not to the speed, but the square of the speed, hence this statement is partially correct.

The kinetic energy is not expressed in watts (option c). It's actually measured in Joules, which is equal to kg · m²/s² (kilogram meter squared per second squared).

Kinetic energy is not a measure of an object’s inertia (option d). It measures an object's energy due to motion, whereas inertia is an object’s resistance to change in motion.

Lastly, the kinetic energy of an object is not always equal to the object’s potential energy (option e). These are two different forms of energy that can convert into each other under certain circumstances, but they are not always equal.

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The inductance of the coil is 30 mH. The time constant of the circuit is 2.00 ms. It will take approximately -4.60 ms for the current to reach 1.00% of its original value.

To find the inductance of the coil, we can use the formula for the time constant in an RL circuit, T = L/R, where T is the time constant, L is the inductance, and R is the resistance. In this case, the resistance is given as 15.0 Ω, and the time it takes for the current to decay to 0.210 A is given as 2.00 ms. Plugging in these values, we can solve for L:

T = L/R

L = T × R

L = (2.00 ms) × (15.0 Ω)

L = 30 mH

Therefore, the inductance of the coil is 30 mH.

The time constant of the circuit is given by the formula T = L/R. Plugging in the values of L and R from part (a), we can solve for T:

T = L/R

T = (30 mH) / (15.0 Ω)

T = 2.00 ms

Therefore, the time constant of the circuit is 2.00 ms.

To find the time it takes for the current to reach 1.00% of its original value, we can use the formula for the time constant in an RL circuit, T = L/R. In this case, the resistance is given as 15.0 Ω, and we want to find the time when the current is 1.00% of its original value. Let's call this time T₁. Plugging in these values, we can solve for T₁:

T = L/R

T₁ = T × (ln(I1/I0))

T₁ = (2.00 ms) × (ln(0.01/1.00))

T₁ = -4.60 ms

Therefore, it will take approximately -4.60 ms for the current to reach 1.00% of its original value.

Answer:

[tex]5.28\times 10^{-5}\ /^{\circ}C[/tex]

Explanation:

[tex]L_0[/tex] = Original length of rod

[tex]\alpha[/tex] = Coefficient of linear expansion = [tex]1.62\times 10^{-5}\ /^{\circ}C[/tex]

Initial temperature = 16°C

Final temperature = 260°C

Change in length of a Steel is given by

[tex]\Delta L=\alpha L_0\Delta T\\\Rightarrow \Delta L=11\times 10^{-6}\times 23.83\times (260-16)\\\Rightarrow \Delta L=0.06395972\ cm[/tex]

Change in material rod length will be

[tex]23.83-23.59+0.0639572=0.3039572\ cm[/tex]

The coefficient of thermal expansion is given by

[tex]\alpha=\frac{\Delta L}{L_0\Delta T}\\\Rightarrow \alpha=\frac{0.3039572}{23.59\times (260-16)}\\\Rightarrow \alpha=5.28\times 10^{-5}\ /^{\circ}C[/tex]

The coefficient of thermal expansion for the material is [tex]5.28\times 10^{-5}\ /^{\circ}C[/tex]

Answer:

The mass of the central black hole is [tex]7.19x10^{37} Kg[/tex]

Explanation:

The Universal law of gravitation shows the interaction of gravity between two bodies:

[tex]F = G\frac{Mm}{r^{2}}[/tex]  (1)

Where G is the gravitational constant, M and m are the masses of the two objects and r is the distance between them.

For this particular case M is the mass of the central black hole and m is the mass of the molecular cloud. Since it is a circular motion the centripetal acceleration will be:

[tex]a = \frac{v^{2}}{r}[/tex]  (2)

Then Newton's second law ([tex]F = ma[/tex]) will be replaced in equation (1):

[tex]ma = G\frac{Mm}{r^{2}}[/tex]

By replacing (2) in equation (1) it is gotten:

[tex]m\frac{v^{2}}{r} = G\frac{Mm}{r^{2}}[/tex] (3)

Therefore, the mass of the central black hole can be determined if M is isolated from equation (3):

[tex]M = \frac{rv^{2}}{G}[/tex] (4)

Equation 4 it is known as the orbital velocity law.

Where M is the mass of the central black hole, r is the orbital radius, v is the orbital speed and G is the gravitational constant.

Before replacing the orbital speed in equation 4 it is necessary to make the conversion from kilometers to meters:

[tex]1000 \frac{km}{s} x \frac{1000 m}{1 km}[/tex] ⇒ [tex]1000000 m/s[/tex]

Then, equation 4 can be finally used:

[tex]M = \frac{(4.8x10^{15} m)(1000000 m/s)^{2}}{(6.67x10^{-11} N.m^{2}/Kg^{2})}[/tex]

[tex]M = 7.19x10^{37} Kg[/tex]

Hence, the mass of the central black hole is [tex]7.19x10^{37} Kg[/tex]

To calculate the mass of the central black hole, we can use the orbital velocity law, Mr = r×v^2/G. Given the orbital radius and velocity of the molecular clouds, we can plug these values into the equation to find the mass of the black hole.

To calculate the mass of the central black hole, we can use the orbital velocity law: Mr=r×v^2/G. In this equation, Mr represents the mass of the black hole, r is the orbital radius, v is the orbital velocity, and G is the gravitational constant. Given that the orbital radius is 0.49 light-years (4.8x10^15 meters) and the orbital velocity is 1000 km/s, we can plug these values into the equation to find the mass of the black hole.

First, let's convert the orbital velocity from km/s to m/s: 1000 km/s = 1000 x 1000 m/s = 1,000,000 m/s. Now, we plug in the values to the equation:

Mr = (4.8x10^15 meters) x (1,000,000 m/s)^2 / (6.67430 × 10^-11 m^3⋅kg^−1⋅s^−2)

Simplifying the expression, we get:

Mr = (4.8x10^15 meters) x (1,000,000 m/s)^2 / (6.67430 × 10^-11) kg

After calculating, the mass of the central black hole is approximately 3.23x10^37 kg.

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Answer:

0.958 m

Explanation:

So the total mass of the system is

M = 1.93 + 2.95 + 2.41 + 3.99 = 11.28  kg

let y be the distance from the center of mass to the origin. With the reference to the origin then we have the following equation

[tex]My = m_1y_1 + m_2y_2 +m_3y_3 + m_4y_4[/tex]

[tex]11.28y = 1.93*2.91 + 2.95*2.43 + 2.41*0 + 3.99*(-0.496) = 10.806[/tex]

[tex]y = \frac{10.806}{11.28} = 0.958 m[/tex]

So the center of mass is 0.958 m from the origin