A meter stick is at rest on frictionless surface. A hockey puck is going towards the 30cm mark on the stick and is traveling perpendicular to the stick. After the collision the puck is deflected 30 degrees from original path and is traveling half its original speed.
~Mass of a meter stick = 0.05 kg~Mass of hockey puck = 0.17 kg~Initial speed of the hockey puck = 9 m/sA) Choosing an origin at the starting position of the meter stick's center of mass, what is the angular momentum of the hockey puck before the collision.B) What is the angular momentum of the hockey puck after the collision? (Use same origin.)C) What is the velocity (direction and speed) of the stick's center of mass after the collision?D) What is the angular velocity of the stick (assume it will rotate about its center of mass)?

Answer:

W=2.2 N

F=1.4 N

W'=0.8 N

Explanation:

Given that

Radius ,r = 0.0135 m

Density of the platinum ,ρ₁ = 2.14 x 10⁴ kg/m³

Density of the mercury ,ρ₂ = 1.36  x 10⁴ kg/m³

The weight of the sphere

W= m g

mass = m = volume x density

[tex]m=\dfrac{4}{3}\pi r^3\times \rho_1\ kg[/tex]

[tex]m=\dfrac{4}{3}\times \pi\times 0.0135^3\times 2.14\times 10^4\ kg[/tex]

m = 0.22 kg

W= 0.22 x 10 = 2.2 N   (↓)           ( take g =10 m/s²)

The buoyant force

[tex]F= \dfrac{4}{3}\pi r^3\times \rho_2\times g[/tex]

[tex]F=\dfrac{4}{3}\times \pi\times 0.0135^3\times 1.36\times 10^4\times 10[/tex]

F= 1.4 N  (↑)

The  apparent weight

W' = 2.2 - 1.4 N

W'= 0.8 N

Answer:

Answer:u=66.67 m/s

Explanation:

Given

mass of meteor m=2.5 gm\approx 2.5\times 10^{-3} kg

velocity of meteor v=40km/s \approx 40000 m/s

Kinetic Energy of Meteor

K.E.=\frac{mv^2}{2}

K.E.=\frac{2.5\times 10^{-3}\times (4000)^2}{2}

K.E.=2\times 10^6 J

Kinetic Energy of Car

=\frac{1}{2}\times Mu^2

=\frac{1}{2}\times 900\times u^2

\frac{1}{2}\times 900\times u^2=2\times 10^6  

900\times u^2=4\times 10^6

u^2=\frac{4}{9}\times 10^4

u=\frac{2}{3}\times 10^2

u=66.67 m/s

Answer:

C. The bullet will hit the monkey because both the monkey and the bullet are falling downward at the same rate due to gravity.

Explanation:

As we know that acceleration due to gravity is equal g m/s².The acceleration due to gravity is in downward direction always and the numerical value is equal to g =10 m/s².

Both the bullet and money moving downward with constant acceleration that is why bullet will hit the money.

Therefore the answer is C.

To solve this problem it is necessary to apply the Discharge of flow equations.

From the theory the flow rate is defined as

Q = AV

Where,

A =Area

V = Velocity

PART A) The question is telling us about the total fluid flow rate then

[tex]Q_T = Q_1+Q_2+Q_3[/tex]

[tex]Q_T = 27+19+12[/tex]

[tex]Q_T = 58L/min[/tex]

PART B) The radius would be given between another pipe with a flow rate of 27L / min.

For proportionality ratio we have to

[tex]\frac{Q_T}{Q'} = \frac{A_TV_T}{A'V'}[/tex]

[tex]\frac{V_T}{V'} = \frac{A_'Q_T}{A_TQ'}[/tex]

[tex]\frac{V_T}{V'} = \frac{(\pi r_T^2)Q_T}{(\pi r'^2)Q'}[/tex]

[tex]\frac{V_T}{V'} = \frac{1.2*^2 58}{1.8^2 27}[/tex]

[tex]\frac{V_T}{V'} = 0.9547[/tex]

Answer:

(a) 7.1 m /sec

(b) 339.9 N/m

(c) 19.91 cm

Explanation:

We have given mass m = 267 gram = 0.267 kg

Time period T = 0.176 sec

Total energy of the oscillating  system = 6.74 J

We know that energy is given by

(a) [tex]Ke=\frac{1}{2}mv_{max}^2[/tex]

[tex]6.74=\frac{1}{2}\times 0.267\times v_{max}^2[/tex]

[tex]v_{max}=7.1m/sec[/tex]

(b) Now [tex]\omega =\frac{2\pi }{T}=\frac{2\times 3.14}{0.176}=35.681rad/sec[/tex]

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

[tex]35.68=\sqrt{\frac{k}{0.267}}[/tex]

[tex]k=339.9N/m[/tex]

(c) We know that energy is given by

[tex]E=\frac{1}{2}KA^2[/tex]

[tex]6.74=\frac{1}{2}\times 339.9\times A^2[/tex]

[tex]A=19.91cm[/tex]

Answer:

(a) Kav Ne = Kav Kr = 7.29x10⁻²¹J

(b) v(rms) Ne= 659.6m/s and v(rms) Kr= 323.7m/s

Explanation:

(a) According to the kinetic theory of gases the average kinetic energy of the gases can be calculated by:

[tex] K_{av} = \frac{3}{2}kT [/tex] (1)        

where [tex] K_{av} [/tex]: is the kinetic energy, k: Boltzmann constant = 1.38x10⁻²³J/K, and T: is the temperature

From equation (1), we can calculate the average kinetic energies for the krypton and the neon:

[tex] K_{av} = \frac{3}{2} (1.38\cdot 10^{-23} \frac{J}{K})(352.2K) = 7.29\cdot 10^{-21}J [/tex]  

(b) The rms speeds of the gases can be calculated by:

[tex] K_{av} = \frac{1}{2}mv_{rms}^{2} \rightarrow v_{rms} = \sqrt \frac{2K_{av}}{m} [/tex]  

where m: is the mass of the gases and [tex]v_{rms}[/tex]: is the root mean square speed of the gases

For the neon:

[tex] v_{rms} = \sqrt \frac{2(7.29 \cdot 10^{-21}J)}{20.1797 \frac{g}{mol} \cdot \frac {1mol}{6.022\cdot 10^{23}molecules} \cdot \frac{1kg}{1000g}} = 659.6 \frac{m}{s} [/tex]          

For the krypton:

[tex] v_{rms} = \sqrt \frac{2(7.29 \cdot 10^{-21}J)}{83.798 \frac{g}{mol} \cdot \frac {1mol}{6.022\cdot 10^{23}molecules} \cdot \frac{1kg}{1000g}} = 323.7 \frac{m}{s} [/tex]  

Have a nice day!

The average kinetic energy and root mean square speed of both neon and krypton gases can be calculated using constants such as the Boltzmann constant, the molecular mass of the gases, and the temperature of the system in Kelvins.

The average kinetic energy (Kav) of a gas molecule, irrespective of the type of gas, is given by the equation Kav = 3/2 kT, where k is Boltzmann's constant and T is the absolute temperature. The root mean square speed (vrms) of the molecules in a gas is given by vrms = sqrt(3kT/m), where m is the molecular mass of the gas.

The temperature should be converted to Kelvins by adding 273.15 to the Celsius temperature, hence T = 79.2°C + 273.15 = 352.35K.

For neon, the molecular mass, m = 20.1797 u. Plugging in these values, we can calculate the average kinetic energy and rms speed for neon molecules in the vessel. Similar calculations are done for krypton with m = 83.798 u.

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Answer:

(a) Tangential velocity will be 38.648 m/sec

(b) Acceleration will be [tex]133.617m/sec^2[/tex]

Explanation:

We have given radius r = 11.2 m

Angular speed [tex]\omega =0.550rev/sec=0.550\times 2\pi =3.454rad/sec[/tex]

(a) We have to find the tangential velocity

We know that tangential velocity is given by  

[tex]v_t=\omega r=3.454\times 11.2=38.684m/sec[/tex]

(b) We know that acceleration is given by

[tex]a=\frac{v^2}{r}=\frac{38.684^2}{11.2}=133.617m/sec^2[/tex]

Answer:

31.66 m/s

Explanation:

mass of player, M = 75 kg

mass of ball, m = 0.45 kg

initial velocity of player, U = + 4 m/s

initial velocity of ball, u = - 24 m/s

Let the final speed of player is V and the ball is v.

use conservation of momentum

Momentum before collision = momentum after collision

75 x 4 - 0.45 x 24 = 75 x V + 0.45 x v

289.2 = 75 V + 0.45 v    .... (1)

As the collision is perfectly elastic, coefficient of restitution,e = 1

So, [tex]e=\frac{v_{1}-v_{2}}{u_{2}-u_{1}}[/tex]

V - v = u - U

V - v = -24 - 4 = - 28

V = v - 28, put this value in equation (1), we get

289.2 = 75 (v - 28) + 045 v

289.2 = 75 v - 2100 + 0.45 v

2389.2 = 75.45 v

v = 31.66 m/s

Thus, the velocity of ball after collision is 31.66 m/.

Answer:

a) 1.875 V b) 5.86 W c) 18.75 V 586 W d) 187.5 V 5.86 kW.

Explanation:

a) As the load is purely resistive, we can get the Irms, applying directly the definition of electrical power, as follows:

I = P / V = 250,000 W / 80,000 V = 3.125 A

Applying Ohm’s Law to the resistance of the conductors in the transmission line, we have:

∆V = I. R = 3.125 A . 0.6 Ω = 1.875 V

b) The rate at which energy is dissipated in the line as thermal energy, can be obtained applying Joule’s law, using the RMS value of the current that we have already got:

Pd = I2. R = (3.125)2 A . 0.6Ω = 5.86 W

c) If Vt were 8,000 V instead of 80,000 V, we would have a different value for I, as follows:

I= 250,000 W / 8,000 V = 31.25 A

So, we would have a different ∆V:

∆V = 31.25 A . 0.6 Ω = 18.75 V

As the RMS current is different, we will have a different value por the dissipated power in the line:

P = (31.25)2 . 0.6 Ω = 586 W

d) As above, we will have a new value for I, as follows:

I = 250,000 W / 800 V = 312. 5 A

The new voltage loss in the transmission line will be much larger:

∆V = 312.5 A . 0.6 Ω = 187.5 V

Consequently, we will have a higher dissipated power:

P = (312.5)2 . 0.6 Ω = 58.6 kW

To find the voltage drop across the transmission line and the power dissipated, we use Ohm's Law and the power equation. The voltage drop across the line is ∆V = I * R, and the power dissipated is Ploss = I² * R. Substituting the values, we can calculate the voltage drop and power dissipated for different values of Vt.

To find the voltage drop across the transmission line, we can use Ohm's Law. The total resistance of the transmission line is 0.3 Ω per cable, so the resistance for the entire line is 0.6 Ω. The voltage drop is given by: ∆V = I * R, where I is the current flowing through the transmission line. Since power P = V * I, we can rewrite the equation as: Vt * I = Pt, which gives us I = Pt / Vt. Substituting the values, we get I = 250,000 watts / 80,000 volts = 3.125 A.

(a) The voltage drop ∆V across the transmission line is therefore: ∆V = I * R = 3.125 A * 0.6 Ω = 1.875 V.

(b) The rate at which energy is dissipated in the line as thermal energy is equal to the power loss due to resistance, given by: Ploss = I² * R = (3.125 A)² * 0.6 Ω = 5.859375 W.

(c) If Vt is 8000 V (rms), the current flowing through the transmission line is: I = Pt / Vt = 250,000 watts / 8000 volts = 31.25 A. The voltage drop across the line is then ∆V = I * R = 31.25 A * 0.6 Ω = 18.75 V. The power dissipated in the line is: Ploss = I² * R = (31.25 A)² * 0.6 Ω = 585.9375 W.

(d) If Vt is 800 V (rms), the current flowing through the transmission line is: I = Pt / Vt = 250,000 watts / 800 volts = 312.5 A. The voltage drop across the line is ∆V = I * R = 312.5 A * 0.6 Ω = 187.5 V. The power dissipated in the line is: Ploss = I² * R = (312.5 A)² * 0.6 Ω = 58593.75 W.

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Answer:

Half as large.

Explanation:

Using Newton's law of universal gravitation, if the mass of the planet is M and of the Moons 1 and 2 is m, them the force exerted by the planet on them will be:

[tex]F_1=\frac{GMm}{r}[/tex]

[tex]F_2=\frac{GMm}{2r}[/tex]

Which clearly shows that the force that the planet exerts on the Moon 2 is half  the force it exerts on the Moon 1.

The gravitational force exerted by the planet on Moon 2, which orbits twice as far from the planet as Moon 1, is one-fourth as large as the force exerted on Moon 1 due to the inverse-square law of gravity.

This question is about the gravitational force that a planet exerts on its moons. Each moon has an identical mass but they differ in their distances from the planet - the radius of their orbits. One moon orbits at a radius of r, while the second moon orbits at a radius of 2r.

The gravitational force that a planet exerts on an object is inversely proportional to the square of the distance between the center of the planet and the object. So, if you double the distance, the gravitational force becomes one-fourth as large. Therefore, the magnitude of the gravitational force exerted by the planet on Moon 2, which is twice as far from the planet as Moon 1, is one-fourth as large as the gravitational force exerted by the planet on Moon 1.

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The number of electrons emitted from the metal per second increases if the intensity of the incident light is increased.

Answer: Option B

Explanation:

As a result of photoelectric effect, electrons are emitted by the light incident on a metal surface. The emitted electrons count and its kinetic energy can measure as the function of light intensity and frequency. Like physicists, at the 20th century beginning, it should be expected that the light wave's energy (its intensity) will be transformed into the kinetic energy of emitted electrons.

In addition, the electrons count emitting from metal must vary with light wave frequency. This frequency relationship was expected because the electric field oscillates due to the light wave and the metal electrons react to different frequencies. In other words, the number of electrons emitted was expected to be frequency dependent and their kinetic energy should be dependent on the intensity (constant wavelength) of light.

Thus, the maximum in kinetic energy of electrons emitted increases with increase in light's frequency and is experimentally independent of light intensity. So, the number of emitted electrons is proportionate to the intensity of the incident light.

Final answer:

If the intensity of the incident light on a metal surface in a photoelectric effect experiment is increased, while keeping frequency constant, the number of electrons emitted per second increases. This happens because a higher intensity means more photons are available to eject electrons. Neither the work function of the metal, the maximum speed of the emitted electrons, nor the stopping potential are affected by changes in light intensity.

Explanation:

The question relates to the effect of increasing the intensity of light in a photoelectric effect experiment while keeping the frequency of the incident light and the temperature of the metal constant. In this setup:

It is important to note that:

To solve this problem it is necessary to apply the concepts related to the drag force.

By definition we know that drag force can be expressed as

[tex]F_D = \frac{1}{2} \rho C_D AV^2[/tex]

Where,

\rho = Density

[tex]C_D =[/tex]Drag Coefficient

A = Area

V = Velocity

Our values are given as

[tex]A = 2.5ft^2[/tex]

[tex]V = 88ft/s[/tex]

[tex]C_D = 0.9[/tex]

[tex]\rho = 0.0765lb/ft^3 \rightarrow[/tex] Air at normal temperature

Replacing at the equation we have,

[tex]F_D = \frac{1}{2} \rho C_D AV^2[/tex]

[tex]F_D = \frac{1}{2} (0.0765lb/ft^3) (0.9) (2.5ft^2) (88ft/s)^2[/tex]

[tex]F_D = 666.468lbf[/tex]

The aerodynamic drag force is 666.468Lbf

This answer provides the equations and steps to calculate the buoyant force, the mass of the barge, the mass of the coal, and how far the barge would be submerged.

a) The equation for the buoyant force on the empty barge can be written as:

FB = ρgV

where FB is the buoyant force, ρ is the density of water, g is the acceleration due to gravity, and V is the volume of the barge submerged in water.

b) To determine the mass of the barge, we can use the equation:

FB = mg

where FB is the buoyant force, m is the mass of the barge, and g is the acceleration due to gravity.

c) The mass of the coal can be found by subtracting the mass of the empty barge from the mass of the fully loaded barge.

d) To find the mass of the coal in kilograms, you can use the equation:

mc = ρcVc

where mc is the mass of the coal, ρc is the density of the coal, and Vc is the volume of the coal.

e) To calculate how far the barge would be submerged with 250,000 kg of coal, you can use the equation:

FB = mg

and solve for H, where H is the height the barge is submerged.

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Answer:

Final velocity at the bottom of hill is 15.56 m/s.

Explanation:

The given problem can be divided into four parts:

1. Use conservation of momentum to determine the speed of the combined mass (Gayle and sled)

From the law of conservation of momentum (perfectly inelastic collision), the combined velocity is given as:  

[tex]p_i = p_f[/tex]  

[tex]m_1u_1 + m_2v_2 = (m_1 + m_2)v[/tex]

[tex]v = \frac{(m_1u_1 + m_2v_2)}{(m_1 + m_2)} [/tex]

[tex]v=\frac{[50.0\ kg)(3.85\ m/s) + 0]}{(50.0\ kg + 5.00\ kg)}= 3.5\ m/s[/tex]  

2. Use conservation of energy to determine the speed after traveling a vertical height of 5 m.

The velocity of Gayle and sled at the instant her brother jumps on is found from the law of conservation of energy:  

[tex]E(i) = E(f)[/tex]  

[tex]KE(i) + PE(i) = KE(f) + PE(f)[/tex]  

[tex]0.5mv^2(i) + mgh(i) = 0.5mv^2(f) + mgh(f)[/tex]  

[tex]v(f) = \sqrt{[v^2(i) + 2g(h(i) - h(f))]}[/tex]

Here, initial velocity is the final velocity from the first stage. Therefore:  

[tex]v(f) = \sqrt{[(3.5)^2+2(9.8)(5.00-0)]}= 10.5\ m/s[/tex]

3. Use conservation of momentum to find the combined speed of Gayle and her brother.  

Given:

Initial velocity of Gayle and sled is, [tex]u_1(i)=10.5[/tex] m/s

Initial velocity of her brother is, [tex]u_2(i)=0[/tex] m/s

Mass of Gayle and sled is, [tex]m_1=55.0[/tex] kg

Mass of her brother is, [tex]m_2=30.0[/tex] kg

Final combined velocity is given as:

[tex]v(f) = \frac{[m_1u_1(i) + m_2u_2(i)]}{(m_1 + m_2)}[/tex]  

[tex]v(f)=\frac{[(55.0)(10.5) + 0]}{(55.0+30.0)}= 6.79 [/tex] m/s  

4. Finally, use conservation of energy to determine the final speed at the bottom of the hill.

Using conservation of energy, the final velocity at the bottom of the hill is:  

[tex]E(i) = E(f)[/tex]  

[tex]KE(i) + PE(i) = KE(f) + PE(f)[/tex]  

[tex]0.5mv^2(i) + mgh(i) = 0.5mv^2(f) + mgh(f)[/tex]  

[tex]v(f) = \sqrt{[v^2(i) + 2g(h(i) - h(f))]} \\v(f)=\sqrt{[(6.79)^2 + 2(9.8)(15 - 5.00)]}\\v(f)= 15.56\ m/s[/tex]

Using principles of kinetic and potential energy, and assuming ideal conditions devoid of friction and air resistance, Gayle and her brother's final speed would be equal to the square root of twice the product of acceleration due to gravity and the total height of the hill.

This question relates to the principles of kinetic energy and potential energy, as well as conservation of energy. When Gayle initially sleds down the hill, she converts potential energy (mgh) into kinetic energy (1/2 mv^2). Taking 'g' as acceleration due to gravity (9.8 m/s^2) and 'h' as the vertical height she descends (5.0 m), the speed she attains at the time her brother joins (v') can be calculated using √(2gh).

The potential energy at that point is their combined mass times gravity and their height, which is transformed to kinetic energy at the bottom. If 'H' is the total height of the hill (15.0 m), their final combined speed is given by √(2gH). Ignoring resistive forces, they maintain this speed for the entire ride down the hill.

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To calculate the force required to hold a hollow ball stationary while it is completely submerged underwater, we can use Archimedes' principle. The buoyant force on the ball is equal to the weight of the water it displaces.

To calculate the force required to hold a hollow ball stationary while it is completely submerged underwater, we can use Archimedes' principle. The buoyant force on the ball is equal to the weight of the water it displaces. Since the ball is completely submerged, it displaces a volume of water equal to its own volume. The formula to calculate the force is:

Force = Density of water × Volume of ball × Acceleration due to gravity

Using the given diameter of the ball, we can calculate its volume by using the formula for the volume of a sphere: Volume = (4/3) × π × (radius)3

Once we have the volume, we can substitute it into the formula to find the force required.

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To develop this problem it is necessary to use the equations of description of the simple harmonic movement in which the acceleration and angular velocity are expressed as a function of the Amplitude.

Our values are given as

[tex]f = 4.11 *10^{12} Hz[/tex]

[tex]A = 1.23 * 10^{-11}m[/tex]

The angular velocity of a body can be described as a function of frequency as

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

[tex]\omega = 2\pi 4.11 *10^{12}[/tex]

[tex]\omega=2.582*10^{13} rad/s[/tex]

PART A) The expression for the maximum angular velocity is given by the amplitude so that

[tex]V = A\omega[/tex]

[tex]V =( 1.23 * 10^{-11})(2.582*10^{13})[/tex]

[tex]V =  = 317.586m/s[/tex]

PART B) The maximum acceleration on your part would be given by the expression

[tex]a = A \omega^2[/tex]

[tex]a =( 1.23 * 10^{-11})(2.582*10^{13})^2[/tex]

[tex]a= 8.2*10^{15}m/s^2[/tex]

To develop this problem it is necessary to apply the concepts related to the Aerodynamic Drag Force.

By definition the Drag Force is defined as

[tex]F_D = \frac{1}{2} \rho A C_D v^2[/tex]

Where,

A = Area

[tex]\rho[/tex]= Density

[tex]C_d[/tex] = Drag coefficient

v = Velocity

According to our values we have,

[tex]A = 1.8*1.8=3.24m^2[/tex]

[tex]C_D = 1.4[/tex]

[tex]V = 5m/s[/tex]

Replacing we have

[tex]F_D = \frac{1}{2} 1.23*3.24*1.4*5^2[/tex]

[tex]F_D = 69.74[/tex]

By definition we know that the thermal energy is given by the force applied in a given displacement then

[tex]W = F*d[/tex]

[tex]W = 69.74*200[/tex]

[tex]W = 13948J[/tex]

Therefore the thermal energy is added to the air by the drag force is 13.9kJ

The thermal energy added to the air by the drag force from a runner's parachute can be calculated by determining the work done against this force during a run. This involves first calculating the drag force with known parameters, then using the force to calculate the work done over the run's distance. This quantity represents the thermal energy imparted to the air.

The calculation for thermal energy added to the air by the drag force requires understanding the basic principles of work and energy. The work done by the drag force is given by the formula W = Fd, where W is work, F is force, and d is distance. In this case, the force is the drag force internalized by the parachute. The drag force is calculated using the equation F = 0.5 * p * v² * C * A, where:

By substituting these values into the formula, we can compute the drag force. Thereafter, we can use this force in our initial work equation to calculate the work done over a 200 m run. It's important to note that the work done against the drag force is transformed into heat (thermal energy). Thus, the work done equals the thermal energy added to the air by the drag force.

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The angular momentum of the particle with respect to the origin is 50 kgm²/s.

The angular momentum of an object is the product of moment of inertia and angular velocity.

L = mvr

where;

r = 5i + 5tj

v = dr/dt

v = 5 m/s

L = m(v x r)

v x r = 5j x (5i + 5(7)j)

v x r = 5j x (5i + 35j)

v x r = -25k

|v x r| = 25

L = m(v x r)

L = 2 x 25

L = 50 kgm²/s

Thus, the angular momentum of the particle with respect to the origin is 50 kgm²/s.

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Final answer:

The magnitude of the angular momentum of the particle with respect to the origin at time 7 s is 50 kg m²/s.

Explanation:

To find the magnitude of the angular momentum of a particle with a certain position vector as a function of time, we use the formula ℒ = r × p, where r is the position vector and p is the linear momentum vector (p = mv, with m as mass and v as velocity).

Given the position vector r(t) = (5 m)î + (5 m/s)tì for a mass of 2 kg, we first find the velocity vector by differentiating the position vector with respect to time, v(t) = dr/dt = 0î + (5 m/s)ì. The linear momentum at time t is p(t) = (2 kg)(v(t)) = (2 kg)(0î + (5 m/s)ì) = 0î + (10 kg m/s)ì.

To find the angular momentum at 7 s, we evaluate r at t = 7 s: r(7 s) = (5 m)î + (5 m/s)(7 s)ì = (5 m)î + (35 m)ì. Then, the angular momentum ℒ(7 s) = r(7 s) × p(7 s) = ((5 m)î + (35 m)ì) × (0î + (10 kg m/s)ì), which must be computed using the cross product. ℒ(7 s) ends up being ((5 m)(10 kg m/s))ê = (50 kg m²/s)ê. Therefore, the magnitude of the angular momentum is 50 kg m²/s.

To solve this problem it is necessary to apply the concepts related to the Gravitational Force, for this purpose it is understood that the gravitational force is described as

[tex]F_g = \frac{Gm_1m_2}{r^2}[/tex]

Where,

G = Gravitational Universal Force

[tex]m_i =[/tex] Mass of each object

To solve this problem it is necessary to divide the gravitational force (x, y) into the required components and then use the tangent to find the angle generated between both components.

Our values are given as,

[tex]m_1 =4.6 kg\\m_2 = 7.2 kg\\m_3 = 2.6 kg\\r_1 = 5.7 m\\r_2 = 4.2 m[/tex]

Applying the previous equation at X-Axis,

[tex]F_x = \frac{Gm_1m_3}{R_{1}^2}\\F_x = \frac{6.67*10^{-11}*4.6*2.6}{5.7^2}\\F_x = 2.46*10^{-11}N[/tex]

Applying the previous equation at Y-Axis,

[tex]F_y = \frac{Gm_2m_3}{R_2^2}\\F_y = \frac{6.67*10^{-11}*7.2*2.6}{4.2^2}\\F_y = 7.08*10^{-11} N[/tex]

Therefore the angle can be calculated as,

[tex]tan\theta = \frac{F_y}{F_x}\\\theta = tan^{-1} \frac{F_y}{F_x}\\\theta = tan^{-1} \frac{7.08*10^{-11}}{2.46*10^{-11}}\\\theta = 71\°[/tex]

Then in the measure contrary to the hands of the clock the Force in the particle 3 is in between the positive direction of the X and the negative direction of the Y at 71 ° from the positive x-axis.

Answer:

Explanation:

Let us first calculate for Virgo

[tex]A= d_{virgo} = 17;M_{pc}, V_{virgo} = 1200 km/s, [/tex]

Using Hubble's law

[tex]v = H_{0}D[/tex] For Virgo

[tex]V_{virgo} = H_{0}D_{virgo}[/tex]

[tex]O = D_{virgo} = \frac{v_{virgo}}{H_{0}} = \frac{1200 km/s}{70 km/s/Mpc}= 17.143 Mpc [/tex]

Percentage difference for the Virgo

[tex]\% = \frac{|A-O|}{A}\times 100 = \frac{|17 Mpc-17.143 Mpc|}{17 Mpc}\times 100 = 0.84 \% [/tex]

Now for calculate for Corona Borealis

[tex]A= d_{Corona Borealis}  = 310 Mpc, v_{Corona Borealis}  = 22000 km/s, [/tex]

Using Hubble's law

[tex]v = H_{0}D[/tex] For Corona Borealis

[tex]v_{Corona Borealis } = H_{0}D_{Corona Borealis } \\O = D_{Corona Borealis } = \frac{v_{Corona Borealis }}{H_{0}} = \frac{22000 km/s}{70 km/s/Mpc}= 314.286 Mpc [/tex]

Percentage difference for the Virgo

[tex]\% = \frac{|A-O|}{A}\times 100 = \frac{|310 Mpc-314.286 Mpc|}{310 Mpc}\times 100 = 1.3825 \% [/tex]

So clearly  Hubble's law is more accurate for the closer objects

Answer:

a. [tex]F=126N[/tex]

b. [tex]E_K=1323J[/tex]

Explanation:

Given:

[tex]m=54kg[/tex]

[tex]v=7 m/s[/tex]

[tex]t= 3s[/tex]

The runner force average to find given the equations

a.

[tex]F=m*a[/tex]

[tex]a=\frac{v}{t}[/tex]

[tex]F=m*\frac{v}{t}=54kg*\frac{7m/s}{3s}[/tex]

[tex]F=126N[/tex]

b.

Work done by the system by this force so

[tex]W=F*d[/tex]

[tex]W=E_K[/tex]

[tex]E_K=\frac{1}{2}*m*v^2[/tex]

[tex]E_K=\frac{1}{2}*54kg*(7m/s)^2[/tex]

[tex]E_K=1323J[/tex]

(a) The average horizontal component of the force that the ground exerts on the runner's shoes is 126 N.

(b) The work done on the point-particle system by this force is 1,323 J.

The net force on the runner is determined from Newton's second law of motion as shown below.

Ff = ma

The average horizontal component of the force that the ground exerts on the runner's shoes is calculated as follows;

Ff = m(v/t)

Ff = 54 x (7/3)

Ff = 126 N

The work done by frictional force is calculated as follows;

W = ¹/₂mv²

W = 0.5 x 54 x 7²

W = 1,323 J

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Answer:

  W = 661.6 N

Explanation:

The weight of a body is the force of attraction of the plant on the body, so we must use the law of gravitational attraction

       F = G m M / r²

Where G is the gravitational attraction constant that values ​​6.67 10-11 N m² / kg², M is the mass of the planet and r is the distance from the center of the planet.

Let's look for the mass of the planet, for this we write Newton's second law for the landing craft

     F = m a

Acceleration is centripetal a = v² / r

     G m M / r² = m (v² / r)

The ship rotates rapidly (constant velocity module), let's use uniform kinematic relationships

    v = d / t

The distance of a circle is

    d = 2π r

    v = 2π r / t

We replace

     G m M / r² = m (4π² r² / t² r)

    G M = 4 π² r³ / t²

    M = 4π² r³ / G t²

The measured distance r from the center of the plant is

     r = R orbit + R planet

     r = 5.90 10⁵ + ½ 9.80 10⁶

     r = 5.49 10⁶ m

    M = 4 π² (5.49 10⁶)³ / (6.67 10⁻¹¹ (5.900 10³)²)

    M = 6,532 10²¹ / 2,321 10⁺³

    M = 2.814 10²⁴ kg

With this data we calculate the astronaut's weight

     W = (G M / R²) m

     W = (6.67 10⁻¹¹ 2,816 10²⁴ /(4.90 10⁶)2)   84.6

     W = 7.82  84.6

     W = 661.57 N

To calculate the weight of the astronaut on the surface of the planet, we first need to find out the gravitational acceleration at the planet's surface using the parameters of the landing craft's orbit. After we have the value of gravitational acceleration, we multiply it by the mass of the astronaut.

To answer your question, we first need to calculate the gravitational acceleration at the surface of the planet, denoted by g. We can use the orbital parameters of the landing craft to find g using the following formula where G represents the universal gravitation constant, M is the mass of the planet, and r is the radius of the planet:

g = GM/r^2

After we solve this, we can then calculate the weight of the astronaut on the surface using W = mg, where m is the mass of the astronaut.

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Answer:

  α  = 17 rad / s² ,   t = 0.4299 s

Explanation:

Let's use Newton's second angular law or torque to find angular acceleration

     τ = I α

     W r = I α

The weight is applied in the middle of the pencil,

    sin 10 = r / (L/2)

    r = L/2  sin 10

The pencil can be approximated by a bar that rotates on one end, in this case its moment of inertia is

     I = 1/3 M L²

Let's calculate

    mg L / 2 sin 10 = (1/3 m L²) α  

    α f = 3/2 g / L sin 10

    α  = 3/2 9.8 / 0.150 sin 10

    α  = 17 rad / s²

If the pencil has a constant acceleration we can use the angular kinematics relationships, as part of the rest wo = 0

     θ = w₀ t + ½ α  t²

     t = RA (2θ / α )

The angle from the vertical to the ground is

    θ = π / 2

    t = √ (2 π / (2 α ))

    t = √ (π / α )

    t = √ (π / 17)

     t = 0.4299 s

The time for the pencil to hit the ground if it falls from standing perfectly vertical and maintains the angular acceleration is 0.4299s.

Newton's second angular law will be used to find the angular acceleration. This will be:

τ = I α

Wr = I α

Here, the weight is applied in the middle of the pencil, therefore,

sin 10 = r / (L/2)

Make r the subject of the formula

r = L/2 sin 10

Then, the moment of inertia will be:

I = 1/3 M L²

mg L / 2 sin 10 = (1/3 m L²) α  

αf = 3/2 g / L sin 10

α  = [3/2 × 9.8] / [0.150 sin10]

α  = 17 rad/s²

The angle from the vertical to the ground will be denoted as:

θ = π / 2

t = √ (2 π / (2 α ))

t = √(π / α )

t = √(π / 17)

t = √3.142 / 17

t = [tex]\sqrt{0.1848}[/tex]

t = 0.4299s

Therefore, the time for the pencil to hit the ground is 0.4299s.

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Answer:

0.15694 m

Explanation:

m = Mass of block = [tex]2.16\times 10^{-2}\ kg[/tex]

v = Velocity of block = 11.2 m/s

k = Spring constant = 110 N/m

Here the kinetic energy of the fall and spring are conserved

[tex]\frac{1}{2}mv^2=\frac{1}{2}kA^2\\\Rightarrow mv^2=kA^2\\\Rightarrow A=\sqrt{\frac{mv^2}{k}}\\\Rightarrow A=\sqrt{\frac{2.16\times 10^{-2}\times 11.2^2}{110}}\\\Rightarrow A=0.15694\ m[/tex]

The amplitude of the resulting simple harmonic motion is 0.15694 m

The force per unit length between the two wires is [tex]6.0\cdot 10^{-5} N/m[/tex]

Explanation:

The magnitude of the force per unit length exerted between two current-carrying wires is given by

[tex]\frac{F}{L}=\frac{\mu_0 I_1 I_2}{2\pi r}[/tex]

where

[tex]\mu_0 = 4\pi \cdot 10^{-7} Tm/A[/tex] is the vacuum permeability

[tex]I_1, I_2[/tex] are the currents in the two wires

r is the separation between the two wires

For the wires in this problem, we have

[tex]I_1 = 1.75 A[/tex]

[tex]I_2 = 3.45 A[/tex]

r = 2.00 cm = 0.02 m

Substituting into the equation, we find

[tex]\frac{F}{L}=\frac{(4\pi \cdot 10^{-7})(1.75)(3.45)}{2\pi (0.02)}=6.0\cdot 10^{-5} N/m[/tex]

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Rowan's feet appear closer than usual when they are under 0.60 m of water due to refraction, making the apparent depth approximately 0.45 m.

When Rowan is looking at her feet under 0.60 m of water, they appear closer to her than usual due to the phenomenon of refraction. Refraction occurs because light travels at different speeds in different mediums. When light passes from water to air, it speeds up and changes direction. This bending of light causes objects under water to look closer to the surface than they really are.

For apparent depth (how far away her feet appear), we can use Snell's law of refraction, which states that n1 × sin(θ_1) = n2 × sin(θ_2), where n1 and n2 are the refractive indices of water and air, respectively, and θ_1 and θ_2 are the angles of incidence and refraction.

Knowing that the refractive index of water is approximately 1.33 and that of air is 1 (approximating a normal incidence angle where sin(θ_1) ≈ sin(θ_2)), we get an apparent depth d’ = d/n, where d is the real depth (0.60 m). Thus, the apparent depth of Rowan's feet would be 0.60 m / 1.33 ≈ 0.45 m.

Answer:

Fundamental frequency= 174.5 hz

Explanation:

We know

fundamental frequency=[tex]\frac{velocity}{2 *length}[/tex]

velocity =[tex]\sqrt{\frac{tension}{mass per unit length} }[/tex]

mass per unit length=[tex]\frac{3.5}{1000*1.22}[/tex]=0.00427[tex]\frac{kg}{m}[/tex]

Now calculating velocity v=[tex]\sqrt{\frac{255}{0.00427} }[/tex]

                                           =244.3[tex]\frac{m}{sec}[/tex]

Distance between two nodes is 0.7 m.

Plugging these values into to calculate frequency

f = [tex]\frac{244.3}{2 *0.7}[/tex] =174.5 hz

To solve this problem it is necessary to apply the concepts related to the kinetic energy expressed in terms of simple harmonic movement, as well as the concepts related to angular velocity and acceleration and linear acceleration and velocity.

By definition we know that the angular velocity of a body can be described as a function of mass and spring constant as

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

Where,

k = Spring constant

m = mass

From the given values the angular velocity would be

[tex]\omega = \sqrt{\frac{277}{0.4}}[/tex]

[tex]\omega = 26.31rad/s[/tex]

The kinetic energy on its part is expressed as

[tex]E = \frac{1}{2} m\omega^2A^2[/tex]

Where,

A = Amplitude

[tex]\omega[/tex] = Angular Velocity

[tex]m = Mass[/tex]

PART A) Replacing previously given values the energy in the system would be

[tex]E = \frac{1}{2} m\omega^2A^2[/tex]

[tex]E = \frac{1}{2} (0.4)(26.31)^2(3*10^{-2})^2[/tex]

[tex]E= 0.1245J[/tex]

PART B) Through the amplitude and angular velocity it is possible to know the linear velocity by means of the relation

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

[tex]v = (3*10^{-2})(26.31)[/tex]

[tex]v = 0.7893m/s[/tex]

PART C) Finally, the relationship between linear acceleration and angular velocity is subject to

[tex]a = A\omega^2[/tex]

[tex]a = (3*10^{-2})(26.31)^2[/tex]

[tex]a = 20.76m/s^2[/tex]

(a) The mechanical energy of the system is 1.245 J (b) The maximum speed of the oscillating mass is 0.6225 m/s (c) The magnitude of the maximum acceleration of the oscillating mass is 20.775m/s².

(a) The mechanical energy of a mass†“spring system in simple harmonic motion (SHM) is given by the equation:

[tex]\[ E = \frac{1}{2}kA^2 \][/tex]

where [tex]$k$[/tex] is the spring constant and [tex]$A$[/tex]is the amplitude of oscillation. Plugging in the given values:

[tex]\[ E = \frac{1}{2} \times 277 \, \text{N/m} \times (0.03 \, \text{m})^2 \][/tex]

[tex]\[ E = \frac{1}{2} \times 277 \times 0.0009 \][/tex]

[tex]\[ E = 1.245 \, \text{J} \][/tex]

(b) The maximum speed of the mass occurs at the equilibrium position (when the displacement is zero) and is given by:

[tex]\[ v_{max} = A\sqrt{\frac{k}{m}} \][/tex]

where[tex]$m$[/tex] is the mass of the object. Using the given values:

[tex]\[ v_{max} = 0.03 \, \text{m} \times \sqrt{\frac{277 \, \text{N/m}}{0.40 \, \text{kg}}} \][/tex]

[tex]\[ v_{max} = 0.03 \times \sqrt{692.5} \][/tex]

[tex]\[ v_{max} = 0.03 \times 27.0676 \][/tex]

[tex]\[ v_{max} = 0.6225 \, \text{m/s} \][/tex]

(c) The maximum acceleration of the mass occurs at the maximum displacement (amplitude) and is given by:

[tex]\[ a_{max} = \frac{F_{max}}{m} \][/tex]

where [tex]$F_{max}$[/tex]is the maximum force exerted by the spring, which is [tex]$kA$[/tex]. Thus:

[tex]\[ a_{max} = \frac{kA}{m} \][/tex]

Using the given values:

[tex]\[ a_{max} = \frac{277 \, \text{N/m} \ times 0.03 \, \text{m}}{0.40 \, \text{kg}} \][/tex]

[tex]\[ a_{max} = \frac{8.31}{0.40} \][/tex]

[tex]\[ a_{max} = 20.775 \, \text{m/s}^2 \][/tex]

Answer:

τext =(M*R*g)*(m₂ - m₁) / (m₁ + m₂)

The wheel is rotating clockwise. Its vector direction is - k  which is perpendicular to the plane of the wheel.

Explanation:

Given info

m₁ = mass of block 1

m₂ = mass of block 2

m₁ < m₂

M = Mass of the wheel

R = Radius of the wheel

I = Moment of inertia of the wheel = M*R²

We assume that the wheel is rotating clockwise since m₁ < m₂

In order to get an expression for the net external torque acting on the system we can apply the following equation

τext = I*α

where α is the angular acceleration of the wheel which can be found as follows

at = R*α   ⇒    α = at / R

then we have to compute the acceleration of the system (at), using the Newton's 2nd Law

Block 1:

∑Fy = m₁*at  (↑+)  ⇒    T - m₁*g = m₁*at    (I)

Block 2:

∑Fy = m₂*at  (↓+)  ⇒    - T + m₂*g = m₂*at    (II)

If  we apply  

(I) + (II)  ⇒        at = g*(m₂ - m₁) / (m₁ + m₂)

Now, we get α:

α = at / R = (g*(m₂ - m₁) / (m₁ + m₂)) / R

⇒     α = (g / R)*(m₂ - m₁) / (m₁ + m₂)

Finally

τext = I*α = (M*R²)*((g / R)*(m₂ - m₁) / (m₁ + m₂))

⇒   τext =(M*R*g)*(m₂ - m₁) / (m₁ + m₂)

The wheel is rotating clockwise. Its vector direction is - k  which is perpendicular to the plane of the wheel.

Answer:

2993 °C

Explanation:

A reaction is spontaneous if the free Gibbs energy (ΔG) is negative. ΔG is related to the enthalpy of the reaction (ΔH) and the entropy of the reaction (ΔS) through the following expression.

ΔG = ΔH - T . ΔS

For 1 mol, if ΔG < 0, then

ΔH - T . ΔS < 0

ΔH < T . ΔS

-320.1 × 10³ J < T . (-98.00 J/K)

T < 3266 K

To convert Kelvin to Celsius we use the following expression.

K = °C + 273.15

°C = K - 273.15 = 3266 - 273.15 = 2993 °C