Linear Thermal Expansion (in one dimension)
1) The change in length ΔL is proportional to the original length L, and the change in temperature ΔT : ΔL = αLΔT, where ΔL is the change in length , and α is the coefficient of linear expansion.
a) The main span of San Francisco’s Golden Gate Bridge is 1275 m long at its coldest (–15ºC). The coefficient of linear expansion, α , for steel is 12×10−6 /ºC. When the temperatures rises to 25 °C, what is its change in length in meters?
2) The change in volume ΔV is very nearly ΔV ≈ 3αVΔT . This equation is usually written as ΔV = βVΔT, where β is the coefficient of volume expansion and β ≈ 3α . V is the original volume. ΔT is the change in temperature. Suppose your 60.0-L (15.9-gal) steel gasoline tank is full of gasoline, and both the tank and the gasoline have a temperature of 15.0ºC . The coefficients of volume expansion, for gasoline is βgas = 950×10−6 /ºC , for the steel tank is βsteel = 35×10−6 /ºC .
a) What is the change in volume (in liters) of the gasoline when the temperature rises to 25 °C in L?
b) What is the change in volume (in liters) of the tank when the temperature rises to 25 °C in L?
c) How much gasoline would be spilled in L?

The gauge pressure at the water/mercury interface is simply the "head" due to the water:

p = ρgh = 1000kg/m³ * 9.8m/s² * 0.19m = 1862 Pa

If we take the specific gravity of mercury to be 13.6, then the difference in height between the water and mercury columns is

h' = h(1 - 1/s.g.) = 19cm * (1 - 1/13.6) = 17.6 cm

Answer:

353 s

Explanation:

Energy               = energy needed to raise the temperature of pot to 100°C +

required              energy needed to raise the temperature of water to 100°C

                    = MCΔT + mcΔT                      

                     = 0.65 kg * 4186 J/kg-°C * 87 °C + 0.36 kg * 900 J/kg-°C * 87 °C

                     = 264 906.3 J

Energy required = energy delivered * time

264 906.3 J = 750 J/s * time

time = 353 s

The study of chemicals and bonds is called chemistry. There are two types of elements these rare metals and nonmetals.

The correct answer is [tex]5787*10^{-6}[/tex]

The balanced reaction is:-

[tex]2H_2O_2(aq) ----> 2H_2O(l) + O_2(g).[/tex]

The rate law of decomposition:-

[tex]rate =-\frac{1}{2}\frac{d[H_2O_2]}{dt}[/tex]

Where d[H₂O₂] is the change in concentration of H₂O₂ (between 0s and 2,16x10⁴s) is (1,000M-0,500M) and dt is (0s-2,16x10^4s). Replacing:

[tex]rate= -\frac{1}{2} \frac{0.5}{2.16*10^4}[/tex]

rate =[tex]1157*10^{-5}[/tex]

As this rate is = d[O₂]/dt(Rate of production of O₂), the rate of production of O₂(g) is 1,157x10⁻⁵M/s

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Answer:

A

Explanation:

v = change of X / change of T

v = 200/19.3

Answer:

A. 10.36 m/s

Explanation:

The Olympic record for running the 200 m dash is 19.3 seconds. What is the average speed for this record?

A. 10.36 m/s

B. 3960 m/s

C. 219.3 m

D. 0.0965 m/s

speed is the change in distance per time

speed is scalar quantity and hence as no direction but only magnitude. time  and distance are also  scalar quantity

200/19.3

speed=10.36m/s

Answer:

A) Bohr’s work with atomic spectra led him to say that the electrons were limited to existing in certain energy levels, like standing on the rungs of a ladder.

Explanation:

The change to the atomic model that helped solve the problem seen in Rutherford's model was the discovery of the strong nuclear force.

Rutherford's model required that the electrons be in motion. Positive and negative charges attract each other, so stationary electrons would fall into the positive nucleus. However, Rutherford's model failed to explain why electrons were not pulled into the atomic nucleus by this attraction. The change to the atomic model that helped solve this problem was the discovery of the strong nuclear force, which is much stronger than electrostatic interactions and holds the protons and neutrons together in the nucleus.

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The angular velocity (ω) of a wheel can be expressed in terms of the force applied (F), wheel's radius (r), and its moment of inertia (I), all of which factor into the wheel's torque and resulting angular acceleration. Through the rotational analog of Newton's second law and kinematic equations, one can derive a formula for ω that connects these physical quantities

To express the angular velocity (ω) of the wheel in terms of displacement (d), magnitude of the force (F), and the moment of inertia of the wheel (Iw), we use physical principles of rotational motion. We can derive this using the definition of torque and the rotational version of Newton's second law, which states that torque (τ) is equal to the moment of inertia (I) times the angular acceleration (α): τ = Iα. For a wheel, torque is also equal to the force applied (F) times the radius (r): τ = Fr.

By equating the two expressions for torque, we get Fr = Iα, and therefore α = F × r/I. With the angular acceleration and knowing the time the force is applied, we can find the final angular velocity using the kinematic equation for rotational motion: ωfinal = ωinitial + α × t. Assuming that initial angular velocity (ωinitial) is zero and time (t) can be deduced from the linear displacement (d) and the linear velocity (v), given that v = ω × r, we can express the final angular velocity in terms of displacement (d), wheel's radius (r), and angular acceleration (α).

To further involve the moment of inertia in this expression, we would substitute the earlier derived expression for α into our final angular velocity equation: ω = (F × r/I) × t. Now, if we relate time (t) to the linear displacement (d) such that the linear velocity (v) = d/t, and since v = ω×r, we can isolate ω to derive a relationship that includes the moment of inertia.

Answer:

(a). The value of d is 0.056 cm and 1.496 cm.

(b). The time period is 1.35 sec.

Explanation:

Given that,

Length = 50.00 cm

Time period = 2.50 s

Time period of pendulum is defined as the time for one complete cycle.

The period depends on the length of the pendulum.

Using formula of time period

[tex]T=2\pi\sqrt{\dfrac{I}{mgh}}[/tex]

Where, I = moment of inertia

We need to calculate the value of d

Using parallel theorem of moment of inertia

[tex]I=I_{cm}+md^2[/tex]

For a meter stick mass m , the rotational inertia about it's center of mass

[tex]I_{cm}-\dfrac{mL^2}{12}[/tex]

Where, L = 1 m

Put the value into the formula of time period

[tex]T=2\pi\sqrt{\dfrac{\dfrac{mL^2}{12}+md^2}{mgd}}[/tex]

[tex]T=2\pi\sqrt{\dfrac{L^2}{12gd}+\dfrac{d}{g}}[/tex]

[tex]T^2=4\pi^2(\dfrac{L^2}{12gd}+\dfrac{d}{g})[/tex]

Multiplying both sides by d

tex]T^2d=4\pi^2(\dfrac{L^2}{12g}+\dfrac{d^2}{g})[/tex]

[tex](\dfrac{4\pi^2}{g})d^2-T^2d+\dfrac{\pi^2L^2}{3g}=0[/tex]

Put the value of T, L and g into the formula

[tex]4.028d^2-6.25d+0.336=0[/tex]

[tex]d = 0.056\ m, 1.496\ m[/tex]

The value of d is 0.056 cm and 1.496 cm.

(b). Given that,

L = 50-5 = 45 cm

We need to calculate the time period

Using formula of period

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

Put the value into the formula

[tex]T=2\pi\sqrt{\dfrac{45\times10^{-2}}{9.8}}[/tex]

[tex]T=1.35\ sec[/tex]

Hence, (a). The value of d is 0.056 cm and 1.496 cm.

(b). The time period is 1.35 sec.

Answer:D

Explanation:

Given

mass of A is twice the mass of B half the velocity of B

Suppose [tex]F_a[/tex] and [tex]F_b[/tex] be the average force exerted on A and B respectively

and According to Newton third law of motion Force on the body A is equal to Force on body B but opposite in direction as they are action and reaction force.

Thus [tex]F_a=-F_b[/tex]  and option d is correct

According to Newton's third law of motion, the forces exerted by each object on the other are the same.

The correct answer is (C) The forces exerted by each object on the other are the same because inter objects cannot exert forces of different magnitudes on each other.

According to Newton's third law of motion, whenever one body exerts a force on a second body, the first body experiences a force that is equal in magnitude and opposite in direction to the force it exerts. This means that the forces exerted by object A on object B and by object B on object A are equal in magnitude.

Here, object A has twice the mass of object B, but because both objects have the same speed before the collision, the forces exerted on each other will be the same.

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Answer:

a) r eq = -a/(2b)

b) k = a/r eq = -2b

Explanation:

since

U(r) = ar + br²

a) the equilibrium position dU/dr = 0

U(r) = a + 2br = 0 →  r eq= -a/2b

b) the Taylor expansion around the equilibrium position is

U(r) = U(r eq) + ∑ Un(r eq) (r- r eq)^n / n!  

,where Un(a) is the nth derivative of U respect with r , evaluated in a

Since the 3rd and higher order derivatives are =0 , we can expand until the second derivative

U(r) = U(r eq) + dU/dr(r eq) (r- r eq) +  d²U/dr²(r eq) (r- r eq)² /2

since dU/dr(r eq)=0

U(r) = U(r eq) + d²U/dr²(r eq) (r- r eq)² /2

comparing with an energy balance of a spring around its equilibrium position

U(r) - U(r eq)  = 1/2 k (r-r eq)² → U(r) = U(r eq)  + 1/2 k (r-r eq)²

therefore we can conclude

k = d²U/dr²(r eq) = -2b , and since r eq = -a/2b → -2b=a/r eq

thus

k= a/r eq

Answer:

N = 5 harmonics

Explanation:

As we know that frequency of the sound is given as

[tex]f = \frac{N}{2L}\sqrt{\frac{T}{\mu}}[/tex]

now we have

[tex]T = 350 N[/tex]

[tex]\mu = 0.0180 kg/m[/tex]

L = 17.43 m

now we have

[tex]f = \frac{N}{2(17.43)}\sqrt{\frac{350}{0.0180}}[/tex]

[tex]f = 4 N[/tex]

if the lowest audible frequency is f = 20 Hz

so number of harmonics is given as

[tex]20 = 4 N[/tex]

N = 5 harmonics

Answer:

Height will be 5.127 m

Explanation:

We have given mass m = 49 kg

Speed over the bar v = 10 m /sec

Acceleration due to gravity [tex]g=9.8m/sec^2[/tex]

Kinetic energy on the ground [tex]=\frac{1}{2}mv^2=\frac{1}{2}\times 49\times 10^2=2450J[/tex]

Potential energy on the ground [tex]=mgh=0[/tex] (as height will be zero )

Speed above the bar = 0.7 m /sec

So kinetic energy above the bar [tex]=\frac{1}{2}mv^2=\frac{1}{2}\times 49\times 0.7^2=12J[/tex]

Potential energy above the bar = mgh

From energy conservation

Total kinetic energy = total potential energy

So [tex]2450+12=0+49\times 9.8\times h[/tex]

[tex]h=5.127m[/tex]

Using the principle of conservation of energy, we can calculate the pole vaulter's height above the bar as being approximately 5.02 meters.

In this question, we're dealing with conservation of energy. The energy of the pole vaulter is kinetic energy when she is running (1/2*mass*speed^2), and as she goes over the bar, her energy becomes potential energy (mass*gravity*height) and a little kinetic energy (1/2*mass*speed^2).

Let's use the following formula: Kinetic Energy initial + Potential Energy initial = Kinetic Energy final + Potential Energy final.

Initially, her kinetic energy is 1/2 * 49 kg * (10 m/s)^2 = 2450 J. She has no potential energy, because she's on the ground (height=0). When she's above the bar, her kinetic energy is 1/2 * 49 kg * (0.7 m/s)^2 = 12.075 J. We don't yet know her final potential energy, because we're trying to find her height. So, we'll call her final potential energy m*g*h, or 49 kg * 9.8 m/s^2 * h.

We then plug these values into our energy equation: 2450 J + 0 J = 12.075 J + 49 kg * 9.8 m/s^2 * h. Solving for h gives us h = (2450 J - 12.075 J) / (49 kg * 9.8 m/s^2) = 5.02 meters, so this would be her altitude as she crosses the bar.

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Answer:

v = 37.4 m/s , h = 71.39m

Explanation:

To find the velocity given:

m = 0.4 kg

F =190.4 N

d = 1.47 m

g = 9.8 m/s^2

So use the equation of work to solve the kinetic energy

W = F *d = 190.4 N * 1.47m

W = 279.88 J

Ke = 1 / 2 * m* v^2

v = √2*Ke / m =√ 2 *279.88 / 0.4 kg

v = 37.4 m/s

Now to find the high to rise can use the conserved law so:

Ke = Pe

279.88 = m*g*h

Solve to h'

h = 279.88 / 0.4kg * 9.8m/s^2

h =71.39 m

Final answer:

The arrow leaves the bow with a speed of approximately 37.49 m/s and, when shot straight up, rises to a maximum height of about 71.4 m.

Explanation:

To determine the speed with which the arrow leaves the bow, we apply the work-energy principle, which states that work done on the arrow is converted into its kinetic energy. The work done W by the bow can be calculated by multiplying the force F exerted by the distance d over which the force is applied: W = F × d. It is given that F is 190.4 N and d is 1.47 m; thus, W = 190.4 N × 1.47 m = 279.888 J.

The kinetic energy KE of the arrow can be given by KE = ½ mv², where m is the mass of the arrow and v is its velocity. Since work done equals the kinetic energy, we get 279.888 J = ½ × 0.4 kg × v². Solving for v gives us a velocity of approximately 37.49 m/s.

To find how high the arrow goes if it's shot straight up, we use the conservation of energy, where the initial kinetic energy is converted to gravitational potential energy at the highest point. The potential energy PE at maximum height can be given by PE = mgh, where g is the acceleration due to gravity (9.8 m/s²) and h is the height. Setting KE equal to PE, we have 279.888 J = 0.4 kg × 9.8 m/s² × h. Solving for h gives us a maximum height of approximately 71.4 m.

Answer:

The answer is C.

Explanation:

For acceleration to be achieved so to speak there must be a force acting on it. The only force on the ball before the air drag increases is gravity. As the air resistance increases the force resisting gravity increases. This means that the forces start to cancel out. Therefore the acceleration must get smaller.

These forces will continue to cancel until it reaches terminal velocity.

What happens to the motion of the ball is, the acceleration of the ball decreases and will become zero when drag force on the ball equals the force of gravity.

The downward motion of the ball is reduced by frictional force opposing the motion. The frictional force opposing the motion is the drag force of the air or air resistance.

The net downward force on the ball is given as;

[tex]W -F_D = ma\\\\[/tex]

when the drag force on the ball equals force of gravity, the acceleration of the ball will be zero.

[tex]W - W = ma\\\\(W = F_D)\\\\0 = ma\\\\a = 0[/tex]

Thus, we can conclude that what happens to the motion of the ball is, the acceleration of the ball decreases and will become zero when drag force on the ball equals the force of gravity.

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Final Answer:

The mass of the water is approximately 48.4 grams.

Explanation:

1. To solve this problem, we can use the principle of conservation of energy, specifically the principle of heat exchange.

2. We can calculate the heat gained by the iron rod and the heat lost by the water, and equate them since there is no heat exchange with the surroundings.

3. The formula for heat exchange is given by: [tex]\[ Q = mc\Delta T \][/tex]

  where Q is the heat exchanged, m is the mass, c is the specific heat capacity, and [tex]\( \Delta T \)[/tex] is the change in temperature.

4. We can set up the equation for heat exchange for the iron rod and the water:

  [tex]\[ mc_{\text{iron}}\Delta T_{\text{iron}} = mc_{\text{water}}\Delta T_{\text{water}} \][/tex]

5. Rearranging the equation and solving for the mass of water:

  [tex]\[ m_{\text{water}} = \frac{m_{\text{iron}}c_{\text{iron}}\Delta T_{\text{iron}}}{c_{\text{water}}\Delta T_{\text{water}}} \][/tex]

6. Substituting the given values, we can calculate the mass of water to be approximately 48.4 grams.

Therefore, the mass of the water is approximately 48.4 grams.

The mass of the water is approximately 199.8 grams.

Here's how to find the mass of the water:

1. Define the variables:

m_iron = mass of iron rod = 32.5 g

c_iron = specific heat capacity of iron = 0.49 J/g°C

T_iron_initial = initial temperature of iron rod = 21.8°C

T_water_initial = initial temperature of water = 63.3°C

T_final = final temperature of the mixture = 58.9°C

m_water = mass of water (unknown)

c_water = specific heat capacity of water = 4.186 J/g°C

2. Apply the principle of heat transfer:

Heat lost by iron = Heat gained by water

3. Set up the equation:

m_iron * c_iron * (T_iron_initial - T_final) = m_water * c_water * (T_final - T_water_initial)

4. Substitute the known values and solve for m_water:

32.5 g * 0.49 J/g°C * (21.8°C - 58.9°C) = m_water * 4.186 J/g°C * (58.9°C - 63.3°C)

5. Solve for m_water:

m_water ≈ 199.8 g

Therefore, the mass of the water is approximately 199.8 grams.

Answer:

The carrier lengthen is 0.08436 m.

Explanation:

Given that,

Length = 370 m

Initial temperature = 2.0°C

Final temperature = 21°C

We need to calculate the change temperature

Using formula of change of temperature

[tex]\Delta T=T_{f}-T_{i}[/tex]

[tex]\Delta T=21-2.0[/tex]

[tex]\Delta T=19^{\circ}C[/tex]

We need to calculate the carrier lengthen

Using formula of length

[tex]\Delta L=\alpha_{steel}\times L_{0}\times\Delta T[/tex]

Put the value into the formula

[tex]\Delta L=1.2\times10^{-5}\times370\times19[/tex]

[tex]\Delta L=0.08436\ m[/tex]

Hence, The carrier lengthen is 0.08436 m.

The new tangential speed of the ball is four times the original speed.

When the length of the string is decreased to R/4, the radius of the circular path becomes R/4. The tension force exerted on the ball remains constant because it only depends on the mass of the ball and the gravitational force acting on it. The tangential speed is given by the formula v = ΩR, where Ω is the angular velocity. Since the radius is now 1/4 of the original radius, the new tangential speed is 4 times the original speed (option a).

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Answer:

Explanation:

Decrease in kinetic energy of the ball

= 1/2 m ( v² - u² )

= 1/2 x .6 ( 8.3² - 7.1² )

= 5.544 J

This energy will be converted into potential energy

mgH = 5.544 ( H is maximum height attained )

H = 5.544 /( .6 x 9.8 )

= .942 m

= 94.2 cm

b ) In a ) we used the relation

mgH = 1/2 m ( v² - u² )

H =  ( 1/2g ) x ( v² - u² )

Here we see H or height attained does not depend upon mass as m cancels out on both sides.

Answer:

50 J

Explanation:

Work = force × distance

W = Fd

W = (5 N) (10 m)

W = 50 J

The amount of work done is 50 J (Joule)

To express this concept mathematically, the work W is equal to the force f times the distance d, or W = fd. If the force is being exerted at an angle θ to the displacement, the work done is W = fd cos θ.

By applying the formula, we get:

Work = force × distance

Force = 5 newtons                       ...(given)

distance = 10 meters                    ...(given)

⇒ W = Fd

⇒ W = (5 N) (10 m)

⇒ W = 50 J

Work is done whenever a force moves something over a distance. You can calculate the energy transferred, or work done, by multiplying the force by the distance moved in the direction of the force.

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Answer:

4.20m/s^2

Explanation:

The mass of 210g stretch the spring by 26.9cm, 9 oscillation took 14.3 s.

Period (T) = time of oscillation / number of oscillation = 14.3/ 9 = 1.59s

Using the formula for period of a spring;

T = 2π√(m/k)

Divide both side by 2π

T/2π = √(m/k)

Square both side

T^2/ (2π)^2 = m/k

Make k subject of the formula

K = 4π^2 * m/ T^2

Also using Hooke's law

Since the spring was suspended from the ceiling,

F (mg) = k * DL ( where g is the new gravity in m/s^2, k is the force constant of the spring in N/m and DL is the change in length in meter

Make k subject of the formula

Mg/DL = k

Since both equation equal to K then

4π^2 * m/ T^2 = mg/DL

Cancel m on both side

4π^2 * DL / T^2 = g

DL = 26.9cm = 26.9/100 m = 0.269 m

T = 1.59s and π = 3.142

Substitute these into the equation

g = ( 4 * 3.142 * 3.142* 0.269) / (1.59^2) = 4.20 m/s^2

Final answer:

To determine the acceleration due to gravity on Planet X, the period T of the spring-mass system's oscillations is used along with the mass m and the spring constant k, which is calculated using the initial stretch of the spring.

Explanation:

To find the acceleration due to gravity (g) on Planet X, we can use the properties of simple harmonic motion for a spring-mass system. The formula for the period T of a mass m attached to a spring with spring constant k is T = 2π√(m/k). Since the spring stretches 26.9 cm when the 210 g mass is attached, the force due to gravity must be equal to the force the spring exerts at that stretch, which is F = kx, where x is the stretch distance. Thus, mg = kx, and we can solve for k using k = mg/x (with g as the gravity on Earth, 9.81 m/s²). Then we can calculate the period using the measured time of oscillations. From the period, we can solve for the gravity on Planet X.

Considering 9 oscillations took 14.3 seconds, the period T for one oscillation is 14.3 s / 9. With T and m known, one can solve for k and then use that to find g. Since the initial stretch was caused by the force of gravity, the spring constant k can be found using Earth's gravity, and subsequently used in the equation T = 2π√(m/g) rearranged to solve for g on Planet X.

Answer:

The magnitude of the net force is 2.48294 N and the angle is 25.02° with respect to the vertical

Explanation:

Consider the gravitational force as [tex]\vec{F_g}=2.25\ N[/tex]

Consider the wind force as [tex]\vec{F_w}=1.05\ N[/tex]

The angle between the gravitational force and wind force is 90°

From the parallelogram law we have

[tex]F=\sqrt{F_g^2+F_w^2+2F_gF_wcos\theta}\\\Rightarrow F=\sqrt{2.25^2+1.05^2+2\times 2.25\times 1.05\times cos90}\\\Rightarrow F=2.48294\ N[/tex]

The angle between the resultant and the gravitational force would be

[tex]tan^{-1}\frac{1.05}{2.25}=25.02^{\circ}[/tex]

The magnitude of the net force is 2.48294 N and the angle is 25.02° with respect to the vertical

Answer:

A)

665.5 W

B)

575.5 m

Explanation:

A)

[tex]S[/tex] = Sound level registered = 128 dB

[tex]I_{o}[/tex] = Intensity at reference level = [tex]1\times10^{-12}[/tex] Wm⁻²

[tex]I[/tex] = Intensity at the location of meter

Sound level is given as

[tex]S = 10 log_{10}\left ( \frac{I}{I_{o}} \right )[/tex]

[tex]128 = 10 log_{10}\left ( \frac{I}{1\times10^{-12}} \right )\\12.8 = log_{10}\left ( \frac{I}{1\times10^{-12}} \right )\\10^{12.8} = \frac{I}{1\times10^{-12}} \right \\I = 6.3[/tex]

[tex]P[/tex] = power output of the speaker

[tex]r[/tex] = distance from the speaker = 2.9 m

Power output of the speaker is given as

[tex]P = I(4\pi r^{2} )\\P = (6.3) (4) (3.14) 2.9^{2}\\P = 665.5 W[/tex]

B)

[tex]S[/tex] = Sound level = 82 dB

[tex]I_{o}[/tex] = Intensity at reference level = [tex]1\times10^{-12}[/tex] Wm⁻²

[tex]I[/tex] = Intensity at the location of meter

Sound level is given as

[tex]S = 10 log_{10}\left ( \frac{I}{I_{o}} \right )[/tex]

[tex]82 = 10 log_{10}\left ( \frac{I}{1\times10^{-12}} \right )\\8.2 = log_{10}\left ( \frac{I}{1\times10^{-12}} \right )\\10^{8.2} = \frac{I}{1\times10^{-12}} \right \\I = 1.6\times10^{-4}Wm^{-2}[/tex]

Power of the speaker is given as

[tex]P = I(4\pi r^{2} )\\665.5 = (1.6\times10^{-4}) (4) (3.14) r^{2}\\r = 575.5 m[/tex]

The power output of the speaker at the rock concert was approximately 137 Watts. The sound level would be 82 dB at a distance of about 103 meters from the speaker.

This problem involves the use of the formula for calculating sound levels in decibels (dB). The formula is L = 10 log(I/I0) where L is the sound level in dB, I is the intensity of the sound wave, and I0 is the reference intensity. Now, let's solve both parts.

A. At the dB meter location, reverse calculate the intensity I: I = I0 * 10^(L/10) = 1.0 ‍ 10^−12 * 10^(128/10) = 1.26 W/m^2. Since the sound spreads spherically, the power P is the intensity times the area of the sphere: P = I * 4πr^2 = 1.26 * 4π*(2.9)^2 = 137 Watts.

B. To find the distance where the sound would register at 82 dB, we first find the intensity I at that level: I = I0*10^(82/10) = 1.0x10^−12 * 10^8.2 = 6.31x10^-5 W/m^2. With I, we can solve for r: r = sqrt(P/(4πI)) = sqrt(137/(4π*(6.31x10^-5))) = 103 meters.

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The angular momentum of the hockey stick-puck system before the collision, with respect to the center of mass of the final system, is given by Lcm = mp * v0 * D, where mp is the puck's mass, v0 is its initial velocity, and D is the distance from the center of the stick.

Angular Momentum of Hockey Stick-Puck System

To find the angular momentum Lcm of the system before the collision with respect to the center of mass of the final system, we follow these steps:

Therefore, the angular momentum of the system before the collision, with respect to the center of mass of the final system, is Lcm = mp * v0 * D.

Answer:

Length, l = 0.866 meters

Explanation:

Given that,

Frequency of sound, f = 99 Hz

Speed of sound in air, v = 343 m/s

To find,

The length of a wind instrument.

Solution,

The standing wave will gets formed in wind instrument. For the closed tube, the closed tube the frequency is given by :

[tex]f=\dfrac{v}{4l}[/tex]

Where

l is the length of the instrument

[tex]l=\dfrac{v}{4f}[/tex]

[tex]l=\dfrac{343\ m/s}{4\times 99\ Hz}[/tex]

l = 0.866 meters

So, the length of a wind instrument is 0.866 meters. Hence, this is the required solution.

Answer: (2) Use the Momentum Principle.

Explanation:

In fact, it is called the Conservation of linear momentum principle, which establishes the initial momentum [tex]p_{i}[/tex] of the asteroids before the collision must be equal to the final momentum [tex]p_{f}[/tex] after the collision, no matter if the collision was elastic or inelastic (in which the kinetic energy is not conserved).

In this sense, the linear momentum [tex]p[/tex] of a body is defined as:

[tex]p=mV[/tex]

Where [tex]m[/tex] is the mass and [tex]V[/tex] the velocity.

Therefore, the useful approach in this situation is option (2).

Answer:before throwing and after catching the ball

Explanation:

When basketball is in the hand of player net force on it zero as holding force is canceled by gravity Force. During its entire motion gravitational force is acting on the ball which is acting downward. Even at highest point gravity is constantly acting downwards.

After catching the ball net force on it zero as holding force is canceled by gravity force and ball is continue to be in stationary motion.          

The point in time where the zero net force acted on the ball is before throwing and after catching the ball

At the time of basketball is in the hand of the player, the net force should be zero since holding force is canceled via gravity Force. At the time of overall motion,  gravitational force should be acted on the ball which is acting downward. After catching the ball net force on it zero since holding force should be canceled via gravity force and ball should be continue in stationary motion.    

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Answer:

from los angeles distance plan a = 1111.08 mi

from los angeles distance plan b =  888.92 mi

Explanation:

given data

new york to los angeles distance = 2000 miles

Plane b speed = 100 mph faster than plane a

Plane b takes off  time = 1 hour after plane a

to find out

How far are they from los angeles when they pass

solution

we consider speed of plan a is = x mph

so speed of plan b will be = x + 100 mph

and we know plan b take here 1 hour less time than plan a so it mean time is distance divide speed i.e

[tex]\frac{2000}{x} - 1 =\frac{2000}{x+100}[/tex]

solve it we get x = 400 mph

it mean here

plan a speed is 400 mph

and plan b speed is 500 mph

and

now we consider they meet at time = t hour  after a take off

then plan a travel  = 400 t

and plan b travel = 500 ( t - 1 )

add both distance that is equal to 2000 mi

so 400 t + 500 ( t -1 ) = 2000

400 t + 500 ( t -1 ) = 2000

400 t + 500 ( t-1) = 2000

solve we get

t = 2.777

so total distance travel plan a = 400 × 2.777 = 1111.08 mi

total distance travel plan b = 2000 - 1111.08

total distance travel plan b = 888.92 mi

Answer:

(a). The normal force exerted by the ground on the front wheel is [tex]4.67\times10^{5}\ N[/tex].

(b). The normal force exerted by the ground on each of the two rear wheels is [tex]7.02\times10^{5}\ N[/tex]

Explanation:

Given that,

Weight of jet [tex]W=1.87\times10^{6}\ N[/tex]

Distance = 16 m

Second distance = 12.0 m

We need to calculate the normal force exerted by the ground on the front wheel

Using formula of torque

[tex]\sum\tau=-Wl_{W}+F_{f}l_{f}=0[/tex]

Where, W = weight of jet

[tex]l_{f}[/tex]=lever arms for the forces [tex]F_{f}[/tex]

[tex]l_{w}[/tex]=lever arms for the forces W

Put the value into the formula

[tex]-(1.87\times10^{6})\times(16.0-12.0)+F_{f}\times16.0=0[/tex]

[tex]F_{f}=\dfrac{(1.87\times10^{6})\times(16.0-12.0)}{16.0}[/tex]

[tex]F_{f}=4.67\times10^{5}\ N[/tex]

(b). We need to calculate the normal force exerted by the ground on each of the two rear wheels

The sum of vertical forces equal to zero.

[tex]\sum F_{y}=F_{f}+2F_{r}-W=0[/tex]

We using 2 for two rear wheels

[tex]\sum F_{y}=0[/tex]

[tex]F_{f}+2F_{r}-W=0[/tex]

[tex]F_{r}=\dfrac{F_{f}-W}{2}[/tex]

Put the value into the formula

[tex]F_{r}=\dfrac{-4.67\times10^{5}+1.87\times10^{6}}{2}[/tex]

[tex]F_{r}=7.02\times10^{5}\ N[/tex]

Hence, (a). The normal force exerted by the ground on the front wheel is [tex]4.67\times10^{5}\ N[/tex].

(b). The normal force exerted by the ground on each of the two rear wheels is [tex]7.02\times10^{5}\ N[/tex]

Answer:

[tex]d=627.9\ m[/tex]  is the distance from the obstacle of reflection.

wavelength [tex]\lamb=0.5279\ m [/tex]

Explanation:

Given that:

We know,

[tex]\rm distance = speed \times time[/tex]

During an echo the sound travels the same distance back and forth.

[tex]2d=v.t[/tex]

[tex]2d=322\times 3.9[/tex]

[tex]d=627.9\ m[/tex]  is the distance from the obstacle of reflection.

Now the wavelength of sound waves:

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

[tex]\lambda=\frac{322}{610}[/tex]

[tex]\lamb=0.5279\ m [/tex]

Answer:

The terminal speed of this object is 12.6 m/s

Explanation:

It is given that,

Mass of the object, m = 80 kg

The magnitude of drag force is,

[tex]F_{drag}=12v+4v^2[/tex]

The terminal speed of an object is attained when the gravitational force is balanced by the gravitational force.

[tex]F_{drag}=mg[/tex]

[tex]12v+4v^2=80\times 9.8[/tex]

[tex]4v^2+12v=784[/tex]

On solving the above quadratic equation, we get two values of v as :

v = 12.58 m/s

v = -15.58 m/s (not possible)

So, the terminal speed of this object is 12.6 m/s. Hence, this is the required solution.

Answer:

[tex]l=0.068 m[/tex]

Explanation:

given,

inductance of solenoid = 2.07 μ H

winding of the wire = 1.19 m

Using formula of inductance

[tex]L = \dfrac{\mu_0N^2A}{l}[/tex]

L is the inductance

N is number of turns of the coil

μ₀ is permeability of free space

L is length of winding

N (2π r) = 1.19          

squaring both side

4π(N²(πr²))=1.19²    

N² A = 0.113            

now                              

[tex]2.07 \times 10^{-6}= \dfrac{4\pi\times 10^{-7}\times 0.113}{l}[/tex]

[tex]l= \dfrac{4\pi\times 10^{-7}\times 0.113}{2.07 \times 10^{-6}}[/tex]

[tex]l=0.068 m[/tex]

This question involves the concepts of the inductance of an inductor and the length of the winding.

The tube should be "6.84 cm" long.

The inductance of an inductor can be given by the following formula:

[tex]L=\frac{N^2\mu_oA}{l}[/tex]

where,

L = inductance = 2.07 μH = 2.07 x 10⁻⁶ H

N = No. of turns in coil = [tex]\frac{length\ of\ wire}{circumference\ of\ tube} = \frac{1..19\ m}{2\pi r}[/tex]

μ₀ = permeability of free space = 4π x 10⁻⁷ Wb/A.m

A = Area of tube = πr²

l = length of tube = ?

Therefore,

[tex]2.07\ x\ 10^{-6}\ H = \frac{(\frac{1.19\ m}{2\pi r})^2(\pi r^2)(4\pi\ x\ 10^{-7}\ Wb/A.m)}{l}\\\\l = \frac{1.4161\ x\ 10^{-7}}{2.07\ x\ 10^{-6}}\\\\[/tex]

l = 0.0684 m = 6.84 cm

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The attached picture shows the inductance formula.

To compensate for the clock's slowness, you should adjust the length of the pendulum to [tex]\( 0.9939 \)[/tex] meters.

To calculate the necessary adjustment to the length of the pendulum to compensate for the clock's slowness, we can start by determining the time period of the pendulum using the given length and the unknown acceleration due to gravity, denoted as [tex]\( g \).[/tex] The time period [tex]\( T \)[/tex] of a simple pendulum is given by the formula:

[tex]\[ T = 2\pi \sqrt{\frac{L}{g}} \][/tex]

where [tex]\( L \)[/tex] is the length of the pendulum.

Given that the clock loses 19 seconds per day, we convert this to seconds per period, as [tex]\( 1 \) day has \( 24 \) hours, \( 60 \)[/tex] minutes per hour, and [tex]\( 60 \)[/tex] seconds per minute:

[tex]\[ \text{Seconds lost per period} = \frac{19 \, \text{s}}{24 \times 60 \times 60 \, \text{s/day}} \][/tex]

Let's calculate the time period [tex]\( T \)[/tex]  and the adjustment needed.

[tex]\[ T = 2\pi \sqrt{\frac{0.9930 \, \text{m}}{g}} \][/tex]

Now, we'll solve for [tex]\( g \):[/tex]

[tex]\[ T = 2\pi \sqrt{\frac{0.9930 \, \text{m}}{g}} \]\[ T^2 = 4\pi^2 \frac{0.9930 \, \text{m}}{g} \]\[ g = 4\pi^2 \frac{0.9930 \, \text{m}}{T^2} \][/tex]

Substitute [tex]\( T \) with \( 24 \, \text{hours} \)[/tex] converted to seconds:

[tex]\[ T = 24 \times 60 \times 60 \, \text{s} = 86,400 \, \text{s} \]\[ g = 4\pi^2 \frac{0.9930 \, \text{m}}{(86,400 \, \text{s})^2} \][/tex]

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

[tex]\[ g = 4\pi^2 \frac{0.9930 \, \text{m}}{7.48224 \times 10^9 \, \text{s}^2} \]\[ g = 0.0000016512 \, \text{m/s}^2 \][/tex]

Now that we have [tex]\( g \)[/tex], we can find the adjusted length of the pendulum. The new time period [tex]\( T' \)[/tex] can be calculated as follows:

[tex]\[ T' = 2\pi \sqrt{\frac{L'}{g}} \][/tex]

Where [tex]\( L' \)[/tex] is the adjusted length. Rearrange to solve for [tex]\( L' \):[/tex]

[tex]\[ L' = \frac{T'^2 \times g}{4\pi^2} \][/tex]

Given that the clock loses [tex]\( 19 \)[/tex] seconds per day and we want it to lose [tex]\( 0 \)[/tex]  seconds, the new time period [tex]\( T' \) is \( 86,400 \)[/tex] seconds. Substitute into the formula:

[tex]\[ L' = \frac{(86,400 \, \text{s})^2 \times 0.0000016512 \, \text{m/s}^2}{4\pi^2} \][/tex]

[tex]\[ L' = 0.9939 \, \text{m} \][/tex]

So, to compensate for the clock's slowness, you should adjust the length of the pendulum to [tex]\( 0.9939 \)[/tex] meters.

Complete Question:
Your grandfather clock's pendulum has a length of 0.9930 m. Part A If the clock runs slow and loses 19 s per day, how should you adjust the length of the pendulum? Note: due to the precise nature of this problem you must treat the constant g as unknown (that is, do not assume it is equal to exactly 9.80 m/s2).