A tall, cylindrical chimney falls over when its base is ruptured. Treat the chimney as a thin rod of length 53.2 m. At the instant it makes an angle of 34.1° with the vertical as it falls, what are (a) the radial acceleration of the top, and (b) the tangential acceleration of the top. (Hint: Use energy considerations, not a torque.) (c) At what angle θ is the tangential acceleration equal to g? Assume free-fall acceleration to be equal to 9.81 m/s2

Answer:

90 N

Explanation:

The force applied to the ball is given by:

[tex]F=\frac{\Delta p}{\Delta t}[/tex]

where

[tex]\Delta p[/tex] is the change in momentum of the ball

[tex]\Delta t[/tex] is the time taken

The change on momentum of the ball is:

[tex]\Delta p=m\Delta v=(0.45 kg)(20 m/s)=9 kg m/s[/tex]

So, the force applied is

[tex]F=\frac{9 kg m/s}{0.10 s}=90 N[/tex]

A) At h=4.0 m

At h=4.0 m, the kinetic energy of the ball is zero, because its velocity is 0, so

[tex]K=\frac{1}{2}mv^2=\frac{1}{2}(0.18 kg)(0)^2=0[/tex]

The gravitational potential energy is instead:

[tex]U=mgh=(0.18 kg)(9.8 m/s^2)(4.0 m)=7.06 J[/tex]

So, the total mechanical energy is

[tex]E=K+U=0+7.06 J=7.06 J[/tex]

B) At h=3.0 m

The gravitational potential energy is now:

[tex]U=mgh=(0.18 kg)(9.8 m/s^2)(3.0 m)=5.29 J[/tex]

Since air resistance is negligible, the total mechanical energy is conserved, so it is still

[tex]E=7.06 J[/tex]

And so we can find the kinetic energy as follows:

[tex]K=E-U=7.06 J-5.29 J=1.77 J[/tex]

C) At h=2.0 m

The gravitational potential energy is now:

[tex]U=mgh=(0.18 kg)(9.8 m/s^2)(2.0 m)=3.53 J[/tex]

Since air resistance is negligible, the total mechanical energy is conserved, so it is still

[tex]E=7.06 J[/tex]

And so we can find the kinetic energy as before:

[tex]K=E-U=7.06 J-3.53 J=3.53 J[/tex]

Answer:

Part a)

[tex]\theta = 45 degree[/tex]

Part b)

[tex]R = 39.2[/tex]

Part c)

[tex]v_x = 13.86 m/s[/tex]

[tex]v_y = 13.86 m/s[/tex]

Explanation:

Since it took 4.0 s to come back at the same position so we can say

[tex]\Delta y = v_y t + \frac{1}{2}at^2[/tex]

[tex]0 = v_y(4.0) - \frac{1}{2}(9.8)(4.0)^2[/tex]

[tex]v_y = 19.6 m/s[/tex]

Part a)

Now we know that horizontal range of projectile is given as

[tex]R = \frac{v^2 sin2\theta}{g}[/tex]

Now this range would be maximum if the angle of the projectile is giving maximum value of sine

so we have

[tex]sin(2\theta) = 1[/tex]

[tex]\theta = 45 degree[/tex]

Part b)

For maximum range we have

[tex]R = \frac{v^2}{g}[/tex]

[tex]R = \frac{19.6^2}{9.8}[/tex]

[tex]R = 39.2[/tex]

Part c)

Since we projected at an angle of 45 degree

so the components are given as

[tex]v_x = 19.6 sin45 = 13.86 m/s[/tex]

[tex]v_y = 19.6 cos45 = 13.86 m/s[/tex]

Answer:

Nail should be located at the bottom of claw as possible And hand should be place at the tip of handle as possible without losing perfect grip.

Load x Load distance = Effort x Effort distance

Effort = Load x load distance / Effort distance.

Final answer:

Resonance is a condition where a system oscillates at its natural frequency, leading to increased amplitude. It is utilized in radio tuning, medical MRI diagnostics, playground swings, and ensuring the stability of structures like bridges.

Explanation:

Resonance is a fundamental concept in physics that refers to the condition when a system oscillates at its natural frequency, leading to an increase in amplitude. This phenomenon has several practical applications in our daily lives. For instance, when tuning a radio, the resonant frequency is adjusted to match the frequency of the desired radio station, ensuring a clear signal. Medical diagnostic tools such as Magnetic Resonance Imaging (MRI) rely on resonance to create detailed images of the human body by making atomic nuclei oscillate using radio waves.

Additionally, children achieve maximum enjoyment on a swing when it is pushed at its natural frequency, embodying the concept of resonance. Even in engineering, understanding resonance is essential to avoid harmful oscillations, as evidenced by historical events like the Tacoma Narrows Bridge collapse or the adjustments made to the Millennium Bridge in London to prevent wobbling.

Answer: radiation from the sun

Explanation: the sun produces ranges if radiation ranging from gamma Ray to Infrared but ultraviolet Ray is the most abundant

She must have been had the sun burn from the head propagated by the ultalraviolet ray from the sunshine

50 years converted into seconds would be equivalent to 1.58×10⁹ seconds.

A unit of measurement is a specified magnitude of a quantity that is established and used as a standard for measuring other quantities of the same kind. It is determined by convention or regulation. Any additional quantity of that type can be stated as a multiple of the measurement unit.

As given in the problem we have to convert 50 years into seconds,

1 year = 365 days

50 years = 50×365 days

50 years = 50×365×24 hours

50 years =50×365 ×24×60 minutes

50 years =50×365 ×60×60 seconds

               =1576800000 seconds

               =1.58×10⁹ seconds

Thus,50 years converted into seconds would be 1.58×10⁹ seconds

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(a) The currents direction must be counterclockwise.

(b) The magnitude of the current in the outer wire is 15.83 A.

The direction of the current will flow in such a way that the magnetic field due to the wires combination will cancel out. Thus, the current will flow in opposite or counterclockwise direction.

The magnitude of the current is calculated using the following formulas;

[tex]\frac{I_1}{D_1} = \frac{I_2}{D_2} \\\\I_2 = \frac{I_1 D_2}{D_1} \\\\I_2 = \frac{10 \times 38}{24} \\\\I_2 = 15.83 \ A[/tex]

Thus, the magnitude of the current in the outer wire is 15.83 A.

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(a) The work done by the worker on the crate is 8,280 J.

(b) The work done by the floor on the crate is 7,920 J.

(c) The net work done on the crate is 360 J.

The total work done by the worker is calculated as follows;

[tex]W = Fd\\\\W = 345 \times 24\\\\W = 8,280 \ J[/tex]

The work done by the floor on the crate is calculated as follows;

[tex]W = F_f d\\\\W = \mu F_n d\\\\W = 0.22 \times 1.5 \times 10^3 \times 24\\\\W = 7,920 \ J[/tex]

The net work done on the crate is calculated as follows;

[tex]W = 8,280 - 7,920\\\\W = 360 \ J[/tex]

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The work done by the worker is 8280 Joules. The work done by the floor (accounting for friction) is -7920 Joules. This makes the net work done on the crate 360 Joules.

The question is asking about the concept of work and net work in physics, specifically in the context of friction and forces. To start, we need to understand that work is defined as the force applied on an object times the distance that object moves in the direction of the force (Work = Force x Distance).

a) The work is done by the worker on the crate can be calculated using the formula Work = Force x Distance. Hence, Work = 345 N x 24.0 m = 8280 Joules.

b) The work done by the floor on the crate is the frictional force times the distance. The frictional force can be calculated using the formula Force = Coefficient of friction x Normal force. In this case, the normal force is the weight of the crate (1.50 x 10^3 N). Hence, Frictional force = 0.220 x (1.50 x 10^3 N). Work done by the floor is Frictional force x Distance, so the work done by the floor = Frictional force x 24.0 m = -7920 Joules (It's negative because the force of friction acts in the opposite direction to motion).

c) The net work done on the crate is the sum of the work done by the worker and the work done by the floor. Hence, Net work = 8280 Joules - 7920 Joules = 360 Joules.

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Carbon-14 dating is used to date organic materials, not inorganic objects like stone tablets, which require other methods like potassium-argon dating.

Carbon-14 dating, or radiocarbon dating, is a method used to date materials that were once living and contains carbon. The process measures the decay rate of carbon-14, a radioisotope, to determine the age of organic artifacts up to approximately 60,000 years old. However, this technique is not suitable for dating stone tablets or any other inorganic objects such as pottery or projectile points because they do not contain carbon from living organisms. To date geological materials like stone, other methods such as potassium-argon dating are used. This method relies on the radioactive decay of potassium-40 to argon-40 to estimate the age of igneous and volcanic rocks.

To calculate the electric field amplitude of light from a Lasik laser, start from the pulse energy and duration to find the power and intensity, then use known constants and the equation I = cε₀E²/2 to derive the amplitude.

The electric field amplitude of the light wave at the cornea from a Lasik laser can be calculated using the energy of light and its other characteristics. Given the pulse energy E = 3.0 mJ and pulse duration t = 10 ns, we first calculate the instantaneous power Pi during one pulse using the relationship: Pi = E / t.

Then, given the diameter of the laser beam d = 0.80 mm, we can calculate the area of the laser beam A = π(d/2)². So, we find the intensity of the pulse by dividing the power by the area: I = Pi / A.

As the intensity of an electromagnetic wave (light wave in this case) is proportional to the square of the electric field amplitude (E), we can deduce the equation I = cε₀E²/2, where c is the speed of light and ε₀ is the permittivity of free space. We solve for E to get the amplitude of the electric field: E = sqrt(2I / cε₀).

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The electric field amplitude of the light wave at the cornea is approximately 3.35 × [tex]10^8[/tex] V/m.

The electric field amplitude of the light wave at the cornea can be calculated using the energy of the light pulse and its other characteristics. Given the pulse energy E = 3.0 mJ and pulse duration t = 10 ns, we first calculate the instantaneous power Pi during one pulse using the relationship:

Pi = E / t = 3.0 mJ / 10 ns = 300 MW

Then, given the diameter of the laser beam d = 0.80 mm, we can calculate the area of the laser beam A = π(d/2)²:

A = π(0.80 mm / 2)² ≈ 0.503 mm² = 5.03 × [tex]10^{-6[/tex] m²

So, we find the intensity of the pulse by dividing the power by the area:

I = Pi / A = 300 MW / 5.03 × [tex]10^{-6[/tex] m² ≈ 5.98 × [tex]10^8[/tex] W/m²

Finally, the electric field amplitude E of the light wave is related to the intensity I by the following equation:

E² = (2I) / (ε₀c)

where ε₀ is the permittivity of free space and c is the speed of light. Substituting the values of I, ε₀, and c, we get:

E² = (2 × 5.98 × [tex]10^8[/tex] W/m²) / ((8.85 × [tex]10^{-12[/tex] F/m) × (3 × [tex]10^8[/tex] m/s)) ≈ 1.12 × [tex]10^{17[/tex]V²/m²

Taking the square root of both sides, we find the electric field amplitude E:

E ≈ 3.35 ×[tex]10^8[/tex] V/m

Therefore, the electric field amplitude of the light wave at the cornea is approximately 3.35 × [tex]10^8[/tex] V/m.

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Answer:

The energy stored in a room is 0.0286 Joules.

Explanation:

Given that,

The dimensions of the room are 3 m by 4 m by 2.4 m

The strength of the Earth's magnetic field, [tex]B=5\times 10^{-5}\ T[/tex]

We need to find the energy stored in a room due to the Earth'a magnetic field. It is given by :

[tex]E=\dfrac{B^2}{2\mu_o}V[/tex]

Here,

V is the volume of the room

[tex]E=\dfrac{(5\times10^{-5})^{2}}{2\times4\pi\times10^{-7}}\times(3\times4\times2.4)\\\\E=0.0286\ J[/tex]

[tex]E=0.0286\ J[/tex]

So, the energy stored in a room is 0.0286 Joules. Hence, this is the required solution.

To determine the reaction orders for CH₃Cl and H₂O in the given reaction, one must analyze rate data showing how changing reactant concentrations affect the reaction rate, and apply algebraic methods to solve for the orders, which we do not have. Without the specific tabulated data, we cannot provide the exact orders for the reaction.

The reaction orders for the given chemical reaction can be determined using the method of initial rates, which requires experimental data that compares how the reaction rate changes with varying concentrations of the reactants while keeping other conditions constant. Unfortunately, the full tabulated data set required to determine the reaction orders is missing in the question; however, the example provided in the reference indicates how one would go about calculating these orders once the appropriate data is available.

If we had a table showing how the reaction rate varies when the concentrations of CH₃Cl and H₂O are changed, we could look for patterns similar to those described. For instance, if doubling the concentration of CH₃Cl leads to a doubling of the reaction rate, this would suggest that the reaction is first-order to CH₃Cl. If changing the concentration of H₂O does not change the rate, then the reaction would be zero-order to H₂O. The overall reaction order is then the sum of the individual orders.

It is important to carefully analyze the tabulated data, comparing the rates and concentrations and using algebraic methods to solve for the reaction orders of each reactant and determine the overall reaction order for the chemical reaction.

The common velocity of the two-car train after the collision, given that the 15000 kg train car was initially moving due east at 23 m/s, is +13.8 m/s

The question given above talks about conservation of linear momentum of colliding objects.

Thus, we can calculate the common velocity of the two-car train after the collision by using the formula of conservation of linear momentum as shown below:

Momentum before = Momentum after

[tex]m_1u_1 + m_2u_2 = v(m_1 + m_2)[/tex]

(15000 × 23) + (10000 × 0) = v(15000 + 10000)

345000 + 0 = 25000v

345000 = 25000v

Divide both sides by 25000

v = 345000 / 25000

= +13.8 m/s

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The critical temperature of a compound is the point at which its gas phase cannot be liquefied by pressure and for binary mixtures, it represents the temperature where two immiscible liquids become fully miscible.

The critical temperature of a compound is the temperature above which the gas phase of the substance cannot be made to liquefy, no matter how much pressure is applied. For a binary mixture, this can also refer to the upper critical temperature where two immiscible liquids become completely miscible in all proportions. An example provided is that of methyl acetate and carbon disufide, which has a critical temperature of approximately 230 K at one atmosphere. A similar behavior occurs with hexane/nitrobenzene mixtures, which have a critical temperature of 293 K.

These temperatures represent the critical point at which there is no distinction between the liquid and the gas phase of a substance. This is shown by the fact that at the critical temperature and the associated critical pressure, the gas does not condense. Understanding and utilizing the critical temperature of compounds is important in various applications, including the development and use of high-temperature superconductors.

Answer:

The purpose of having a control sample is the possibility of having something to compare the results of your experiment.

For example with medicine, the sample that received the medicine are in X state now, this has not enough information really, you need a control sample that is in a state Y, and now you can compare the states X and Y and see how your medicine really affects the patients.

Other example is how a substance X changes the color of something, if you do not have a control sample, at the end of the experiment you can't se the actual color change, so your really need a control sample.

The control sample is called the "zero" or starting point in an experiment.

The Pearson correlation coefficient, r, measures the strength and direction of the linear relationship between two variables. In this case, a correlation coefficient of r = 0.43 was obtained for a sample of 20 participants. The correlation is not statistically significant at both alpha levels of 0.01 and 0.05.

The Pearson correlation coefficient, r, measures the strength and direction of the linear relationship between two variables. In this case, the researcher obtained a correlation coefficient of r = 0.43 for a sample of n = 20 participants. To determine the significance of the correlation, a hypothesis test can be conducted at different alpha levels. At an alpha level of 0.01, the critical value for a two-tailed test with 18 degrees of freedom is 2.898.

The calculated t-value using the formula r * sqrt((n-2)/(1-r^2)) is compared to the critical value. If the t-value is less than the critical value, the correlation is not statistically significant. In this example, the calculated t-value is less than 2.898, indicating that the correlation is not statistically significant at the 0.01 level.

At an alpha level of 0.05, the critical value for a two-tailed test with 18 degrees of freedom is 2.101. Since the calculated t-value is less than 2.101, the correlation is not statistically significant at the 0.05 level as well.

To determine if the Pearson correlation of r = 0.43 is significant for n = 20, one must calculate the t-statistic and compare it to the critical t values from a t distribution table for df = 18 for each alpha level (α). These critical values determine whether the correlation is significant at the .05 or .01 level.

To determine the significance of a Pearson correlation coefficient r for a given alpha level (α), one can use the t distribution as a reference. In the given scenario, the researcher has obtained a Pearson correlation of r = 0.43 with a sample size of n = 20 participants. Before evaluating significance, one must calculate the t-statistic using the formula given:

t = r√n-2/√(1-r²)

The degrees of freedom (df) for this calculation could be determined by df = n - 2, which here would be 20 - 2 = 18. Upon comparing the calculated t value to the critical t value from a t distribution table for df = 18, one can establish the significance of the correlation coefficient.

Looking at a standard t distribution table or using statistical software, one would find the critical t values for a two-tailed test. For α = 0.05, the critical t value for df = 18 would typically be around 2.101. For α = 0.01, this value would be higher, representing a stricter criterion for significance. If the calculated t value is greater than the critical value, the null hypothesis, which proposes no significant correlation, is rejected.

Given the details of the Pearson correlation in the question, without the exact t value calculated, it is not possible to definitively say whether the r = 0.43 is significant at the α = 0.05 or 0.01 level. However, provided that the calculated t value is larger than the critical t value at either α level, the null hypothesis would be rejected, indicating a significant correlation.

The index of refraction of the medium is 1.5.

The speed of light in a medium is given by the equation c = v/n, where c is the speed of light in a vacuum, v is the speed of light in the medium, and n is the index of refraction of the medium. Rearranging the equation, we can find the index of refraction as n = c/v. Substituting the given values, the index of refraction is n = (3.00 x 10^8 m/s) / (2.00 x 10^8 m/s) = 1.5.

The rpm of the second pulley is 15.

To find the rpm of the second pulley, we can use the formula:

Angular velocity of first pulley * diameter of first pulley = angular velocity of second pulley * diameter of second pulley

Plugging in the given values:

10 rpm * 15.0 cm = rpm of second pulley * 10.0 cm

Dividing both sides by 10.0 cm:

Rpm of second pulley = (10 rpm * 15.0 cm) / 10.0 cm = 15 rpm

The magnitude of the angular deceleration of the cylinder is -10.0 rad/s^2, and the magnitude of the force of friction applied by the brake shoe is -8.50 N.

(a) To find the magnitude of the angular deceleration of the cylinder, we can use the formula:

angular deceleration = (final angular velocity - initial angular velocity) / time

Given that the initial angular speed is 88.0 rad/s, the final angular speed is half of that, and the time is 4.40 s, we can calculate:

angular deceleration = (44.0 rad/s - 88.0 rad/s) / 4.40 s = -10.0 rad/s^2

(b) To find the magnitude of the force of friction applied by the brake shoe, we can use the formula:

force of friction = moment of inertia * angular deceleration

Given that the moment of inertia is 0.850 kg·m², and the angular deceleration is -10.0 rad/s², we can calculate:

force of friction = 0.850 kg·m² * -10.0 rad/s² = -8.50 N

Answer:

diameter is decreased by factor 0.91

Explanation:

As we know by the equation of trajectory

[tex]y = xtan\theta - \frac{gx^2}{2v^2cos^2\theta}[/tex]

here as per first given situation we know that

x = 36.9 m

y = 8.85 m

[tex]\theta = 45^0[/tex]

now from above equation we have

[tex]8.85 = 36.9 tan45 - \frac{(9.8)(36.9)^2}{2(v^2)cos^245}[/tex]

[tex]8.85 = 36.9 - \frac{13343.8}{v^2}[/tex]

[tex]\frac{13343.8}{v^2} = 28.05[/tex]

[tex]v = 21.8 m/s[/tex]

now similarly after nozzle is adjusted we have

y = 17.9 m

x = 36.9 m

[tex]\theta = 45^0[/tex]

now again from equation we have

[tex]17.9 = 36.9 tan45 - \frac{(9.8)(36.9)^2}{2(v'^2)cos^245}[/tex]

[tex]17.9 = 36.9 - \frac{13343.8}{v'^2}[/tex]

[tex]\frac{13343.8}{v'^2} = 19[/tex]

[tex]v' = 26.5 m/s[/tex]

Now by equation of continuity we can find the change in diameter

as we know that

[tex]A_1v_1 = A_2v_2[/tex]

now we have

[tex]\pi d_1^2 v_1 = \pi d_2^2 v_2[/tex]

[tex]d_1^2 (21.8) = d_2^2(26.5)[/tex]

[tex]\frac{d_1}{d_2} = \sqrt{\frac{26.5}{21.8}}[/tex]

[tex]\frac{d_1}{d_2} = 1.10[/tex]

so we have

[tex]\frac{d_2}{d_1} = 0.91[/tex]

so diameter is decreased by factor of 0.91

The nozzle diameter changed by a factor of 0.907

[tex]\texttt{ }[/tex]

Acceleration is rate of change of velocity.

[tex]\large {\boxed {a = \frac{v - u}{t} } }[/tex]

[tex]\large {\boxed {d = \frac{v + u}{2}~t } }[/tex]

a = acceleration (m / s²)v = final velocity (m / s)

u = initial velocity (m / s)

t = time taken (s)

d = distance (m)

Let us now tackle the problem!

[tex]\texttt{ }[/tex]

Given:

horizontal distance = x = 36.9 m

angle of projection = θ = 45°

initial height = y₁ = 8.85 m

final height = y₂ = 17.9 m

Asked:

ratio of nozzle diameter = d₂ : d₁ = ?

Solution:

The motion of the water is a parabolic motion.

Firstly, we will calculate the time taken for the water to reach the hotspot:

[tex]x = (u \cos \theta) t[/tex]

[tex]t = x \div ( u \cos \theta )[/tex]

[tex]t = x \div ( u \cos 45^o )[/tex]

[tex]\boxed {t = \frac{\sqrt{2}x}{u}}[/tex]

[tex]\texttt{ }[/tex]

Next , we could calculate the initial speed (u) of the water as it leaves the nozzle:

[tex]y = (u \sin \theta) t - \frac{1}{2}gt^2[/tex]

[tex]y = (u \sin 45^o)( \frac{\sqrt{2}x}{u} ) - \frac{1}{2}g ( \frac{\sqrt{2}x}{u} )^2[/tex]

[tex]y = x - \frac{gx^2}{u^2}[/tex]

[tex]\frac{gx^2}{u^2} = x - y[/tex]

[tex]u^2 = \frac{gx^2}{x - y }[/tex]

[tex]u = \sqrt{ \frac{gx^2}{x - y } }[/tex]

[tex]\boxed {u = x \sqrt{ \frac{g}{x - y} }}[/tex]

[tex]\texttt{ }[/tex]

Finally , we could find the ratio of the diameter by using Continuity Equation as follows:

[tex]u_1 A_1 = u_2 A_2[/tex]

[tex]u_1 \frac{1}{4} \pi (d_1)^2 = u_2 \frac{1}{4} \pi (d_2)^2[/tex]

[tex](d_2)^2 : (d_1)^2 = u_1 : u_2[/tex]

[tex](d_2)^2 : (d_1)^2 = x \sqrt{ \frac{g}{x - y_1} } : x \sqrt{ \frac{g}{x - y_2} }[/tex]

[tex](d_2)^2 : (d_1)^2 = \sqrt { x - y_2 } : \sqrt { x - y_1}[/tex]

[tex]\frac {d_2}{d_1} = \sqrt[4] { \frac {x - y_2} {x - y_1} }[/tex]

[tex]\frac {d_2}{d_1} = \sqrt[4] { \frac {36.9 - 17.9} {36.9 - 8.85} }[/tex]

[tex]\frac {d_2}{d_1} \approx 0.907[/tex]

[tex]d_2 \approx 0.907 \times d_1[/tex]

[tex]\texttt{ }[/tex]

[tex]\texttt{ }[/tex]

Grade: High School

Subject: Physics

Chapter: Kinematics