Sound waves are Sound waves are
(a) transverse waves characterized by the displacement of air molecules.
(b) longitudinal waves characterized by the displacement of air molecules.
(c) longitudinal waves characterized by pressure differences.
(d) Both (b) and (c). (e) (a), (b) and (c).

Answers:

a) 13 s

b) 0.362 m/s

Explanation:

We have the following data:

[tex]m=660000 kg[/tex] is the mass of the mass damper

[tex]L=42 m[/tex] is the length of the pendulum

[tex]A=75 cm \frac{1m}{100 cm}=0.75 m[/tex] is the amplitude

This can be solved by the following equation:

[tex]T=2 \pi \sqrt{\frac{L}{g}}[/tex] (1)

Where:

[tex]T[/tex] is the period

[tex]g=9.8 m/s^{2}[/tex] is the acceleration due gravity

[tex]T=2 \pi \sqrt{\frac{42 m}{9.8 m/s^{2}}}[/tex] (2)

[tex]T=13 s[/tex] (3)

The velocity in a pendulum is maximum [tex]V_{max}[/tex] when the pendulum is in its mean position and the amplitude is maximum. So, the equation in this case is:

[tex]V_{max}=A \frac{2 \pi}{T}[/tex] (4)

[tex]V_{max}=0.75 m \frac{2 \pi}{13 s}[/tex] (5)

[tex]V_{max}=0.362 m/s[/tex]

Answer: 0.4911 kg

Explanation:

We have the following data:

[tex]\rho_{0\°C}= 730 kg/m^{3}[/tex] is the density of gasoline at [tex]0\°C[/tex]

[tex]\beta=9.60(10)^{-4} \°C^{-1}[/tex] is the average coefficient of volume expansion

We need to find the extra kilograms of gasoline.

So, firstly we need to transform the volume of gasoline from gallons to [tex]m^{3}[/tex]:

[tex]V=8.50 gal \frac{0.00380 m^{3}}{1 gal}=0.0323 m^{3}[/tex] (1)

Knowing density is given by: [tex]\rho=\frac{m}{V}[/tex], we can find the mass [tex]m_{1}[/tex] of 8.50 gallons:

[tex]m_{1}=\rho_{0\°C}V[/tex]

[tex]m_{1}=(730 kg/m^{3})(0.0323 m^{3})=23.579 kg[/tex] (2)

Now, we have to calculate the factor [tex]f[/tex] by which the volume of gasoline is increased with the temperature, which is given by:

[tex]f=(1+\beta(T_{f}-T_{o}))[/tex] (3)

Where [tex]T_{o}=0\°C[/tex] is the initial temperature and [tex]T_{f}=21.7\°C[/tex] is the final temperature.

[tex]f=(1+9.60(10)^{-4} \°C^{-1}(21.7\°C-0\°C))[/tex] (4)

[tex]f=1.020832[/tex] (5)

With this, we can calculate the density of gasoline at [tex]21.7\°C[/tex]:

[tex]\rho_{21.7\°C}=730 kg/m^{3} f=(730 kg/m^{3})(1.020832)[/tex]

[tex]\rho_{21.7\°C}=745.207 kg/m^{3}[/tex] (6)

Now we can calculate the mass of gasoline at this temperature:

[tex]m_{2}=\rho_{21.7\°C}V[/tex] (7)

[tex]m_{2}=(745.207 kg/m^{3})(0.0323 m^{3})[/tex] (8)

[tex]m_{2}=24.070 kg[/tex] (9)

And finally calculate the mass difference [tex]\Delta m[/tex]:

[tex]\Delta m=m_{2}-m_{1}=24.070 kg-23.579 kg[/tex] (10)

[tex]\Delta m=0.4911 kg[/tex] (11) This is the extra mass of gasoline

Final answer:

To find the extra kilograms of gasoline received when buying at 0°C instead of 21.7°C, we can use the formula for volume expansion and the density of gasoline.

Explanation:

To find the extra kilograms of gasoline received when buying at 0°C instead of 21.7°C, we need to calculate the difference in volume and then convert it to kilograms using the density of gasoline. The average coefficient of volume expansion is given as 9.60 x 10^-4 (°C)^-1. We can calculate the change in volume using the formula ΔV = V₀ * β * ΔT, where ΔV is the change in volume, V₀ is the initial volume, β is the coefficient of volume expansion, and ΔT is the temperature difference in Celsius. In this case, the initial volume is 0.00380 m³ and the temperature difference is 21.7 - 0 = 21.7 °C. Substituting the values, we get ΔV = 0.00380 * 9.60 x 10^-4 * 21.7. Now, to convert the change in volume to kilograms, we multiply it by the density of gasoline, which is 730 kg/m³. So, the extra kilograms of gasoline received is ΔV * 730.

Answer:

[tex]\frac{dm}{dt} = 10352 kg/s[/tex]

Explanation:

As we know that thermal energy given to the water to raise its temperature is given as

[tex]Q = ms\Delta T[/tex]

now rate of energy given to the system is

[tex]\frac{dQ}{dt} = \frac{dm}{dt} s \Delta T[/tex]

so we have

[tex]\frac{dQ}{dt} = 1.3 \times 10^8[/tex]

s = 4186 J/kg C

now we have

[tex]1.3 \times 10^8 = \frac{dm}{dt} (4186) 3[/tex]

[tex]\frac{dm}{dt} = 10352 kg/s[/tex]

With the given data, the mass of the warmed water is approximately 1.0×10⁴ kg each second.

To determine the mass of water that is warmed as it flows through the power plant, we can use the following steps:

Calculate the amount of heat transferred to the water:

The power station dissipates 1.3×10⁸ J of heat each second.

Use the specific heat capacity formula:

The formula to calculate heat transfer is

where:

Rearrange the formula to solve for mass (m):

Plugging in the values, we get:

Therefore, the mass of warmed water flowing through the plant each second is approximately 1.0×104 kg.

Answer:

maximum amplitude  = 0.13 m

Explanation:

Given that

Time period T= 0.74 s

acceleration of gravity g= 10 m/s²

We know that time period of simple harmonic motion given as

[tex]T=\dfrac{2\pi}{\omega}[/tex]

[tex]0.74=\dfrac{2\pi}{\omega}[/tex]

ω = 8.48 rad/s

ω=angular frequency

Lets take amplitude = A

The maximum acceleration given as

a= ω² A

The maximum acceleration should be equal to g ,then block does not separate

a= ω² A

10= 8.48² A

A=0.13 m

maximum amplitude  = 0.13 m

The maximum amplitude of the motion for which the block does not separate from the plate is : 0.14 m

Given data :

period ( T ) = 0.74 secs

acceleration due to gravity ( g ) = 9.8 m/s62

Time period of simple harmonic motion ( T ) = [tex]\frac{2\pi }{w}[/tex]

First step : solve for w

w = 2π / T

   = 2π / 0.74

   = 8.49 rad/sec

Next step : determine the maximum amplitude ( A )

a = w² A

where ;  a = maximum acceleration = 9.8 , w = 8.49

therefore

A = a / w²

   = 9.8 / ( 8.49 )²

   = 0.136  ≈ 0.14 m

Hence we can conclude that The maximum amplitude of the motion for which the block does not separate from the plate is : 0.14 m.

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Answer:

Glycosidic Bonds

Explanation:

When two adjacent mono-saccharide units link to form di-saccharides and this repeated chain link forms poly-saccharide. These reactions are called dehydration or condensation reactions because upon the formation of the bond the water molecule is released in the form of a byproduct. Glycosidic bonds are covalent bonds that link ring-shaped sugar molecules to other molecules.

Final answer:

Both Technician A and Technician B are correct.

Explanation:

Technician A is correct. Boring and honing the cylinders is done to achieve the proper size, shape, and finish required for the installation of the pistons. This ensures a precise fit between the cylinders and pistons, allowing for optimal performance and reduced wear.

Technician B is also correct. Chamfering the bolt holes helps to ensure smooth insertion of the bolts and prevents damage to the threads. Cleaning with a thread chaser removes any debris or contaminants that could compromise the integrity of the assembly.

Therefore, the correct answer is C) Both technicians A and B.

Answer:

h should become four times.

Explanation:

The speed at the bottom depends only on the vertical height difference from the top ... if there is no friction ... then the path you take from top to bottom doesn't matter in any way, only the difference in height between the two

mgh at top = 1/2 mv^2 at bottom, so that v=sqrt[2gh]

You see, the final velocity, v, depends on the square root of h, so double v, quadruple h, or four times higher

Final answer:

The acceleration of the crate on the frictionless ramp is 2.39 m/s², while the acceleration on the ramp with friction is 0.49 m/s².

Explanation:

To find the acceleration of the crate, we first need to determine the force component parallel to the incline. This can be done using the formula F_parallel = F * sin(theta), where F is the applied force and theta is the angle of the ramp. Substituting the values, we have F_parallel = 7.0 N * sin(20°) = 2.39 N. The acceleration can then be found using the equation a = F_parallel / m, where m is the mass of the crate. Substituting the values, we have a = 2.39 N / 1.0 kg = 2.39 m/s².

If the ramp has a friction force of 1.9 N, it will oppose the motion of the crate. In this case, we need to subtract the friction force from the parallel force to find the effective force. The effective force, Feff, is given by Feff = F_parallel - friction force. Substituting the values, we have Feff = 2.39 N - 1.9 N = 0.49 N. The acceleration can then be found using the equation a = Feff / m. Substituting the values, we have a = 0.49 N / 1.0 kg = 0.49 m/s².

Answer:

6.62 m

About 44.14 grapefruits lined in a single line would be the length of the diameter of star Aldebaran

Explanation:

First we know,

Diameter of the Sun - 1.39 million km (Ds)

Diameter of Aldbaran - 61.40 million km (Da)

Thus,

[tex]\frac{Da}{Ds}[/tex] =  [tex]\frac{61.40}{1.39}[/tex] ≈ 44.14

We know that in this scale the size of the sun is that of a grapefruit.

The average diameter of a grapefruit is about 15cm

Thus  Ds in this scale is 15cm

Thus Da in this scale,

Da = Ds *44.14

Da = 0.15 * 44.14 ≈ 6.62 m

Thus the diameter of Aldebaran in this scale is 6.62 m

Answer:

[tex]x = 6.7 cm[/tex]

Explanation:

Here we can use energy conservation for two positions of the gymnast

When she is at the lowest position of the trampoline then its potential energy will convert into gravitational potential energy at the top

So we will have

[tex]\frac{1}{2}kx^2 = mg(H + x)[/tex]

so we have

[tex]\frac{1}{2}k(0.61^2) = 51 \times 9.81(2.1 + 0.61)[/tex]

[tex]k = 7287.5 [/tex]

now when she stands on it

then by force balance we will have

[tex]mg = kx[/tex]

[tex]51 \times 9.81 = 7287.5 \times x[/tex]

[tex]x = 6.7 cm[/tex]

The size of the real heart is B) 11.7 cm

Explanation:

In lenses, mirror and other optical systems, the magnification factor is given by:

[tex]M=\frac{y'}{y}[/tex]

where

y' is the size of the image

y is the size of the real object

In this problem, we have:

M = 1.3 is the magnification factor

y' = 15.2 cm is the size of the image (the image of the heart)

Solving for y, we find the actual size of the heart:

[tex]y=\frac{y'}{M}=\frac{15.2 cm}{1.3}=11.7 cm[/tex]

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A molecule of glucose is an example of potential energy rather than kinetic energy.

Potential energy is a form of stored energy that is dependent on the relationship between different system components. Example are when a spring is compressed or stretched, its potential energy increases. If a steel ball is raised above the ground as opposed to falling to the ground, it has more potential energy.

The energy an object has as a result of motion is known as kinetic energy. A force must be applied to an object in order to accelerate it. We must put in effort in order to apply a force. After the work is finished, energy is transferred to the item, which then moves at a new, constant speed.

From the given option, crawling beetle foraging for food, light flashes emitted by firefly water, rushing over Niagara Falls and a molecule of glucose. Molecule of glucose have potential rather than kinetic energy

because there is no motion in the molecule of glucose.

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A molecule of glucose is an example of potential energy because it has energy stored within its chemical bonds waiting to be used, whereas the other examples are instances of kinetic energy in action.

Potential vs. Kinetic Energy

When discussing energy, it is vital to understand the difference between potential energy and kinetic energy. Potential energy is stored energy, waiting to be released, whereas kinetic energy is the energy of motion. An example of potential energy is a molecule of glucose, which contains energy in its chemical bonds that is not currently in motion but has the possibility to be used in the future. This contrasts with kinetic energy, which would include objects or substances that are actively moving or releasing energy, such as a crawling beetle, light flashes from a firefly, or water rushing over Niagara Falls.

To answer the student's question, the example representing potential energy rather than kinetic energy is:

The examples of a crawling beetle, light from a firefly, and rushing water over Niagara Falls all represent kinetic energy as they are all in motion and actively using or releasing energy.

Answer:

Velocity of ball1 after the collision is 0.35 m/s at 53.87° due south of east.

Explanation:

By conservation of the linear momentum:

[tex]m*V_{1o}+m*V_{2o}=m*V_{1f}+m*V_{2f}[/tex]  Since both masses are the same, and expressing the equation for each axis x,y:

[tex]V_{1ox}+V_{2ox}=V_{1fx}[/tex]      eqX

[tex]0=V_{1fx}+V_{2fx}[/tex]                 eqY

From eqX:  [tex]V_{1fx}=1m/s[/tex]

From eqY:  [tex]V_{1fy}=-1.37m/s[/tex]

The module is:

[tex]V_{1f}=\sqrt{1^2+(-1.37)^2}=0.35m/s[/tex]

The angle is:

[tex]\theta=atan(-1.37/1)=-53.87\°[/tex]   This is 53.87° due south of east

The total force on the spring is the sum of the weight of the 5-kg object and the additional force due to the elevator's upward acceleration. This force leads to a stretch in the spring, calculated using Hooke's law, of 0.064 m or 6.4 cm.

To solve this problem we'll use Hooke's law (F = -kx), where F is the force, k is the spring constant (in our case, 802 N/m), and x is the displacement or stretch of the spring.

Here, the force isn't just from the weight of the 5.0-kg object (mg), but also from the additional force due to the upward acceleration of the elevator (ma). By adding these two forces together, we can find the total force on the spring (F = mg + ma).

First, let's calculate the weight of the 5-kg object: F_weight = mg = 5.0 kg * 9.81 m/s^2 = 49.05 N.

Next, let's calculate the additional force due to the elevator's acceleration: F_acceleration = ma = 5.0 kg * 0.41 m/s^2 = 2.05 N.

Adding these two forces together: F_total = F_weight + F_acceleration = 49.05 N + 2.05 N = 51.1 N.

Now, we can use Hooke's law to find the stretch of the spring x = F_total / k = 51.1 N / 802 N/m = 0.064 m, or 6.4 cm when converted to centimeters.

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The spring stretches by approximately 0.020 meters relative to its unstrained length when the elevator is accelerating upward at 0.41 m/s^2.

The amount by which the spring stretches can be determined using Hooke's Law, which states that the force exerted by a spring is directly proportional to the displacement of the spring. In this case, the force exerted by the spring is equal to the weight of the object attached to it. The equation for this can be written as F = kx, where F is the force, k is the spring constant, and x is the displacement. Rearranging this equation, we can solve for x:

x = F / k

Plugging in the values given in the question, we have:

x = (m * a) / k

Substituting the values m = 5.0 kg, a = 0.41 m/s^2, and k = 802 N/m:

x = (5.0 kg * 0.41 m/s^2) / 802 N/m

x ≈ 0.020 m

Therefore, the spring stretches by approximately 0.020 meters relative to its unstrained length when the elevator is accelerating upward at 0.41 m/s^2.

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Meaning of geosynchronous satellite:

A satellite above Earth placed in a geosynchronous circular orbit, inclined at an angle with the equatorial plane of Earth and orbiting at an orbital period equal to that of earth’s rotational period is termed as a geosynchronous satellite. Geosynchronous orbit is a High Earth orbit at a distance of 42,164 kilometres from the centre of earth such that the gravitational pull from earth is fair enough for the object in orbital motion to match with the speed of earth’s rotational period (24 hrs)

As an observer from ground, a geosynchronous satellite appears to be at a stationary point, at any part of the day; since it's velocity synchronizes with the angular velocity that of earth. A geosynchronous satellite orbiting in an equatorial plane without any inclinations is termed as a geostationary satellite.  

Answer:

n=5, l=1, m(l) = -1, m(s)= + 1/2

Explanation:

Quantum number are used to describe the position and spin of an electron inside an atom. There are four types of quantum number for describing an electron inside an atom. They are: the principal quantum number, spin quantum number, magnetic quantum number and angular momentum quantum number.

(1).PRINCIPAL QUANTUM NUMBER: denoted by n, and has possible values of n= 1,2,3,4,.... IN HERE, n= 5

(2).ANGULAR MOMENTUM QUANTUM NUMBER: it is denoted by l, and has possible values of l= 0,1,2,3,...,(n-1).

Our l here is one( that is, s-orbital=0, p-orbital=1, d-orbital= 3 and so on)

(3).MAGNETIC QUANTUM NUMBER: The magnetic quantum number, which is denoted by m subscribt l, specifies the exact orbital in which you can find the electron. It has values ranging from -l,...,-1,0,1,...,l.

Here, our value is -1 that is m(l)= -1

(4).SPIN QUANTUM NUMBER: describes the orientation of electrons. Electrons can only have two values here, either a positive one and the half(+1/2) that is the spin up electron or the negative one and half(-1/2) that is the spin down electron.

A set of quantum numbers for an electron in a 5p orbital could be: principal quantum number (n) = 5, angular momentum quantum number (l) = 1, magnetic quantum number (m) = 0, spin quantum number (m_s) = -1/2.

An electron in a 5p orbital can be described using four quantum numbers. Knowing that p orbitals correspond to the angular momentum quantum number l=1, and the principal quantum number n for the fifth energy level is 5, we can deduce the following. The magnetic quantum number, m, ranges from -l to +l, and thus it can have the values -1, 0, or +1 representing different p orbitals in the same shell. The spin quantum number, m_s, can either be -1/2 or +1/2, indicating the orientation of the electron's spin. Thus, an example of a set of four quantum numbers that could be assigned to an electron occupying a 5p orbital could be: (5, 1, 0, -1/2).

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Answer:

Explained

Explanation:

Potential difference in a battery across its terminals is the force that pushes the charge to move around the circuit. This means that work id neede to be done to move the charge. And potential difference is the work done per unit charge. The unit of potential difference is V  and defined as one joule per coulomb.

V= W/q J/C

Answer:

    [tex]x= 32.544[/tex]

Explanation:

given,

mean weight of bag (μ) = 32

standard deviation (σ) = 0.32

percentage of bag heavier = 4.5%

weight of the bag less than 4.5 % = 100 - 4.5

                                                        = 95.5%

we have to determine the z- value according to 95.5% or 0.955

using z-table

     z-value = 1.70

now, using formula

       [tex]Z = \dfrac{x-\mu}{\sigma}[/tex]

       [tex]1.70 = \dfrac{x-32}{0.32}[/tex]

       [tex]x-32 = 1.70\times {0.32}[/tex]

      [tex]x= 32.544[/tex]

The most a bag of baby carrots can weigh and not need to be repackaged is approximately 32.53 ounces.

To determine the most a bag of baby carrots can weigh and not need to be repackaged, we need to find the weight corresponding to the upper 4.5% of the normal distribution.

The weights are normally distributed with a mean of 32 ounces and a standard deviation (σ) of 0.32 ounces.

We first find the z-score that corresponds to the top 4.5% of the distribution.

Using a z-table or calculator,

Next, we use the z-score formula:

Solving for X (the weight we need):

Multiplying both sides by 0.32:

Adding 32 to both sides:

Thus, the most a bag of baby carrots can weigh and not need to be repackaged is approximately 32.53 ounces.

Answer:

[tex]\Delta T = 52.6 ^o C[/tex]

Explanation:

As we know that radius of the brass ring is given as

[tex]R_{brass} = 1.0000 cm[/tex]

radius of the sphere is given as

[tex]R_{sphere} = 1.0010 cm[/tex]

now by thermal expansion formula we know that

[tex]L = L_o(1 + \alpha \Delta T)[/tex]

so we will have

[tex]1.0010 = 1.0000(1 + (19\times 10^{-6})\Delta T)[/tex]

so we have

[tex]\Delta T = 52.6 ^o C[/tex]

Answer:

b) a horse pulls a wagon at a constant velocity

Explanation:

As we know that ,work done is the dot product of force vector and displacement vector.

W= F.d

W=work

F=Force

d=Displacement

We also know that

F = m a

m= mass ,a = acceleration

When velocity is constant then rate of change in the velocity will be zero,then we can say that acceleration will be zero.

When a= 0  Then F= 0

W= F.d  ( F=0)

W = 0

Therefore option b i correct because horse is going with constant velocity.

Zero net work is done when a bunch of bananas is placed on a spring scale in the supermarket, as there is no displacement of the bananas despite the force exerted by the scale.

The situation in which zero net work is done is when a bunch of bananas is placed on a spring scale in the supermarket. In this scenario, the force exerted by the scale is upward and counteracts the weight of the bananas, but since there is no displacement or movement of the bananas, the work done is zero. The concept of work in physics is defined as the product of the force applied to an object and the displacement of the object in the direction of the force. Since the bananas don't move, the displacement is zero, resulting in zero work.

In the other scenarios provided:

Answer:

The Ring of Fire

Explanation:

The ring of fire is also called the Circum-Pacific Belt, it is a path along the pacific ocean consisting of active volcanoes and frequent earthquakes.

It has a length of approximately 40,000 kilometers. It lies on the edge of tectonic plates where the in-earth vibrations and geothermal energies are prone to erupt out.

Ring of fire inhibits about 75% o the earth's volcanoes and 95% of earthquakes occur in this region.

Answer:

a) Space = [tex] 6.05 x 10^{-3} m = 0.605 cm [/tex]

b) Stress= [tex] -100.8 x 10^{6} Pa [/tex]

Explanation:

1) Data Given

[tex] L = 12 m , T_i = -9 C \degree, T_f = 33 C \degree [/tex]

2) Calculate the space using Linear thermal expansion formula

We need to use Linear thermal expansion formula since the space created would be a change on 1 dimension, the increase of the temperature will increase the length of the steel.  The formula is given by:

[tex] \Delta L = L_i \alpha_{steel} \Delta T [/tex]

We have everything except the [tex] \alpha_{steel} [/tex] , so we look for this on a book and we find that [tex] \alpha_{steel} = 1.2 x 10^{-5} C^{-1} [/tex], so we can replace.

[tex] \Delta L = 12 m (1.2 x 10^{-5} C^{-1}) (33 C \degree -(-9 C \degree)) = 12 m (1.2 x 10^{-5} C^{-1}) 42 C \degree =6.048 x 10^{-3} m = 0.6048cm [/tex]

3) Calculate the stress of the steel

The Stress is the ratio of applied force F to a cross section area - defined as

[tex] \sigma = \frac{F_n}{A} [/tex]

Since we don't have the force and the Area, we need to look for another way to find the stress.

For this we can use the concept called Young's Modulus, defined as : "the mechanical property that measures the stiffness of a solid material", and the formula for this is given by:

[tex] Y =\frac{F L}{A \Delta L} [/tex] (1)

Solving [tex] \frac{F}{A} [/tex] from the previous formula we have this:

[tex] \frac{F}{A} [/tex]  = (Y  Δ L)/L  (2)

From the Linear thermal expansion formula we can solve like this

[tex] \frac{\Delta L}{L} [/tex] =  α  ΔT  (3)

And replacing equation (3) into equation (2) we have:

[tex] \frac{F}{A} [/tex]  = Y α ΔT (4)

We have that the Young's Modulus for the steel is 20x10^{10} Pa, so replacing into equation (4)

[tex] \frac{F}{A} [/tex] = [tex] 20x10^{10} [/tex] Pa (1.2x10^-5 C^-1) (42C) = [tex] 100.8 *10^{6} [/tex] Pa  

That represent the absolute value for the Stress, the sign on this case would be negative since there is a compression.

The rail segment's length change due to thermal expansion can be calculated using the formula ∆L = αL0∆T, which gives the space to be left between rails. When the rails are constrained, thermal stress can be determined using the formula σ= Eα∆T, demonstrating the stress in the steel rails during the summer.

To solve this physics problem, one must first calculate the change in length of the steel rails due to thermal expansion, which introduces the concept of thermal stress—a stress created through thermal expansion or contraction of materials. The formula used is ∆L = αL0∆T, where ∆L is the change in length, α is the coefficient of linear expansion for steel (around 12 × 10⁻⁶ °C⁻¹), L0 is the initial length of the steel rails (12 meters), and ∆T is the change in temperature (33 - (-9) = 42 degrees Celsius). The calculation will yield the amount of space to be left between the rails, providing the answer to the first part of the question.

Next, to determine the stress in the rails when no gap is left for expansion (assuming they are constrained), the formula is σ= Eα∆T, where E represents Young's modulus (200 × 10⁹ N/m² for steel), and the other variables remain as earlier defined. After calculation, the result will exhibit the thermal stress exerted on the steel rails during the summer. Here, Young's modulus reflects the relationship between stress and strain in the material, indicating how much deformation will happen within the steel rails due to thermal stress.

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Answer:

A)

<17, - 1, 0>

B)

0 Ns

C)

<17, - 1, 0>

D)

<4, - 5, 0>

Explanation:

A)

[tex]p_{A,i}[/tex] = initial momentum of object A = 15 i - 8 j + 0 k

[tex]p_{B,i}[/tex] = initial momentum of object B = 2 i + 7 j + 0 k

Total initial momentum of the system is given as the sum of initial momenta of A and B , hence

[tex]p_{i}[/tex] = [tex]p_{A,i}[/tex] + [tex]p_{B,i}[/tex]

[tex]p_{i}[/tex] = (15 i - 8 j + 0 k) + (2 i + 7 j + 0 k)

[tex]p_{i}[/tex] = 17 i - j + 0 k

B)

[tex]F_{ext}[/tex] = Net external force on the two objects = 0 N

[tex]t[/tex] = duration of the collision

[tex]I[/tex] = Impulse

Impulse is given as

[tex]I = F_{ext} t \\I = (0) t \\I = 0 Ns[/tex]

C)

[tex]p_{f}[/tex] = final momentum of the system

we know that the Impulse is nothing but change in momentum of the system of objects, hence

[tex]I = p_{f} - p_{i}\\0 = p_{f} - p_{i}\\p_{f} = p_{i}[/tex]

[tex]p_{f}[/tex] = 17 i - j + 0 k

D)

[tex]p_{A,f}[/tex] = final momentum of object A = 13 i + 4 j + 0 k

[tex]p_{B,f}[/tex] = final momentum of object B = ?

Total final momentum of the system is given as

[tex]p_{f} = p_{A,f} + p_{B,f}[/tex]

17 i - j + 0 k = ( 13 i + 4 j + 0 k ) + [tex]p_{B,f}[/tex]

[tex]p_{B,f}[/tex] = 4 i - 5 j + 0 k

So final momentum of the object is <4, - 5, 0>

The total initial momentum of this system just before the collision is 17 i - j + 0 k.

This is defined as a term which quantifies the overall effect of a force acting over time.

Initial momentum of A = 15 i - 8 j + 0 k

initial momentum of B = 2 i + 7 j + 0 k

Total = sum of initial momentum of A and B

= (15 i - 8 j + 0 k) + (2 i + 7 j + 0 k)

= 17 i - j + 0 k

Impulse = Ft where F is force and t is time

Net external force on the two objects = 0 N

Impulse = (0)t

              = 0Ns

Impulse = change in momentum

= 17 i - j + 0 k

Final momentum of object A = 13 i + 4 j + 0 k

Final momentum of object B = ?

Total final momentum = Final momentum of object A + Final momentum of object B

17 i - j + 0 k = ( 13 i + 4 j + 0 k ) + Final momentum of object B

= 4 i - 5 j + 0 k

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Answer:

E = 2.6411 J

Explanation:

we know that elastic potential energy could be calculated by:

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

where k is the constant of the spring and x is the deform of the spring.

Now x can be calculated as:

x = 3.52m-2.54m

x = 0.98 m

Finally, replacing the data, we get:

[tex]E = \frac{1}{2}(5.5 N/m)(0.98m)^2[/tex]

E = 2.6411 J

Answer:

[tex]B = 1.37 \times 10^{-10} T[/tex]

Explanation:

As we know that in electromagnetic waves the two fields will induce each other

So here we have

[tex]B = \frac{E}{c}[/tex]

we have

[tex]E = 0.041 V/m[/tex]

[tex]c = 3\times 10^8 m/s[/tex]

so we have

[tex]B = \frac{0.041}{3 \times 10^8}[/tex]

[tex]B = 1.37 \times 10^{-10} T[/tex]

Answer:

The rise in temperature is [tex]43.98^{\circ}C[/tex]

Solution:

As per the question:

Mass of hammer, M = 1.30 kg

Speed of hammer, v = 7.3 m/s

Mass of iron, [tex]m_{i} = 14\ g[/tex]

No. of blows, n = 8

Specific heat of iron, [tex]C_{i} = 450\ J/kg.^{\circ}C = 0.45\ J/g^{\circ}C[/tex]

Now,

To calculate the temperature rise:

Transfer of energy in a blow = Change in the Kinetic energy

[tex]\Delta KE = \frac{1}{2}Mv^{2} = \frac{1}{2}\times 1.30\times 7.3^{2} = 34.64\ J[/tex]

For 8 such blows:

[tex]\Delta KE = n\Delta KE = 8\times 34.64\ = 277.12 J[/tex]

Now, we know that:

[tex]Q = m_{i}C_{i}\Delta T[/tex]

[tex]\Delta T= \frac{\Delta KE}{m_{i}C_{i}}[/tex]

[tex]\Delta T= \frac{277.12}{14\times 0.45} = 43.98^{\circ}C[/tex]

The equation for the speed of a satellite in a circular orbit around the Earth depends on mass. Which mass?

a. The mass of the sun

b. The mass of the satellite

c. The mass of the Earth

The equation for the speed of a satellite orbiting in a circular path around the earth depends upon the mass of Earth.

Option c

Any particular body performing circular motion has a centripetal force in picture. In this case of a satellite revolving in a circular orbit around the earth, the necessary centripetal force is provided by the gravitational force between the satellite and earth. Hence [tex]F_{G} = F_{C}[/tex].

Gravitational force between Earth and Satellite: [tex]F_{G} = \frac{G \times M_e \times M_s}{R^2}[/tex]

Centripetal force of Satellite :[tex]F_C = \frac{M_s \times V^2}{R}[/tex]

Where G = Gravitational Constant

[tex]M_e[/tex]= Mass of Earth

[tex]M_s[/tex]= Mass of satellite

R= Radius of satellite’s circular orbit

V = Speed of satellite

Equating  [tex]F_G = F_C[/tex], we get  

Speed of Satellite [tex]V =\frac{\sqrt{G \times M_e}}{R}[/tex]

Thus the speed of satellite depends only on the mass of Earth.

The two ladybugs have same rotational (angular) speed

Explanation:

The rotational (angular) speed of an object in circular motion is defined as:

[tex]\omega=\frac{\theta}{t}[/tex]

where

[tex]\theta[/tex] is the angular displacement

t is the time interval considered

Here we have two ladybugs, which are located at two different distances from the axis. In particular, ladybug 1 is halfway between ladybug 2 and the axis of rotation. However, since they rotate together with the disk, and the disk is a rigid body, every point of the disk cover the same angle [tex]\theta[/tex] in the same time [tex]t[/tex]: this means that every point along the disk has the same angular speed, and therefore the two ladybugs also have the same angular speed.

On the other hand, the linear speed of the two ladybugs is different, because it follows the equation:

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

where r is the distance from the axis: and since the two ladybugs are located at different [tex]r[/tex], they have different linear speed.

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The rotational speed of ladybug 1 is slower than that of ladybug 2.

When a rotating disk slows down at a constant rate, the angular velocity decreases over time. In this scenario, the ladybugs are at rest with respect to the surface of the disk, which means they are not experiencing any relative motion with respect to the disk.

Since ladybug 1 is halfway between ladybug 2 and the axis of rotation, it is closer to the center of the disk than ladybug 2. As the disk slows down, the distance from the axis of rotation to ladybug 1 decreases at a faster rate compared to the distance from the axis of rotation to ladybug 2.

As the distance from the axis of rotation decreases for ladybug 1 at a faster rate than for ladybug 2, ladybug 1 needs to decrease its rotational speed more to maintain the constant product of angular velocity and moment of inertia.

Thus, ladybug 1 has a slower rotational speed compared to ladybug 2.

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The acceleration at the bottom of the loop is 34m/[tex]\bold{s^2}[/tex]

Explanation:

Given:  

Mass=52kg

Force=1750N

To calculate:

The acceleration at the bottom of the loop

Formula:

Force=Mass x Acceleration

1750=52 x Acceleration

1750/52=Acceleration

Therefore acceleration at the bottom of the loop is 34m/[tex]s^2[/tex]

Roller coasters are mainly based upon acceleration theory they have two types of acceleration one is at the top of the loop and the other is at the bottom of the loop.

Then the net forces and the values are given. In many problems the roller coaster concept is included and it gives another level of clarity to the problems including the net forces

Answer:

0.929%

Explanation:

[tex]m[/tex] = mass of water evaporated

[tex]M[/tex] = mass of the lady

[tex]c[/tex] = specific heat of human body = 4200 JK⁻¹kg⁻¹

[tex]L[/tex] = Latent heat of evaporation of water = 2260000 Jkg⁻¹

[tex]\Delta T[/tex] = Drop in temperature = 5 C

Using conservation of heat

[tex]m L = Mc \Delta T \\m (2260000) = M (4200) (5)\\m (2260000) = M (21000)\\m = 0.00929 M \\\frac{m}{M} = 0.00929\\\frac{m(100)}{M} = (0.00929) (100)\\\\\frac{m(100)}{M} = 0.929[/tex]