Your companion on a train ride through Illinois notices that telephone poles near the tracks appear to be passing by very quickly, while telephone poles in the distance are passing by much more slowly. This is an example of

A) Angular speed: 0.42 rev/s

B) Frequency: 0.42 Hz

C) Tangential speed: 26.4 cm/s

D) Distance travelled: 528 cm

Explanation:

A)

In this problem, the ladybug is rotating together with the record.

The angular velocity of the ladybug, which is defined as the rate of change of the angular position of the ladybug, in this problem is

[tex]\omega = 25 rpm[/tex]

where here it is measured in revolutions per minute.

Keeping in mind that

1 minute = 60 seconds

We can rewrite the angular speed in revolutions per second:

[tex]\omega = 25 \frac{rev}{min} \cdot \frac{1}{60 s/min}=0.42 rev/s[/tex]

B)

The relationship between angular speed and frequency of revolution for a rotational motion is given by the equation

[tex]\omega = 2 \pi f[/tex] (1)

where

[tex]\omega[/tex] is the angular speed

f is the frequency of revolution

For the ladybug in this problem,

[tex]\omega=0.42 rev/s[/tex]

Keeping in mind that [tex]1 rev = 2\pi rad[/tex], the angular speed can be rewritten as

[tex]\omega = 0.42 \frac{rev}{s} \cdot 2\pi = 2\pi \cdot 0.42[/tex]

And re-arranginf eq.(1), we can find the frequency:

[tex]f=\frac{\omega}{2\pi}=\frac{(2\pi)0.42}{2\pi}=0.42 Hz[/tex]

And the frequency is the number of complete revolutions made per second.

C)

For an object in circular motion, the tangential speed is related to the angular speed by the equation

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

where

[tex]\omega[/tex] is the angular speed

v is the tangential speed

r is the distance of the object from the axis of rotation

For the ladybug here,

[tex]\omega = 2\pi \cdot 0.42 rad/s[/tex] is the angular speed

r = 10 cm = 0.10 m is the distance from the center of the record

So, its tangential speed is

[tex]v=(2\pi \cdot 0.42)(0.10)=0.264 m/s = 26.4 cm/s[/tex]

D)

The tangential speed of the ladybug in this motion is constant (because the angular speed is also constant), so we can find the distance travelled using the equation for uniform motion:

[tex]d=vt[/tex]

where

v is the tangential speed

t is the time elapsed

Here we have:

v = 26.4 cm/s (tangential speed)

t = 20 s

Therefoe, the distance covered by the ladybug is

[tex]d=(26.4)(20)=528 cm[/tex]

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Explanation:

We have equation of motion v = u + at

     Initial velocity, u = 33 m/s

     Final velocity, v = 29 m/s    

     Time, t = 2.50 s

     Substituting

                      v = u + at  

                      29 = 33 + a x 2.50

                      a = -1.6 m/s²

We have equation of motion v² = u² + 2as

Initial velocity, u = 33 m/s  

Acceleration, a = -1.6 m/s²  

Final velocity, v = 29 m/s  

Substituting  

v² = u² + 2as

29² = 33² + 2 x -1.6 x s

s = 77.5 m  

Distance traveled during that time is 77.5 m  

Final answer:

The car will travel 77.5 meters while decelerating from 33.0 m/s to 29.0 m/s over a period of 2.50 seconds by using the average speed during the deceleration period.

Explanation:

The question deals with calculating the distance covered by a car while decelerating from 33.0 m/s to 29.0 m/s over 2.50 seconds. We can solve this by finding the car's average speed during this period and multiplying it by the time duration. Here's how we can solve it:

Step-by-Step Calculation

Find the average speed: (Initial speed + Final speed) / 2 = (33.0 m/s + 29.0 m/s) / 2 = 31.0 m/s.

Multiply the average speed by the time taken to decelerate to find the distance covered: Distance = Average speed * Time = 31.0 m/s * 2.50 s = 77.5 meters.

Therefore, the car will travel 77.5 meters during that time.

Answer:

The movement of the plates sometimes leads to earthquakes and tsunamis.

The movement of the plates leads to lot of changes across the globe, and they might be positive or negative, though it depends from which point of view the things are seen.

On the boundaries of the plates, there's constant adjustment deep inside the crust because of their movement. Those adjustments cause parts inside the crust to break, crumble, producing a lot of force, transmitted as quick, strong vibrations, which we know as earthquakes.

The earthquakes can appear on land and in the sea/ocean. Both of them can be devastating and cause a lot of damage, be it human lives, destruction of plants, killing off animals, material damage. The ones in the seas'oceans, also lead to the formation of tsunami, which essentially is a huge, powerful, quick moving  waves that wipe out everything on their way.

Explanation:

Answer:

a) [tex]h_m=74625\ m[/tex]

b) [tex]t=265.55\ s[/tex]

Explanation:

Given:

During the phase of fuel exhaustion:

velocity attained by the rocket just as the fuel ends:

[tex]v_f=u+a.t_f[/tex]

where:

[tex]u=[/tex] initial velocity of the rocket = 0

[tex]v_f=0+30\times 35[/tex]

[tex]v_f=1050\ m.s^{-1}[/tex] this will be the initial velocity for the phase of ascending of the rocket's height under the influence of gravity.

height at which the fuel finishes:

[tex]v_f^2=u^2+2a.h_f[/tex]

[tex]1050^2=0^2+2\times 30\times h_f[/tex]

[tex]h_f=18375\ m[/tex]

During the phase of ascend in height of rocket after the fuel is over:

Time taken to reach the top height after the fuel is over:

[tex]v=v_f+g.t'[/tex]

at top v = (final velocity during this course of motion )= 0 [tex]m.s^{-1}[/tex]

[tex]0=1050-9.8\times t'[/tex]

[tex]t'=107.1429\ s[/tex]

Height ascended by the rocket after the fuel is over:

[tex]v^2=v_f^2+2g.h'[/tex]

at the top height the velocity is zero

[tex]0^2=1050^2-2\times 9.8\times h'[/tex] (-ve sign denotes that the direction of motion is opposite to that of acceleration)

[tex]h'=56250\ m[/tex]

Therefore the maximum altitude attained by the rocket:

[tex]h_m=h_f+h'[/tex]

[tex]h_m=18375+56250[/tex]

[tex]h_m=74625\ m[/tex]

b)

time taken by the rocket to fall back to the earth:

[tex]h_m=v.t_m+\frac{1}{2} g.t_m^2[/tex]

where:

[tex]v=[/tex] initial velocity of the rocket during the course of free fall from the top height.

[tex]74625=0+4.9\times t_m^2[/tex]

[tex]t_m=123.41\ s[/tex]

Now the total time for which the rocket is in the air:

[tex]t=t_f+t'+t_m[/tex]

[tex]t=35+107.1429+123.41[/tex]

[tex]t=265.55\ s[/tex]

Answer: Option (2) is the correct answer.

Explanation:

It is known that tap water acts as an electrolyte because there are a number of ions present in it. As electricity is the flow of ions or electrons therefore, a solution which contains the ions is able to conduct electricity.

For example, ions present in tap water are [tex]Mg^{2+}[/tex], [tex]Na^{+}[/tex], [tex]Ca^{2+}[/tex] etc.

Thus, we can conclude that tap water conducts electricity because tap water has many ions dissolved in it, which are responsible for the conduction of electricity.

Answer:

It is called the elementary charge

Explanation:

Charge is quantized; it comes in integer multiples of individual small units called the elementary charge, e, about 1.602×10−19 coulombs, which is the smallest charge which can exist freely

Answer:

range becomes 4 times

Explanation:

We know that the range of a projectile is given as:

[tex]R=\frac{u^2.\sin(2\theta)}{g}[/tex]

where:

[tex]R=[/tex] range of the projectile

[tex]u=[/tex] initial velocity of projectile

[tex]\theta=[/tex] initial angle of projection form the horizontal

g = acceleration due to gravity

When the initial velocity of launch is doubled:

[tex]R'=\frac{(2u)^2.\sin(2\theta)}{g}[/tex]

[tex]R'=\frac{4u^2.\sin(2\theta)}{g}[/tex]

[tex]R'=4R[/tex]

range becomes 4 times

Doubling the initial launch speed of a projectile, while keeping the angle of launch fixed, results in the range being quadrupled.

The range R of a projectile motion is given by the formula R = ((v^2)*sin(2*theta))/g, where v is the initial launch speed, theta is the launch angle, and g is the acceleration due to gravity. If the initial launch speed v is doubled, the new range R' would be R' = ((2v)^2)*sin(2*theta))/g, which simplifies to R' = 4R.

Thus, if you double the initial launch speed, the range of the projectile is quadrupled, assuming the launch angle is fixed.

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Answer:

Both balls struck the ground at the same time

Explanation:

At the moment when both balls fall from the top of the tower, they will reach the ground at the same time, as speed is not a function of mass. All the kinematics equations deduced by Isaac Newton, in his studies show that the velocity of a body in free fall is not in function of its mass.

The only way a body falls faster than another is when one of these bodies has a geometry in which air influences its velocity of fall. For example, when comparing the fall of a steel ball with a feather. But in the same way in modern experiments in vacuum chambers with limited air presence, it has been shown that a ball of steel and a Feather Fall at the same speed. Thus, the mass of a body does not influence its rate of fall.

Answer:

The unit of measurement that defines the space available in a rack are  the inches for the width and for the height the units of Rack (U), this unit is equivalent to 44.45 mm.  Standard racks are 19 inches wide and 42U high. A rack is a type of shelf where servers can be stacked on top of each other.

The unit of measurement that defines rack space is 'rack unit' or 'U', with each unit equal to 1.75 inches or 44.45 millimeters. Standard racks can have different heights, with 42U being the most common.

The unit of measurement that defines the space available in a rack is called 'rack unit' or 'U'. Each rack unit is equal to 1.75 inches or 44.45 millimeters. So, when we talk about the height of a standard rack, we refer to the number of rack units it can accommodate.

The height of standard racks can vary, but the most common height for a rack is 42U, which means it can accommodate equipment up to 73.5 inches or 1866.9 millimeters tall. Other common rack heights include 48U and 45U.

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Answer:

107401 N/m²

Explanation:

Pressure inside the hose = (ρ × h × g) + atmospheric pressure = (1000kg/m³ × 0.62m × 9.8m/s² ) + 101325 N/m² = 107401 N/m²

where ρ = density of water = 1000kg/m³, g = acceleration due to gravity = 9.8 m/s²

The pressure inside the hose is calculated as an additional fluid pressure due to the height of the water column leaking out. With given information, it is computed to be around 6076 Pa. This value is added to the atmospheric pressure to get the absolute pressure inside the hose, which is approximately 107.4 kPa.

The question is about determining the pressure inside the hose when the nozzle is turned off but the water faucet is turned on. We should consider this situation as a case of fluid statics as the water is not flowing due to the closed nozzle. In fluid statics, the pressure in a fluid column, like the water in the hose, is given by the equation P = ρgh, where P is the pressure, ρ is the fluid density, g is the gravitational acceleration, and h is the height of the fluid column above the point in question. Here, ρ is the density of water (about 1000 kg/m³), g is the acceleration due to gravity (about 9.8 m/s²), and h is the height to which the water leaks out, which is 0.62 m.

So, we can calculate the pressure inside the hose as P = (1000 kg/m³)(9.8 m/s²)(0.62 m) = 6076 Pa. This is the additional pressure due to the water column inside the hose. To get the absolute pressure inside the hose, we should add this value to the atmospheric pressure (which is about 101325 Pa at sea level). Therefore, the pressure inside the hose is approximately 101325 Pa + 6076 Pa = 107401 Pa or about 107.4 kPa.

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Answer:

Explanation:

Given

magnitude of centripetal acceleration is twice the magnitude of tangential acceleration

Suppose [tex]\theta [/tex] is theta angle rotated by electric drill

it is given that it starts from rest i.e. [tex]\omega _0=0[/tex]

suppose [tex]\omega [/tex] and [tex]\alpha [/tex] is the final angular velocity and angular acceleration

using rotational motion equation

[tex]\omega ^2-\omega _0^2=2\times \alpha \times \theta [/tex]

where [tex]\theta [/tex]=angle turned by drill

[tex]\omega _0[/tex]=initial angular velocity

[tex]\omega [/tex]=final angular velocity

[tex]\alpha [/tex]=angular acceleration

[tex]\omega ^2-0=2\times \alpha \times \theta [/tex]

[tex]\omega ^2=2\alpha \theta ---1[/tex]

It is also given that centripetal acceleration is twice the magnitude of tangential i.e.

[tex]\omega ^2r=\alpha \times r[/tex]

where r=radial distance of any point from axis of drill

i.e. [tex]\omega ^2=\alpha [/tex]

substitute this value to equation 1

we get

[tex]\theta =\frac{\omega ^2}{2\alpha }[/tex]

[tex]\theta =1\ rad[/tex]

Answer:

a) 24.0 N.m b) 3.6*10⁻² rad/s² c) 1.07 m/s²

Explanation:

a) If the force that produces the torque is perpendicular to the tethering wire, we can determine its magnitude just as follows:

τ = F*r = 0.800 N * 30.0 m = 24.0 N*m (1)

b)  We can express the torque we found above, using the rotational form of Newton´s 2nd Law, as follows:

τ = I* α (2)

where I is the rotational inertia regarding an axis passing through the center of the circle and α is the angular acceleration of the airplane.

If we consider the airplane as a point mass, the rotational inertia I can be calculated as follows:

I = m*r² = 0.750 Kg * (30.0)² m² = 675 Kg*m²

From (1) and (2), we can solve for α, as follows:

[tex]\alpha = \frac{T}{I} = \frac{24.0 N*m}{675.0 kg*m2} = (3.6e-2) rad/s2[/tex]

⇒α = 3.6*10⁻² rad/s²

c) Applying the definition of the angular velocity, and the definition of an angle, we can find the following realtionship between the linear and angular velocity:

v = ω*r

Dividing both sides by Δt, we can extend this relationship to the linear and angular acceleration, as follows:

a = α*r  

a = 3.6*10⁻² rad/s²* 30.0 m = 1.07 m/s²

a. The torque that the net thrust produces about the center of the circle is 24 Newton.

b. The angular acceleration of the airplane is equal to [tex]0.036 \;rad/s^2[/tex]

c. The linear acceleration of the airplane tangent to its flight path is[tex]1.08 \;m/s^2[/tex]

Given the following data:

a. To find the torque that the net thrust produces about the center of the circle:

Mathematically, the torque produced by a perpendicular force is given by:

[tex]T = Fr[/tex]

Where:

Substituting the given parameters into the formula, we have;

[tex]T = 0.8 \times 30[/tex]

Torque, T = 24 Newton

b. To find the angular acceleration of the airplane, we would use Newton's Second Law of rotational motion:

[tex]T = I\alpha[/tex]

But, [tex]I = mr^2[/tex]

[tex]I = 0.750 \times 30^2\\\\I = 0.750 \times 900[/tex]

Moment of inertia, I =  [tex]675\;Kgm^2[/tex]

[tex]\alpha = \frac{T}{I} \\\\\alpha = \frac{24}{675}\\\\\alpha = 0.036 \;rad/s^2[/tex]

c. To find the linear acceleration of the airplane tangent to its flight path:

[tex]a = r\alpha \\\\a = 30 \times 0.036[/tex]

Linear acceleration, a = [tex]1.08 \;m/s^2[/tex]

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C. added an electron to the outer electron shell.

Explanation:

Atoms consist of three particles:

- Protons: they are located in the nucleus, they have positive charge of [tex]+e[/tex], and mass of [tex]1.67\cdot 10^{-27}kg[/tex]

- Neutrons: they are also located in the nucleus, they have no electric charge, and mass similar to that of the proton

- Electrons: they orbit around the nucleus, they have negative charge of [tex]-e[/tex], and mass around 1800 smaller than the proton

Normally, atoms are neutral (no electric charge), because they have an equal number of protons and electrons.

However, sometimes atoms can give off or take electrons from other atoms. We have two cases:

Therefore, in order to form a negative ion, the atom must have

C. added an electron to the outer electron shell

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Answer:

Weathering

Rocks break down as they interact with air and water.

Hydro thermal Energy

Sea water heats up from contact with hot rock and metals dissolve in it.

Volcanic activity

Density differences and other factors cause metals to concentrate in igneous rocks.

Answer:

Explanation:

Answer:

Speed s= 19

Explanation:

Take note of the following parameters:

Speed= s,

Number of cars= N,

Number N of cars that can pass a given spot per minute=

N(s)=88s/16+16(s19)2

The principle of differentiation is  here:

We let N(s) = N

N = 86s / (17 + 17((s/19)^2))

17N = 86s /(1 + s²/19²)

17N = 361* 86s /(361 + s²)

17N = 31046s /(361 + s²)

Next step;

Differentiate with respect to s

17N = 31046s /(361 + s²)

Remember the quotient rule [u/v]’ = (vu’ - uv’) / v²

Therefore,

u = 31046s =====> du/ds = u’ = 31046

v = (361 + s²) ====>dv/ds = v’ = 2s

17N = 31046s /(361 + s²)

17 dN/ds = ( 31046(361 + s²) - 31046s(2s) ) / (361 + s²)²

The maximum when dN/ds = 0

17 dN/ds = ( 31046(361 + s²) - 31046s(2s) ) / (361 + s²)²

17 * 0 = ( 31046(361 + s²) - 31046s(2s) ) / (361 + s²)²

( 31046(361 + s²) - 31046s(2s) ) = 0

31046 [ (361 + s²) - 2s² ] = 0

31046 (361 - s²) = 0

(361 - s²) = 0

(19 - s)(19 + s) = 0

either s = -19 or s = 19, but s > 0

s = 19

The speed at which the greatest number of cars travel safely on that road is; s = 19

We are given the function to represent Number N of cars that can pass a given spot per minute as;

N(s) = 88s/(16 + 16(s/19)²)

where;

s is Speed

N is number of cars

Differentiating the function gives;

N' = -3971(s² - 361)/(2(s² + 361)²

Now, the speed at the greatest number of cars would be gotten when N' = 0. Thus;

-3971(s² - 361)/(2(s² + 361)² = 0

Cross multiply to get;

-3971(s² - 361) = 0

divide both sides by -3971 to get;

s² - 361 = 0

s = √361

s = 19

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Answer:

The time taken for a signal to reach the Earth after it is transmitted from the spacecraft is (500x) seconds

Explanation:

Distance of spacecraft from Earth = x AU = x × 149.9×10^6 Km = (149.9×10^6x) Km

Speed of light = 299,800Km/s

Time taken for a signal to reach the Earth after it is transmitted from the spacecraft = distance of spacecraft from Earth ÷ speed of light = (149.9×10^6x)Km ÷ 299,800Km/s = (500x) seconds

Answer:

Explanation:

compressibility = 1 / bulk modulus of elasticity ( B )

B = 1 / 4.8 x 10⁻⁵ = Δp / Δv /v ( Δp is change in pressure , Δv is change in volume )

1 / 4.8 x 10⁻⁵ = 429  / Δv /v

Δv /v = 429 x 4.8 x 10⁻⁵

= 2059.2 x 10⁻⁵

= .021

v = m / d ( d is density and m is mass of the water taken )

taking log and then differentiating

Δv /v  = - Δd / d

- .021 = - Δd / d

Δd =.021  x d

= .021 x 998

= 20.9

new density

= 998 + 20.9

1018.9 kg/m3

The increase in the density of water when it is compressed from 1 atm to 430 atm pressure isothermally is 20.6 kg/m³.

In this problem, we are asked to find the increase in the density of water when it is compressed from 1 atm to 430 atm pressure isothermally. The isothermal compressibility of water is given as 4.80 × 10−5 atm−1. The density of water at 1 atm pressure and 20°C is given as rho1 = 998 kg/m^3.

The formula that connects these quantities is Δρ = - ρ * β * ΔP, where Δρ is the change in density, β is the isothermal compressibility (in atm^−1), ΔP is the change in pressure (in atm), and ρ is the initial density (in kg/m^3).

Substituting the given values into the formula we get Δρ = - (998 kg/m³) * (4.80 × 10−5 atm−1) * (430 atm - 1 atm) = - ((998 kg/m³) * (4.80 × 10−5 atm−1) * (429 atm)) = -20.6 kg/m^3. The negative sign indicates an increase in density due to compression. So the increase in density of the water is 20.6 kg/m³.

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Explanation:

The force of a spring is described by Hooke's law:

F = kx

where k is the spring stiffness in N/m, and x is the displacement in m.

A spring force vs displacement graph is a line passing through the origin with a slope of k.

Final answer:

A spring force vs. displacement graph represents Hooke's law, showing the restorative force exerted by a spring as it is deformed. The slope of the straight line on the graph is the spring's force constant 'k'. The area under the line represents the work done on the spring.

Explanation:

Physical Meaning of Spring Force vs. Displacement

The physical meaning of a spring force vs. displacement graph is that it shows how the force exerted by a spring changes as the spring is stretched or compressed. This relationship is governed by Hooke's law, which states that the force (F) exerted by a spring is directly proportional to the displacement (x) from its equilibrium position, given by the equation F = -kx. Here, 'k' is the spring's force constant, representing its stiffness, with units of newtons per meter (N/m). The negative sign indicates that the force is restorative, meaning it acts in the opposite direction of displacement.

The graph typically portrays a straight line passing through the origin, with the slope equivalent to the spring's force constant 'k'. The steeper the slope, the greater the force constant, and vice versa. Additionally, the work done on the spring, which is equal to the energy stored in it, can be calculated from the area under the force versus displacement graph, given by W = (1/2)kx², where W is the work done and x is the displacement.

Galileo's work differed from his predecessors by relying on experimental evidence and mathematical analysis.

The work of Galileo differed from his predecessors in a fundamental way in terms of his approach to studying the sky. Unlike his predecessors who relied on pure reasoning and philosophical arguments, Galileo used experimental evidence and mathematical analysis to understand the natural phenomena.

For example, Galileo used a telescope to observe the moons of Jupiter, which challenged the prevailing geocentric model of the universe.

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Galileo used telescopes for detailed astronomical observations, challenging prevailing geocentric views, while predecessors relied on eye observations and theories.

Galileo's work differed fundamentally from his predecessors in astronomy by introducing a new method of observation. While his predecessors primarily relied on eye observations and abstract reasoning to understand celestial phenomena, Galileo pioneered the use of telescopes for systematic astronomical observations. By observing celestial objects through telescopes, he made groundbreaking discoveries such as the moons of Jupiter, the phases of Venus, and the mountains and craters on the Moon.

These observations provided concrete evidence that supported the heliocentric model proposed by Copernicus, challenging the geocentric worldview dominant at the time. Galileo's empirical approach and emphasis on experimental evidence marked a departure from traditional reliance on philosophical arguments. His method paved the way for modern observational astronomy, emphasizing the importance of empirical data and observation through telescopes, revolutionizing our understanding of the cosmos.

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Answer:

specific gravity = 0.8

specific gravity of  solution  = 2

Explanation:

given data

rectangular block above water  = 0.400 in

rectangular block below water = 1.60 in

material floats below water = 0.800 in

solution

first we get here specific gravity of block  that is

specific gravity = block vol below ÷ total block vol × specific gravity  water   ..............1

put here value we get

specific gravity =  [tex]\frac{1.60}{1.60+0.400}[/tex]  × 1

specific gravity = 0.8

and now we get here specific gravity of  solution  that is express as

specific gravity of  solution  = total block vol ÷ block vol below × specific gravity  block   ........................2

put here value we get

specific gravity of  solution  = [tex]\frac{1.60+0.400}{0.800}[/tex] × 0.8

specific gravity of  solution  = 2

Answer:

The potential energy can be given as

E = mgh. m is mass, g = acceleration due to gravity = 9.8m/s, h is the heigh, given as 100.0m

E = m x 9.8 x 100 = (980m)J

E = (980m)/10^9GJ = (0.000000980m)GJ to 3 significant figures

Explanation:

Hydroelectric dams exploit storage of gravitational potential energy. A mass, m, raised a height, h against gravity, g = 9.8 m/s², is given a potential energy E = mgh. The result will be in Joules if the input is expressed in meters, kilograms, and seconds (MKS, or SI units).

Answer: [tex]f_{r}[/tex] = 16.49N

Explanation: The object is placed on an inclined plane at an angle of 37° thus making it weight have two component,

[tex]W_{x}[/tex] = horizontal component of the weight = mgsinФ

[tex]W_{y}[/tex] = vertical component of weight = mgcosФ

Due to the way the object is positioned, the horizontal component of force will accelerate the object thus acting as an applied force.

by using newton's law of motion, we have that

mgsinФ - [tex]f_{r}[/tex] = ma

where m = mass of object=5kg

a = acceleration= unknown

Ф = angle of inclination = 37°

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

[tex]f_{r}[/tex] = frictional force = unknown

we need to first get the acceleration before the frictional force which is gotten by using the equation below

[tex]v^{2} = u^{2} + 2aS[/tex]

where v = final velocity = 2m/s

u = initial velocity = 0m/s (because the object started from rest)

a= unknown

S= distance covered = length of plane = 5m

[tex]2^{2} = 0^{2} + 2*a*5\\\\4= 10 *a\\\\a = \frac{4}{10} \\a = 0.4m/s^{2}[/tex]

we slot in a into the equation below to get frictional force

mgsinФ - [tex]f_{r}[/tex] = ma

3 * 9.8 * sin 37 - [tex]f_{r}[/tex] = 3* 0.4

17.9633 - [tex]f_{r}[/tex] =  1.2

[tex]f_{r}[/tex] = 17.9633 - 1.2

[tex]f_{r}[/tex] = 16.49N

Answer: noise that you hear in passing

Explanation:

Explanation:

Azeotrope is defined as a mixture consisting of two liquids which has constant boiling point and a constant composition throughout the distillation.

For example, the boiling point of water is [tex]100^{o}C[/tex] and the boiling point of ethanol is [tex]78.3^{o}C[/tex]. So, a mixture of ethanol and water will act as a binary azeotrope as there are two components.

As the boiling point of this mixture is below the boiling point of either of the two components then it is also a minimum boiling azeotrope.

Thus, we can conclude that distillation of mixtures of ethanol and water cannot increase the ethanol content of the mixture above 95% because this solution boils at a lower temperature than either pure ethanol or pure water. The term which describes this lower boiling mixture is azeotrope.

Answer:

Option (B)

Explanation:

A heterogeneous mixture is usually defined as a combination of two or more chemical substances. It can be also elements as well as compounds. These contrasting components can be easily separated from one another by means of physical process. These are comprised of substances that are not even everywhere. There is variation in it.

Thus, the correct answer is option (B).

Answer:

B: a heterogeneous mixture

Explanation:

If the components of a substance can be separated by the physical means then the substance is a mixture and its component have not undergone any kind of chemical change with their original molecular structure.

The mixtures are of two types homogeneous and heterogeneous.

The final velocity of the train after 8.3 s on the incline will be 12.022 m/s.

Answer:

Explanation:

So in this problem, the initial speed of the train is at 25.8 m/s before it comes to incline with constant slope. So the acceleration or the rate of change in velocity while moving on the incline is given as 1.66 m/s². So the final velocity need to be found after a time period of 8.3 s. According to the first equation of motion, v = u +at.

So we know the values for parameters u,a and t. Since, the train slows down on the slope, so the acceleration value will have negative sign with the magnitude of acceleration. Then

v = 25.8 + (-1.66×8.3)

v =12.022 m/s.

So the final velocity of the train after 8.3 s on the incline will be 12.022 m/s.

Answer:

Explanation:

consider the principle of moment

when a system is in equilibrium, the clockwise moment (torque) about the pivot is equal to the counterclockwise moment ( torque). Since the plank is uniform the weight of the plank act at the middle which = 6.1 m / 2 = 3.05 m

the distance that can support the weight of the man = d

mass of the man = 70

70 × d = 33 × ( 3.05 - 1.6)

d = 47.85 / 60 = 0.798 m, if the man work beyond this point he will fall.

Answer:c

Explanation:

Although the Carnot engine is the most efficient it is not feasible in real life. Carnot engine contains processes that are reversible which makes it difficult to build in real life.

For a system to be in equilibrium a system must be in equilibrium with its surroundings in each step and every step which allows it to be infinitely slow.

Due to this slow rate Power of the engine is zero as the work is being done at an infinitely slow rate.

Answer:

good

Explanation:

Answer:

b. The Moon once rotated faster, but tidal friction slowed the rotation period until it matched the orbital period

Explanation:

Almost all moons orbiting large planets still show the same face to the earth for the same fundamental reason.Due to the Earth's effect on the moon, tidal forces are always facing the Earth from the same moon's side. The time of the rotation and the moon's orbit are the same. Earth-bound observers can thus never get to see the "moon's far side."

The Moon's orbital period and rotation period are the same due to the law of conservation of angular momentum.

The correct option that explains why the Moon's orbital period and rotation period are the same is d. The law of conservation of angular momentum ensured that the Moon must have the same amount of rotational angular momentum as it has of orbital angular momentum. This principle states that the total angular momentum of a system remains constant unless acted upon by an external torque. In the case of the Moon, the gravitational interaction between the Earth and the Moon caused the Moon's rotation to gradually slow down until it matched its orbital period over millions of years.

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Answer:

(A) 89.442 rad/sec

(B) 0.070 sec

(C) [tex]C=25\times 10^{-6}F[/tex]

Explanation:

We have given mass of the body m = 0.50 kg

It is given that spring is extended 2 mm from its equilibrium position

So x = 2 mm = 0.002 m

Resistive force is given F = 8 N

We know that force is equal to F = Kx, here K is spring constant and x is elongation in spring

So [tex]8=K\times 0.002[/tex]

K = 4000 N/m

(A) Angular frequency is equal to [tex]\omega =\sqrt{\frac{K}{m}}[/tex]

So angular frequency [tex]\omega =\sqrt{\frac{4000}{0.50}}=89.442rad/sec[/tex]

(b) Time period of oscillation is equal to [tex]T=\frac{2\pi }{\omega }=\frac{2\times 3.14}{89.442}=0.070sec[/tex]

So time period will be 0.070 sec

(c) Inductance is given L = 5 H

Angular frequency is equal to [tex]\omega =\frac{1}{\sqrt{LC}}[/tex]

[tex]89.442 =\frac{1}{\sqrt{5C}}[/tex]

Squaring both side

[tex]8000=\frac{1}{5C}[/tex]

[tex]C=25\times 10^{-6}F[/tex]