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Complete Question

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Answer:

The uncertainty in inverse frequency is  [tex]\Delta [\frac{1}{w} ]= \frac{3}{2000} \ s[/tex]

Explanation:

From the question we are told that

   The value of the proportionality constant is  [tex]k = 5 \frac{Hz }{T}[/tex]

   The strength of the magnetic field is  [tex]B = 20 \ T[/tex]

   The change in this strength of magnetic field is  [tex]\Delta B = 3 \ T[/tex]

The magnetic field is given as

           [tex]B = \frac{k}{\frac{1}{w} }[/tex]

Where [tex]w[/tex] is frequency

The uncertainty or error of the field is given as

         [tex]\Delta B = \frac{k }{[\frac{1}{w}^]^2 } \Delta [\frac{1}{w} ][/tex]

The uncertainty in inverse frequency is given  as

           [tex]\Delta [\frac{1}{w} ] = \frac{\Delta B}{k [\frac{1}{w^2} ]}[/tex]

                    [tex]\Delta [\frac{1}{w} ]= \frac{\Delta B}{k (B)^2 }[/tex]

substituting values

                  [tex]\Delta [\frac{1}{w} ]= \frac{3}{5 (20)^2 }[/tex]

               [tex]\Delta [\frac{1}{w} ]= \frac{3}{2000} \ s[/tex]

Answer:

The time when the ball strikes the ground is closest to  [tex]t_t = 9.4 \ s[/tex]

Explanation:

From the question we are told that

  The time of projection is t = 0.0 s

   The  distance of the point  from the ground  is  [tex]d = 90 \ m[/tex]

    The  initial velocity of the ball is  [tex]v _i = 36 .2 \ m/s[/tex]

generally the time required to reach maximum height is  

      [tex]t_r = \frac{g}{v}[/tex]

Where is the acceleration due to gravity  with value  [tex]g = 9.8 \ m/s^2[/tex]

Substituting values

        [tex]t_r = \frac{36.2}{9.8}[/tex]

        [tex]t_r = 3.69 s[/tex]

when returning the time and velocity at the roof level is  t =  3.69 s and  u = 36.2 m/s this due to the fact that  air resistance is negligible

   The final velocity at which it  hit the ground is

      [tex]v_f^2 = u^2 + 2ag[/tex]

So  

    [tex]v_f = \sqrt{ u^2 + 2gs}[/tex]

substituting values

    [tex]v_f = \sqrt{ 3.69^2 + 2* 9.8 * 90}[/tex]

     [tex]v_f = 55.45 \ m/s[/tex]

The time taken for the ball to move from the roof level to the ground is  

     [tex]t_g = \frac{v-u}{a}[/tex]

substituting values

    [tex]t_g = \frac{55.45 -36.2}{9.8}[/tex]

     [tex]t_g = 1.96 \ s[/tex]

The total time for this travel is  

    [tex]t_t = t_g + 2 t_r[/tex]

     [tex]t_t = 1.96 + 2(3.69)[/tex]

      [tex]t_t = 9.4 \ s[/tex]

Answer:

(b) Use approximate relationships to find theelectric field at a point 4.00 cm from the axis, measured radiallyoutward from the midpoint of the shell.

Answer:

4.93 m

Explanation:

According to the question, the computation of the height is shown below:

But before that first we need to find out the speed which is shown below:

As we know that

[tex]Speed = \frac{Distance}{Time}[/tex]

[tex]Speed = \frac{5}{0.504}[/tex]

= 9.92 m/s

Now

[tex]v^2 - u^2 = 2\times g\times h[/tex]

[tex]9.92^2 = 2\times 9.98 \times h[/tex]

98.4064 = 19.96 × height

So, the height is 4.93 m

We simply applied the above formulas so that the height i.e H could arrive

Answer:

normal force = 10 N

Explanation:

Given data

frictional force = 0.400 N

coefficient of kinetic friction = 0.04

Solution

we get here normal force that is express as

normal force = [tex]\frac{Frictional\ force}{coefficient\ of\ friction}[/tex]        ............1

put here value and we will get value

normal force = [tex]\frac{0.400}{0.04}[/tex]  

solve it we get

normal force = 10 N

The correct answer is D. electric current being forced uphill by the battery back to the positive terminal.

What is Electric Current?

Electric current is the flow of electric charge through a conducting medium, such as a wire, due to the movement of electrons or ions. The flow of charge is typically caused by the presence of an electric field that creates a potential difference (voltage) between two points in a circuit

In an electrical circuit, pipes are analogous to wires or conductive paths that allow the flow of electric current. The flow of electric current is from the positive terminal of the battery to the negative terminal, which is opposite to the direction of conventional current flow. Therefore, option C is incorrect.

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Answer:

17.85°

Explanation:

To find the angle to the normal in which the light travels in the aqueous fluid you use the Snell's law:

[tex]n_1sin\theta_1=n_2sin\theta_2[/tex]

n1: index of refraction of Sophia's cornea = 1.387

n2: index of refraction of aqueous fluid = 1.36

θ1: angle to normal in the first medium = 17.5°

θ2: angle to normal in the second medium

You solve the equation (1) for θ2, next, you replace the values of the rest of the variables:

[tex]\theta_2=sin^{-1}(\frac{n_1sin\theta_1}{n_2})\\\\\theta_2=sin^{-1}(\frac{(1.387)(sin17.5\°)}{1.36})=17.85\°[/tex]

hence, the angle to normal in the aqueous medium is 17.85°

Answer:

25m

Explanation:

Let's assume the Jeep attains a velocity of 36km/h ; a constant speed same with that of the car.

While the Jeep is accelerating to that speed, the car with that speed passes it.

Now we can calculate the time taken for the Jeep to attain the velocity of 36km/h on her constant acceleration.

This time is t = v/a; from Newton's Law of Motion:

a = V-U / t ; a-acceleration

V is final velocity = 36km/h

U is initial velocity 0 since the body starts from rest.

Hence t = 36000/3600 ÷ 4 = 2.5s

Note conversting from km/h to m/s we multiply by 1000/3600.

But the distance covered by the car while the Jeep just accelerates is

S = U × t = 10× 2.5 = 25m.

Note From Newton's law of Motion, distance for constant speed is defined as: U × t

Hence the Car would be 25m off the starting point just as the Jeep accelerates. It would overtake the Jeep when it just covers 25m from the Jeep starting point.

Answer:

2500 N/m²

Explanation:

Pressure: This can be defined as the force acting normally on a surface per unit area.

The expression for pressure is give as

P = F/A...................... Equation 1

Where P = pressure (N/m²), F = force (N), A = Contact area (m²)

Given: F = 500 N, A = 2000 cm² = (2000/10000) m = 0.2 m.

Substitute into equation  1

P = 500/0.2

P = 2500 N/m²

Hence the pressure exerted to the surface is 2500 N/m²

Answer: [tex]2m/s^2[/tex]

Explanation:

[tex]Formula: F=ma[/tex]

Where;

F = force

m = mass

a = acceleration

Solve for a;

[tex]a=\frac{F}{m}[/tex]

[tex]a=\frac{30N}{15kg}\\ a=2m/s^2[/tex]

Explanation:

Momentum = mass × velocity

p = mv

p = (1000 kg) (14 m/s)

p = 14000 kg m/s

The momentum of the car as it moves down the highway at the given speed is 14000-kg.m/s

Given the data in the question

Momentum is the product of the mass of a particle and its velocity.

Momentum = Mass × Velocity

[tex]P = m \ * \ v[/tex]

We substitute our given values into the equation

[tex]P = 1000kg \ * \ 14m/s\\\\P = 14000kg.m/s[/tex]

Therefore, the momentum of the car as it moves down the highway at the given speed is 14000-kg.m/s

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Answer:

Letter C) and D) is the correct answer.

Explanation:

We know that the a is an acceleration's magnitude, so the units of a are m/s².

Now, let's analyze each terms. If we want that each term will be dimensionally homogeneous, all of them must have the same units of a.

[tex][\frac{x}{t}]=[\frac{m}{s}][/tex]

[tex][vt]=[m][/tex]

[tex][\frac{F}{m}]=[\frac{N}{kg}]=[kg\frac{m}{s^{2}kg}]=[\frac{m}{s^{2}}][/tex]

Therefore, the term F/m is the correct answer.

I hope it helps you!

We can see that  a and F/M are dimensionally homogeneous.

In solving dimensions, we try to express a quantity in terms of the fundamental quantities;

For the term a, its dimension is LT^-2

For the term F/m, its dimension is LT^-2

Hence, it follows that a and F/M are dimensionally homogeneous.

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Answer:

= 8.2*10⁻¹²

Explanation:

Probability of finding an electron to occupy a state of energy, can be expressed by using Boltzmann distribution function

[tex]f(E) = exp(-\frac{E-E_f}{K_BT} )[/tex]

From the given data, fermi energy lies half way between valence and conduction bands, that is half of band gap energy

[tex]E_f = \frac{E_g}{2}[/tex]

Therefore,

[tex]f(E) = exp(-\frac{E-\frac{E_g}{2} }{K_BT} )[/tex]

Using boltzman distribution function to calculate the ratio of number of electrons in the conduction bands of those electrons in the valence bond is

[tex]\frac{n_{con}}{n_{val}} =\frac{exp(-\frac{[E_c-E_g/2]}{K_BT} )}{exp(-\frac{[E_v-E_fg/2}{K_BT} )}[/tex]

[tex]= exp(\frac{-(E_c-E_v}{K_BT} )\\\\=exp(\frac{-(0.66eV)}{(8.617\times10^-^5eV/K)(300K)} )\\\\=8.166\times10^-^1^2\approx8.2\times10^{-12}[/tex]

A line of best fit expresses the relationship between the points.

Explanation:

It does not go through all the points but goes through most of them and it is like a hardrawn curve

Answer:

☟

Explanation:

Fuels made from oil mixtures containing large hydrocarbon molecules are not efficient as they do not flow easily and are difficult to ignite. Crude oil often contains too many large hydrocarbon molecules and not enough small hydrocarbon molecules to meet demand. This is where cracking comes in.

Cracking allows large hydrocarbon molecules to be broken down into smaller, more useful hydrocarbon molecules. Fractions containing large hydrocarbon molecules are heated to vaporise them. They are then either:

heated to 600-700°C

passed over a catalyst of silica or alumina

These processes break covalent bonds in the molecules, causing thermal decompositionreactions. Cracking produces smaller alkanesand alkenes (hydrocarbons that contain carbon-carbon double bonds). For example:

hexane → butane + ethene

C6H14 → C4H10 + C2H4

Some of the smaller hydrocarbons formed by cracking are used as fuels, and the alkenes are used to make polymers in plastics manufacture. Sometimes, hydrogen is also produced during cracking.

Fractional distillation of crude oil

Fractional distillation separates a mixture into a number of different parts, called fractions.

A tall fractionating column is fitted above the mixture, with several condensers coming off at different heights. The column is hot at the bottom and cool at the top. Substances with high boiling points condense at the bottom and substances with lower boiling points condense on the way to the top.

Crude oil is a mixture of hydrocarbons. The crude oil is evaporated and its vapours condense at different temperatures in the fractionating column. Each fraction contains hydrocarbon molecules with a similar number of carbon atoms and a similar range of boiling points.

Oil fractions

The diagram below summarises the main fractions from crude oil and their uses, and the trends in properties. Note that the gases leave at the top of the column, the liquids condense in the middle and the solids stay at the bottom.

As you go up the fractionating column, the hydrocarbons have:

lower boiling points

lower viscosity (they flow more easily)

higher flammability (they ignite more easily).

Other fossil fuels

Crude oil is not the only fossil fuel.

Natural gas mainly consists of methane. It is used in domestic boilers, cookers and Bunsen burners, as well as in some power stations.

Coal was formed from the remains of ancient forests. It can be burned in power stations. Coal is mainly carbon but it may also contain sulfur compounds, which produce sulfur dioxide when the coal is burned. This gas is a cause of acid rain. Also, as all fossil fuels contain carbon, the burning of any fossil fuel will contribute to global warming due to the production of carbon dioxide.

In fractional distillation, the crude oil is added to the chamber and heated. The components with the highest boiling point will condense in the lower part of the column and the components with the lower boiling point will condense at the top of the column. Petrol with a low boiling point is collected from the top of the column.

Fractional distillation can be described as the separation of a mixture into its component fractions. The chemical compound is separated by heating them to a temperature at which fractions of the mixture will vaporize.

Generally, the components have boiling points that differ by less than 25 °C  from each other under one atmosphere. When the mixture is heated, the component with the lower boiling point boils and changes to vapours.

The more volatile component remains in a vapour state and repeated distillations are used in the process, and the mixture is separated into component parts.

Therefore, the petrol from the crude oil can easily be separated as it has a boiling point of about 25-60°C.

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Answer:idk

Explanation:idk

Answer:

shortage of 50 water bottles

$2

30

Explanation:

Explanation:

Complete the first and second sentences, choosing the correct answer from the given ones.

1. T = 100 K

[tex]^{\circ}C=K-273[/tex]

Put T = 100 K

[tex]T=100-273=-173^{\circ} C[/tex]

A temperature of 100 K corresponds on a Celsius scale to (-173 °C)

2. T = 50 °C

[tex]K=^{\circ}C+273\\\\K=50+273\\\\T=323\ K[/tex]

So, At 50 °C, it corresponds to a Kelvin scale of 323 K.

Answer:

Explanation:

The work required to push will be equal to work done by friction . Let  d be the displacement required .

force of friction = mg x μ where m is mass of the suitcase , μ be the coefficient of friction

work done by force of friction

mg x μ x d   = 660

80 x 9.8 x .272 x d = 660

d = 3 .1 m .

Answer:

E) True. The girl has a larger tangential acceleration than the boy.

Explanation:

In this exercise they do not ask us to say which statement is correct, for this we propose the solution to the problem.

Angular and linear quantities are related

          v = w r

          a = α r

the boy's radius is r₁ = 1.2m the girl's radius is r₂ = 1.8m

as the merry-go-round rotates at a constant angular velocity this is the same for both, but the tangential velocity is different

          v₁ = w 1,2 (boy)

          v₂ = w 1.8 (girl)

whereby

          v₂> v₁

reviewing the claims we have

          a₁ = α 1,2

          a₂ = α 1.8

          a₂> a₁

A) False. Tangential velocity is different from zero

B) False angular acceleration is the same for both

C) False. It is the opposite, according to the previous analysis

D) False. Angular acceleration is equal

E) True. You agree with the analysis above,

Answer:

* air resistance.

*the direction of the rotation of the Earth

rotation of the thrown body

Explanation:

The projectile launch is described by the expressions

x-axis         x = v₀ₓ t

y-axis         y = [tex]v_{oy}[/tex] t - ½ gt²

When the things that affect this movement are analyzed, in order of importance we have:

* air resistance. This significantly changes the body's horizontal position, so it introduces a horizontal acceleration that is not contained in the equations.

* air resistance. At the height that the body reaches, since air resistance has the same direction as the gravity of gravity and therefore the relationship is more challenging.

* to a lesser extent the direction of launch, in the direction of the rotation of the Earth against. Since this creates an operational on the x and y axis that changes the initial assumption

* The possible rotation of the thrown body, since this rotation creates a lift that is not taken in the equations, this value is more noticeable the lighter the body, this effect has to keep the body longer in the air achieving more reach and height

Before she jumped from the cliff, her gravitational potential energy was

GPE = (her weight) x (height of the cliff) .

That's exactly the GPE that she lost on the way down to the water.  So we can write

24,500 J = (her weight) x (44.0 m)

Divide each side by 44.0 m:

Her weight = 24,500 J / 44 m

Her weight = 556.8 Newtons

(about 125 pounds)

Answer:

5.1 rad/s

Explanation:

Mechanical energy of the system is conserved since no external work is done on the sphere.

[tex]mgh = mv^2/2 + I\omega^2/2[/tex]

Substituting v = ωr and I = 2 m r^2/5, we get,

=> [tex]mgh=m(\omega r)^2/2 + (2\omega r^2/5)\omega^2/2[/tex]

=> [tex]mgh = m\omega^2r^2/2 + m\omega^2r^2/5[/tex]

=> [tex]gh =\omega^2r^2/2+\omega^2r^2/5[/tex]

=>  [tex]gh = 7\omega^2 r^2/10[/tex]

=>  [tex]\omega r = (10gh/7)^{1/2}[/tex]

=> [tex]\omega = (1/r)(10gh/7)^{1/2} = (1 / 1.7)(10\times 9.8\times 5.3 / 7)^{1/2}[/tex] = 5.1 rad/s

Sinkholes formation is one feature of karst topography. The top of a cave falls if it develops large enough and its top extends near enough to the surface.

Karst topography is a type of natural environment formed mostly by chemical weathering by water, resulting in caves, sinkholes, cliffs, and steep-sided hills known as towers.

The top of a cave falls if it develops large enough and its top extends near enough to the surface. Sinkholes are formed as a result of this, and they are one of the most distinguishing aspects of karst terrain.

When water absorbs carbon dioxide from the atmosphere and ground, it becomes carbonic acid.

Hence, sinkholes  formation is one feature of karst topography

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Answer: A) Caves

Explanation:

Answer:

The number of molecules displaced in a vibration makes the amplitude of a sound.

Answer:

a) The spring constant of the spring is [tex]28.381\,\frac{N}{m}[/tex], b) The angular frequency of the block is [tex]7.038\,\frac{rad}{s}[/tex].

Explanation:

This question is incomplete and complete version will be presented herein:

A spring is hung from the ceiling. A 0.573-kg block is then attached to the free end of the spring. When released from rest, the block drops 0.198 m before momentarily coming to rest, after which it moves back upward. (a) What is the spring constant of the spring (b) Find the angular frequency of the block 's vibrations.

a) Since spring is hung from the ceiling and is stretched by action of gravity on 0.573 kilogram block. According to the Hooke's Law, force experimented by the spring is directly proportional to elongation. An expression describing the phenomenon is presented and described below: (System at equilibrium - Newton's Second Law)

[tex]m\cdot g = k\cdot \Delta x[/tex]

Where:

[tex]m[/tex] - Mass, measured in kilograms.

[tex]g[/tex] - Gravitational constant, measured in meters per square second.

[tex]k[/tex] - Spring constant, measured in newtons per meter.

[tex]\Delta x[/tex] - Spring linear deformation, measured in meters.

Now, the spring constant is cleared in this equation and outcome is computed: ([tex]m = 0.573\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex] and [tex]\Delta x = 0.198\,m[/tex])

[tex]k = \frac{m\cdot g}{\Delta x}[/tex]

[tex]k = \frac{(0.573\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}{0.198\,m}[/tex]

[tex]k = 28.381\,\frac{N}{m}[/tex]

The spring constant of the spring is [tex]28.381\,\frac{N}{m}[/tex].

b) Let suppose that mass-spring system is experimenting a simple harmonic motion, so that angular frequency is equal to:

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

Given that [tex]k = 28.381\,\frac{N}{m}[/tex] and [tex]m = 0.573\,kg[/tex], the angular frequency, measured in radians per second, of the block is:

[tex]\omega = \sqrt{\frac{28.381\,\frac{N}{m} }{0.573\,kg} }[/tex]

[tex]\omega = 7.038\,\frac{rad}{s}[/tex]

The angular frequency of the block is [tex]7.038\,\frac{rad}{s}[/tex].

Answer:

About 3.06 seconds

Explanation:

[tex]v_f=v_o+at[/tex]

Since at the peak of its trajectory, the ball will have no velocity, you can write the following equation:

[tex]0=30+(-9.81)t\\\\-30=-9.81t\\\\t\approx 3.06s[/tex]

Hope this helps!

Explanation:

It is given that,

Mass of a car is 900 kg

Radius of curve is 600 m

Speed of the car in the curve is 25 m/s

We need to find the centripetal force exerted on a car. The formula used to find the centripetal force is given by :

[tex]F=\dfrac{mv^2}{r}\\\\F=\dfrac{900\times (25)^2}{600}\\\\F=937.5\ N[/tex]

So, the centripetal force exerted on a car is 937.5 N.

Static friction prevents the car from slipping. It means that the magnitude of centripetal force is balanced by the frictional force. So, the frictional force of 937.5 N is acting on the car.  

Answer:

Explanation:

A )

At the bottom of the circle , the potential energy of the stopper is converted into kinetic energy

1/2 m V² = mg x 2r + 1/2 mv²

m is mass of stopper , V is velocity at the bottom , r is radius of the circular path which is length of the string , v is velocity at the top

1/2  V² = g x 2r + 1/2 v²

 V² = g x 4r +  v²

 V² = 9.8 x 4 +  8²

V² = 103.2

V = 10.16 m/s

B )

If T be the tension at the top

Net downward force

= mg + T . This force provides centripetal force for the circular motion

mg +T = mv² / r

T =   mv²/r -mg

= m ( v²/r - g )

= .005 ( 8²/1 -g )

= .005 x 54.2

= .27 N .

C ) At the bottom

Net force = T  - mg , T is tension at the bottom , V is velocity at bottom

T-mg = mV²/r

T = m ( V²/r +g )

= .005 ( 10.16²/1 +9.8)

= .005 x 113

= .56 N .

Answer:

ΔE = ‒0.271 kJ

Explanation:

Let's begin by listing out the given variables:

q = -0.653 kJ, w = 0.389 kJ

Using the formula ΔE = q + w

ΔE = -0.653 + 0.388

ΔE = (‒0.655 + 0.382) kJ

ΔE = ‒0.271 kJ

Therefore, the change in internal energy is -271 J or -0.271 kJ which implies that the system is exothermic

Answer:

d = 80 m

its vertical displacement (in m) after 4 s is 80 m

Explanation:

From the equation of motion;

d = vt + 0.5at^2 ......1

Where;

d = displacement

v = initial velocity = 0 (dropped with no initial speed)

t = time of flight = 4s

a = g = acceleration due to gravity = 10 m/s^2

Substituting the given values into equation 1;

d = 0(4) + 0.5(10 × 4^2)

d = 0.5(10×16)

d = 80 m

its vertical displacement (in m) after 4 s is 80 m