A person sits on a freely spinning lab stool that has no friction in its axle. When this person extends her arms, A. her moment of inertia decreases and her angular speed decreases B. her moment of inertia decreases and her angular speed increases C. her moment of inertia increases and her angular speed decreases D. her moment of inertia increases and her angular speed decreases E. her moment of inertia increases and her angular speed remains the same.

Answers

Answer 1

Answer:

  C. her moment of inertia increases and her angular speed decreases

  D. her moment of inertia increases and her angular speed decreases

Explanation:

The moment of inertia of a body is the sum of the products of an increment of mass and the square of its distance from the center of rotation. When a spinning person extends her arms, part of her mass increases its distance from the center of rotation, so increases the moment of inertia.

The kinetic energy of a spinning body is jointly proportional to the moment of inertia and the square of the angular speed. Hence an increase in moment of inertia will result in a decrease in angular speed unless there is a change in the rotational kinetic energy.

This effect is used by figure skaters to increase their spin rate by drawing their arms and legs closer to the axis of rotation. Similarly, they can slow the spin by extending arms and legs.

When the person extends her arms, her moment of inertia increases and her angular speed decreases.

_____

Note to those looking for a letter answer

Both choices C and D have identical (correct) wording the way the problem is presented here. You will need to check carefully the wording in any problem you may think is similar.


Related Questions

1. An object with a mass of 15 kilograms is pushed by a force of 30 Newtons. How much does
it accelerate?

Answers

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]

A ball is projected upward at time t = 0.0 s, from a point on a roof 90 m above the ground. The ball rises, then falls and strikes the ground. The initial velocity of the ball is 36.2 m/s if air resistance is negligible. The time when the ball strikes the ground is closest to

Answers

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]

 

Which of the following is analogous to the pipes in an electrical circuit?

A. capacitors storing the incoming charge from the battery
B. large resistors causing restrictions to the flow of charge
C. electric current flowing “downhill” from the negative electrode to the positive electrode in a battery electric current being forced uphill by the battery
D. electric current being forced uphill by the battery back to the positive terminal

Answers

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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wha is amplitde in sound

Answers

Answer:

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

A boy and a girl are on a spinning merry-go-round. The boy is at a radial distance of 1.2 m from the central axis; the girl is at a radial distance of 1.8 m from the central axis. Which is true?A- Boy and girl have zero tangential and angular accelerations.B- The girl has a larger angular acceleration than the boy.C- The boy has a larger tangential acceleration than the girl.D- The boy has a larger angular acceleration than the girl.E- The girl has a larger tangential acceleration than the boy.

Answers

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,

A cylindrical shell of radius 7.00 cm and length 2.44 m has its charge uniformly distributed on its curved surface. The magnitude of the electric field at a point 21.9 cm radially outward from its axis (measured from the midpoint of the shell) is 36.0 kN/C.(a) Use approximate relationships to find thenet charge on the shell.
(b) Use approximate relationships to find theelectric field at a point 4.00 cm from theaxis, measured radiallyoutward from the midpoint of the shell.

Answers

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.

Which formation is one feature of karst topography?


Answers

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.

What is karst topography?

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:

Use the Lab screen to expand your ideas about what affects the landing location and path of a projectile. List any discoveries you made to identify additional things that affect the landing site of a projectile and/or path of a projectile. Next to each item, briefly explain why you think the motion of the projectile is affected..

Answers

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

A 1000-kg car is moving down the highway at 14m/s. What is the momentum?

Answers

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

Mass of the car; [tex]g = 1000kg[/tex]Velocity of the car; [tex]v = 14m/s[/tex]Momentum; [tex]p = ?[/tex]

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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A 1.0 m string with a 5 g stopper on the end is whirled in a vertical circle. The speed of the stopper is 8 m/s at the top of the circle. (A) What is the speed of the stopper at the bottom of the circle? (HINT: Use energy conservation principles!) (10.2 m/s) (B) What is the tension in the string when the stopper is at the top of the circle? (0.27 N) (C) What is the tension in the string when the stopper is at the bottom of the circle?

Answers

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 .

What's a line of best fit? Will give BRAINLIEST

Answers

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

The coefficient of kinetic friction between a suitcase and the floor is 0.272. You may want to review (Pages 196 - 203) . Part A If the suitcase has a mass of 80.0 kg , how far can it be pushed across the level floor with 660 J of work

Answers

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 .

A water slide is constructed so that swimmers, starting from rest at the top of the slide, leave the end of the slide traveling horizontally. One person hits the water 5.00 m from the end of the slide in a time of 0.504 s after leaving the slide. Ignore friction and air resistance. Find the height H.

Answers

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

Crystalline germanium (Z=32, rho=5.323 g/cm3) has a band gap of 0.66 eV. Assume the Fermi energy is half way between the valence and conduction bands. Estimate the ratio of electrons in the conduction band to those in the valence band at T = 300 K. (See eq. 10-11) Assume the width of the valence band is ΔΕV ~ 10 eV.

Answers

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]

Concerned with citizen complaints of price gouging during past hurricanes, Florida's state government passes a law setting a price ceiling for a bottle of water equal to the market equilibrium price during normal times. After all, it seems unfair that sellers of water gain because of a hurricane.

Answers

Answer:idk

Explanation:idk

Answer:

shortage of 50 water bottles

$2

30

Explanation:

A student has derived the following nondimensionally homogeneous equation: a=x/t2-vt+F/m where v is a velocity's magnitude , a is an acceleration's magnitude, t is a time, m is a mass, F is a force's magnitude , and x is a distance (or length). Which terms are dimensionally homogeneous? .
a) x/t
b) vt
c) a
d) F/m

Answers

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;

MassLengthTime

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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Problem 3A solid uniform sphere of mass 120 kg and radius 1.7 m starts from rest and rolls without slipping down an inclined plane of vertical height 5.3 m. What is the angular speed of the sphere at the bottom of the inclined plane

Answers

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

Complete the first and second sentences, choosing the correct answer from the given ones.
1. A temperature of 100 K corresponds on a Celsius scale to 100 ° C / 0 ° C / 173 ° C / -173 ° C.
2. At 50 ° C, it corresponds to a Kelvin scale of 150 K / 323 K / 273 K / 223 K.

Answers

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.

A solid exerts a force of 500 N. Calculate the pressure exerted to the surface where area

of contact is 2000 cm2.

Answers

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

1. Calculate the centripetal force exerted on a 900kg900kg car that rounds a 600m600m radius curve on horizontal ground at 25.0m/s25.0m/s. 2. Static friction prevents the car from slipping. Find the magnitude of the frictional force between the tires and the road that allows the car to round the curve without sliding off in a straight line.

Answers

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.  

Consider a system of a cliff diver and the earth. The gravitational potential energy of the system decreases by 24,500 J as the diver drops to the water from a height of 44.0 m. Determine her weight in newtons.

Answers

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)

If you were to drop a rock from a tall building, assuming that it had not yet hit the ground, and neglecting air resistance. What is its vertical displacement (in m) after 4 s? (g = 10 m/s2)

Answers

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

A tuning fork is held over a resonance tube, and resonance occurs when the surface of the water is 12 cm below the top of the tube. Resonance occurs again when the water is 34 cm below the top of the tube. If the air temperature is 23 degrees Celsius, find the frequency of the tuning fork.

Answers

Answer:

786 Hz

Explanation:

Recall, the speed of sound is

v = 332 + 0.6t

Where t = 23°

v = 332 + 0.6(23)

v = 332 + 13.8

v = 345.8 m

Also, it is known that distance between two consecutive resonance length is half of the wavelength.

L2 - L1 = λ/2

34 - 12 = λ/2

λ/2 = 22

λ = 44 cm

Finally, remember that also

Frequency = speed/ wavelength

Frequency = 345.8/0.4

Frequency = 786 Hz

Therefore, the frequency of the tuning fork is 786 Hz

The frequency of the applied RF signal used to excite spins is directly proportional to the magnitude of the static magnetic field used to align the spins, with proportionality constant 5 hz/T. If the strength of the applied field is known to be 20 T plus or minus 3 T, which of the following correctly describes the uncertainty in the INVERSE frequency (1/frequency)?
A. 3/2000s
B. 3/5s
C. 1/15s
D. 1/4

Answers

Complete Question

The complete question is shown on the first uploaded image  

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]

Lebron James and Stephen Curry are playing an intense game of minigolf. The final(18th) hole is 8.2 m away from the tee box (starting location) at an angle of 20◦ east of north. Lebron’s first shot lands 8.6 m away at an angle of 35.2◦ east of north and Steph’s first shot lands 6.1 m away at an angle of 20◦ east of north. Assume that the minigolf course is flat.
(A) Which ball lands closer to the hole?
(B) Each player sunk the ball on the second shot. At what angle did each player hit their ball to reach the hole?

Answers

Answer:

A. we will see that the notion [tex]\mathbf{|r ^ \to - r_2 ^\to| = 2.10006 \ m}[/tex] which denotes Stephen Curry  illustrates that Stephen Curry minigolf ball shot is closer

B.  Lebron James hits at an angle of 17.48° North -East.

The direction of Stephen is   = 20° due to East of North

Explanation:

Let [tex]r ^ {\to[/tex] represent the position vector of the hole;

Also; using  the origin as starting point. Let the east direction be along the positive x axis and the North direction  be + y axis

Thus:

[tex]r ^ {\to[/tex]  = [tex]8.2 \ sin 20^0 \hat i + 8.2 \ cos 20 \hat j[/tex]

[tex]r ^ {\to[/tex]  = [tex](2.8046 \hat i + 7.7055 \hat j ) m[/tex]

Let [tex]r_1 ^ \to[/tex] be the position vector for Lebron James's first shot

So;

[tex]r_1 ^ \to[/tex] = [tex](8.6 \ sin \ 35.2 )^0 \hat i + 8.6 \ cos \ ( 35.2)^0 \hat j[/tex]

[tex]r^ \to = (4.9573 \hat i + 7.02745 \hat j) m[/tex]

Let [tex]r_2 ^ \to[/tex] be the position vector for Stephen Curry's shot

[tex]r_2 ^ \to[/tex]  [tex]=6.1 \ sin 20^0 \hat i + 6.1 \ cos \ 20 \hat j[/tex]

[tex]r_2 ^ \to[/tex]  = [tex](2.0863 \hat i + 5.7321 \hat j )m[/tex]

However;

[tex]r ^ \to - r_1 ^\to = (-2.1527 \hat i + 0.67805 \hat j) m[/tex]

[tex]\mathbf{|r ^ \to - r_1 ^\to| =2.25696 \ m }[/tex]

Also;

[tex]r ^ \to - r_2 ^\to = (0.71013 \hat i - 1.9734 \hat j) m[/tex]

[tex]\mathbf{|r ^ \to - r_2 ^\to| = 2.10006 \ m}[/tex]

Thus; from above ; we will see that the notion [tex]\mathbf{|r ^ \to - r_2 ^\to| = 2.10006 \ m}[/tex] which denotes Stephen Curry  illustrates that Stephen Curry minigolf ball shot is closer

B .

For Lebron James ;

The angle can be determine using the trigonometric function:

[tex]tan \theta = ( \dfrac{0.67805}{-2.1527}) \\ \\ tan \theta = -0.131498 \\ \\ \theta = tan ^{-1} ( -0.31498) \\ \\ \mathbf{\theta = -17.48^0}[/tex]

Thus  Lebron James hits at an angle of 17.48° North -East.

For Stephen Curry;

[tex]tan \theta = ( \dfrac{-1.9734}{0.7183}) \\ \\ tan \theta = -2.74732 \\ \\ \theta = tan ^{-1} ( -2.74732) \\ \\ \mathbf{\theta = -70.0^0}[/tex]

The direction of Stephen is  = 90° - 70° = 20° due to East of North

PLS HELP,WILL GIVE BRAINLIEST + 30 POINTS
Describe how fractional distillation and cracking are used so that sufficient petrol is produced from crude oil to meet demand.

Answers

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.

What is fractional distillation?

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.

Learn more about fractional distillation, here:

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A jeep starts from rest with a constant acceleration of 4m/s2.At the same time a car travels with a constant speed of 36km/h overtake and passes the jeep how far beyond the starting point will the jeep overtakes the car?

Answers

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.

The index of refraction of Sophia's cornea is 1.387 and that of the aqueous fluid behind the cornea is 1.36. Light is incident from air onto her cornea at an angle of 17.5° from the normal to the surface. At what angle to the normal is the light traveling in the aqueous fluid?

Answers

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°

Calculate the change in internal energy of the following system: A balloon is cooled by removing 0.653 kJ of heat. It shrinks on cooling, and the atmosphere does 389 J of work on the balloon. Express your answer to three significant figures and include the appropriate units.

Answers

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

A physics major is cooking breakfast when he notices that the frictional force between the steel spatula and the Teflon frying pan is only 0.400 N. Knowing the coefficient of kinetic friction between the two materials (0.04), he quickly calculates the normal force. What is it (in N)? N

Answers

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

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