A boat radioed a distress call to a Coast Guard station. At the time of the call, a vector A from the station to the boat had a magnitude of 45.0 km and was directed 15.0° east of north. A vector from the station to the point where the boat was later found is B = 30.0 km, 15.0° north of east.
How far did the boat travel from the point where the distress call was made to the point where the boat was found? In other words, what is the magnitude of vector C?
A)65.3 km
B)39.7 km
C)26.5 km
D)54.0 km
E)42.5 km

Answer:

Option b. it has the same position and the same acceleration as A

Explanation:

Let's analyze every statement:

a. it has the same position and the same velocity as A

In the instant where B passes A, they Do have the same position. Velocity however, cannot be the same because if they were, ball B would never pass ball A. So, this is false.

b. it has the same position and the same acceleration as A

As we said in the previous option, the position is the same. The acceleration is gravity for both balls, so this is true.

c. it has the same velocity and the same acceleration as A

Acceleration is the same but velocities are not, so this is false.

d. it has the same displacement and the same velocity as A

The distance they have traveled is the same, so the displacement is the same, but the velocity is not, so this is false.

e. it has the same position, displacement and velocity as A

The position and displacement is the same but not velocity, so this is false.

Only option b is true.

Answer:

The answer to your question is:

Explanation:

Data

mass = 4.33 kg

E = 41.7 J

v = ?

Formula

Ke = (1/2)mv²

Clear v from the equation

v = √2ke/m

Substitution

v = √2(41.7)/4.33

v = 19.26 m/s          Result

Answer:

Part a)

[tex]\omega = 5 ft/s[/tex] counterclockwise

Part b)

[tex]\omega = 2.5 ft/s[/tex] counterclockwise

Part c)

[tex]\omega = 2.5 ft/s[/tex] clockwise

Explanation:

Let the cart has radius R = 1 ft

so here we have speed of the center of the cart is

[tex]v_c = 2.5 ft/s[/tex]

let the angular speed is given as

[tex]\omega[/tex] counter clockwise

Part a)

if the top most point of the rim has same speed as that of speed of the center but it is towards left

so we have

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

[tex]-2.5 = 2.5 + 1(\omega)[/tex]

so we have

[tex]\omega = -5 ft/s[/tex]

Part b)

if the speed of the top point on the rim is zero

[tex]v = v_c + 1(\omega)[/tex]

[tex]0 = 2.5 + \omega[/tex]

[tex]\omega = -2.5 ft/s[/tex]

Part c)

if the speed at the top position on the rim is 5 ft/s

[tex]5 ft/s = 2.5 ft/s + 1(\omega)[/tex]

[tex]\omega = 2.5 ft/s[/tex]

Explanation:

It is given that,

Mass of the aircraft 1, m₁ = 1400 kg

Mass of aircraft 2, m₂ = 1500 kg

Initially, the aircraft 2 is at rest, u₂ = 0

Initial speed of the aircraft 1, u₁ = 40 m/s

After the collision, the total speed of the system, V = 19.3 m/s

Time, t = 8.2 s

Since, two objects stick together it is a case of inelastic collision. The acceleration of the system is given by :

[tex]a=\dfrac{V}{t}[/tex]

[tex]a=\dfrac{19.3\ m/s}{8.2\ s}[/tex]

[tex]a=2.35\ m/s^2[/tex]

Distance covered by the system before stopping is given by :

[tex]x=\dfrac{1}{2}\times a\times t^2[/tex]

[tex]x=\dfrac{1}{2}\times 2.35\times (8.2)^2[/tex]

x = 79.007 meters

Hence, this is the required solution.

Answer:

c)7 mph

Explanation:

Going downstream the speed of the boat is added to the speed of the water, on the way back it is opposite to it, then the speed of the water is substracted to that of the boat, and we have the total time of the trip, then we can write down a system of equations as follows:

x:=Speed of water that is being asked

t1:= time going downstream

t2:= time going upstream

(21+x)*t1=14  (1)

(21-x)*t2=14  (2)

t1+t2=1.5       (3)

then t1=1.5-t2, replacing this in (1) we have  (21+x)*(1.5-t2)=14, then t2=1.5-(14/(21+x))  (4). Doing the same in (2) we get t2=14/(21-x)  (5), this way we can use (4) and (5) and get that:

1.5-(14/(21+x))=14/(21-x), doing the operations we get that 588/(441-x^2)=1.5 or equivalently 1.5x^2-73.5=0 or x^2-49=0 or (x-7)(x+7)=0, then the possible answers are x=7 or x= -7, here we consider the positive answer, that in magnitude is the same as the other, because we already considered the direction of the water into the equations. Then the answer is c) 7mph.

Final answer:

The speed of the current is 3 miles per hour, obtained by setting up an equation based on the total time taken for an upstream and downstream trip and solving for the current's speed. So the correct option is a.

Explanation:

To find the speed of the current, we can set up an equation considering the speed of the boat in still water (which we know is 21 miles per hour) and the speed of the current, which we will call 'c'.

When the boat goes upstream, the effective speed is reduced by the speed of the current, so it's 21 - c miles per hour. Similarly, when the boat goes downstream, the effective speed is increased by the speed of the current, making it 21 + c miles per hour.

Let's calculate the time taken to travel upstream and downstream using the distance and the effective speed. For upstream, it is 14 / (21 - c) hours, and for downstream, it is 14 / (21 + c) hours. The total time for the trip is given as 1.5 hours, so we sum the upstream and downstream times to set up the equation:

14 / (21 - c) + 14 / (21 + c) = 1.5

By solving this equation for 'c', we find that the speed of the current is 3 miles per hour, which corresponds to option (a).

Answer:

They are the same (assuming there is no air friction)

Explanation:

Take a look at the picture.

When the first ball (the one thrown upward) gets to the point marked as A, the speed will has the exact same value V but the velocity will now point downward (just like the second ball).

So if you think about it, the first ball, from point A to the ground, will behave exactly like the second ball (same initial speed, same height).

That is why the speeds will be the same when they reach the ground.

Answer:

Velocity is same

Explanation:

Case I:

When the ball throws upwards

Let the velocity of the ball as it hits the ground is V'.

Initial velocity, u = V

Final velocity, v = V'

height = h

acceleration due to gravity = g

Use third equation of motion

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

By substituting the values

[tex]V'^{2}=V^{2}+2(-g)(-h)[/tex]

[tex]V'=\sqrt{V^{2}+2gh}[/tex]      .... (1)

Case II:

When the ball throws downwards

Let the velocity of the ball as it hits the ground is V''.

Initial velocity, u = V

Final velocity, v = V''

height = h

acceleration due to gravity = g

Use third equation of motion

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

By substituting the values

[tex]V''^{2}=V^{2}+2(-g)(-h)[/tex]

[tex]V''=\sqrt{V^{2}+2gh}[/tex]      .... (2)

By comparing the equation (1) and equation (2), we get

V' = V''

Thus, the velocity of balls in both the cases is same as they strikes the ground.

Answer: Ok, if the position of the bicycle is negative, then you can think is in the -x range, lets call it -r.

And the velocity is positive, you know that the movement equation of something is x= v*t + x0

where v is te velocity, x0 the position and t the time.

so in our case x = v*t  - r.

so when v*t = r, for some time, the bicycle will be on your position.

So the bicycle was getting closer to you.

Answer:

The answer to your question is: 15 m/s2

Explanation:

Equation    x = at3 - bt2 + ct

a = 4.1 m/s3

b = 2.2 m/s2

c = 1.7 m/s

First we find  x at t = 4.1 s

x = 4.1(4.1)3 - 2.2(4.1)2 + 1.7(4.1)

x = 4.1(68.921) - 2.2(16.81) + 6.97

x = 282.58 - 36.98 + 6.98

x = 252.58 m

Now we find speed

v = x/t = 252.58/ 4.1 = 61.6 m/s

Finally

acceleration = v/t = 61.6/4.1 = 15 m/s2

The acceleration of the object is the change in the velocity of the object within given time interval.

The acceleration of the object at time t = 4.1 seconds is 15.26 m/s2.

Given that the position of an object is x = at3 - bt2 + ct, where a = 4.1 m/s3, b = 2.2 m/s2, c = 1.7 m/s.

The position of the object at time t = 4.1 s is calculated as given below.

[tex]x = at^3-bt^2+ct\\[/tex]

[tex]x = 4.1\times (4.1)^3 - 2.2 \times (4.1)^2 + 1.7\times 4.1[/tex]

[tex]x = 252.56\;\rm m[/tex]

The velocity v of the object at time t = 4.1 s is given below.

[tex]v = \dfrac{x}{t}[/tex]

[tex]v = \dfrac {252.56}{4.1}[/tex]

[tex]v = 62.67\;\rm m/s[/tex]

The acceleration of the object at time t =4.1 s is given below.

[tex]a = \dfrac {v}{t}[/tex]

[tex]a = \dfrac {62.67}{4.1}[/tex]

[tex]a = 15.26\;\rm m/s^2[/tex]

Hence we can conclude that the acceleration of the object at time t = 4.1 seconds is 15.26 m/s2.

To know more about the acceleration, follow the link given below.

https://brainly.com/question/12134554.

Answer:

[tex]v_f=8.4m/s[/tex]    

Explanation:

We need to use the conservation of momentum Law, the total momentum is the same before and after the kid leaves the skate.

In the axis X:

[tex](m_{kid}+m_{skate})*v_{o}=m_{kid}*v_f[/tex]     (1)

[tex]v_f=(m_{kid}+m_{skate})*v_{o}/m_{kid}=(44+2.2)*8/44=8.4m/s[/tex]  

Answer:

46 meters

Explanation:

In the figure you can see law of cosines.

In your case

α = 22.5° + 42 ° = 64.5 °

b = 15 m (distance between your's tent and Joe's tent)

c = 42 m (distance between your's tent and Karl's tent)

a is the distance between Joe's tent and Karl's tent

Replacing in the first equation:

[tex]a^2 = 15^2 m^2 + 42^2 m^2 - 2 \times 15 \times 42 \times cos(64.5) [/tex]

[tex] a = \sqrt{2111.46 m^2} [/tex]

[tex] a = 46 m [/tex]

Answer:

10 m/s

Explanation:

u = 20 m/s

v = 0 m/s

t = 5.8 s

Let a be the acceleration and s be the distance traveled by the car in time t.

use first equation of motion

v = u + a t

0 = 20 + a x 5.8

a = -3.45 m/s^2

Use third equation of motion

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

[tex]0^{2}=20^{2}-2 \times3.45 \times s[/tex]

s = 58 m

The average speed is defined as the ratio of distance traveled to the time taken.

[tex]Average speed = \frac{58}{5.8} = 10 m/s[/tex]

thus, the average speed of the car is 10 m/s.  

Answer:

The engine would be warm to touch, and the exhaust gases would be at ambient temperature. The engine would not vibrate nor make any noise. None of the fuel entering the engine would go unused.

Explanation:

In this ideal engine, none of these events would happen due to the nature of the efficiency.

We can define efficiency as the ratio between the used energy and the potential generable energy in the fuel.

n=W, total/(E, available).

However, in real engines the energy generated in the combustion of the fuel transforms into heat (which heates the exhost gases, and the engine therefore transfering some of this heat to the environment). Also, there are some mechanical energy loss due to vibrations and sound, which are also energy that comes from the fuel combustion.

Final answer:

This answer addresses the implications of a hypothetical 100%-efficient engine in an automobile, discussing its heat production, exhaust impact, noise and vibration levels, and fuel usage.

Explanation:

If an automobile had a 100%-efficient engine:

Answer:

Explanation:

When a parachutist jump from an aeroplane, there are two forces acting on the parachutist.

One is the force called gravitational force which is acting downwards.

Other is the force of air resistance which is acting upwards.

When the parachute is not opened, the gravitational force is more than the air resistance force due to which the parachutist experiences a force with which he is coming down and thus , there is an acceleration in the body and the velocity of the body goes on increasing.

When the parachute is opened, the gravitational force is balanced by the air resistance force due to which there is no net force acting on the body of parachutist and thus, he comes down with uniform velocity as acceleration in the body is zero.

In this condition, the magnitude of gravitational force is equal to the air resistance force.

Answer:

Explanation:

Since it only mentions the height we only have to concern ourselves with the y direction, up and down, so that's nice.  Also, acceleration due to gravity is -9.8 m/s^2

First we want the formula for the position of the object, for that we need the initial speed.  We have an initial speed but we need the speed in the y direction.  To do that we break the speed we have into its x and y components.  Imagine a right triangle where the hypotenuse is 100 and the angle between that and the horizontal leg is 42.8 degrees.  Notice that the legs of the right triangle are vertical and horizontal, they represent the y and x components of the speed respectively.  Anyway, we want the y component, so the vertical.  Relative to the given angle  that is the "opposite" side, when we use SOH CAH TOA.  Let me know if this doesn't make sense and I can draw it out.

Now, using trig, we can find the y component  with SOH.  so sin(42.8)=y/100.  Make sure your calculator is in degree mode of course.  we get

y = 100*sin(42.8) = 67.94 so the vertical speed is 67.94 m/s.

With that now we can use the displacement formula s = vi*t + .5a*t^2 if you don't remember this formula you definitely should have some kind of paper or something with all your physics formula.  Anyway, we can plug in all values except for t, which is our variable.

s = 95.5

vi=67.94

a= -9.8

95.5 = 67.94t + .5(-9.8)t^2

Then if you subtract 95.5 from both sides you have a traditional quadratic equation you can solve with the quadratic formula, or completing the square or however you solve them.  If you need help with this step let me know, or if something else didn't make sense.  

Answer:

The answer to your question is: 1.19 gr/ml

Explanation:

empty flask weighs = 45.4 g

filled with water = 121.8 g

filled with a different substance = 91.39 g

density of the second liquid =

First we must calculate the mass of water in the flask by subtracting the flask filled with water and the flask empty.

mass =  flask filled with water - the flask empty.

mass = 121.8 - 45.4 gr

mass = 76.4 gr

The density of water is always 1 gr/ml

and density is mass/volume

from here we calculate the volume = mass/density

                                             volume = 76.4/1 = 76.4 ml

Now we calculate the density of the other liquid

           density = 91.39/76.4 = 1.19 g/ml

The hypothesis could be stated as follows: "Since there is no direct sunlight at night, the temperature of the air blown by the bedroom air conditioner is colder at night than it is during the day."

You may devise an experiment to gauge the temperature of the air being blasted by the air conditioner both at night and during the day in order to verify this theory. This is a general description of the experiment.

Place a thermometer or temperature sensor close to the air conditioner's outlet, which is where the room's cool air is blown. Make sure the sensor is in the same spot for the measurements during the day and at night.

The exact relationship between the air temperature blasted by the air conditioner and the presence or absence of direct sunshine is what the experiment attempts to investigate. According to the theory, cooler air may be blasted by the air conditioner during the daytime when there is sunlight but warmer air during the night when there is no direct sunlight.

Hence, the hypothesis could be stated as follows: "Since there is no direct sunlight at night, the temperature of the air blown by the bedroom air conditioner is colder at night than it is during the day."

To learn more about the hypothesis, here:

https://brainly.com/question/35154833

#SPJ12

Answer:

B

Explanation:

In a common rail system, the fuel is distributed to the injectors from the rail where there is high pressure accumulation.High pressure fuel pump feds the rail and with the help of the start and end signals the injector is activated.For example in diesel common rail direct injection systems, the rail is connected to injectors using individual pipes in that the injectors work hand in hand with the fuel pump which ensures fuel injection timing and amount. Comparing this with early models, the fuel pump was responsible for timing, quantity and pressure.So power stroke in a modern common rail system does not only rely on single injection of pump but also has multiple injection events during combution process.

Answer:

d = .076 m

Explanation:

The time for frog A can be calculated from  equation of motion

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

where v_f is final velocity, a is acceleration due to gravity

so from given data we have

[tex]-0.551= 0.551 + (-9.8)(t)[/tex]

t = 0.112 sec

Now we will use that time for frog B

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

[tex]v_f = 1.23 + (-9.8)(0.112)[/tex]

[tex]v_f = 0.128 m/s [/tex](Note its positive)

For the displacement

[tex]s = v_o t + 0.5at^2[/tex]

[tex]s = (1.23)(0.112) + (.5)(-9.8)(0.112)^2[/tex]

d = .076 m

Answer:

The mole fraction of hexane is 0.67

Explanation:

This problem can be solved using Dalton´s law and Raoult´s law.

The vapor pressure of a mixture of gases is the sum of the partial pressures of  each gas (Dalton´s law).

Example:

In a mixture of gases A and B

Pt = PA + PB

where:

Pt = total pressure

PA = partial pressure of A

PB = partial pressure of B

In an ideal solution, the vapor pressure of each component is equal to the vapor pressure of the pure component times the mole fraction of the component in the solution (Raoult´s law).

Example:

In a solution containing A and B, the vapor pressure of A in the solution will be:

PA = P(pure A) * Xa

Where

PA = vapor pressure of A in the solution

P(pure A) = vapor pressure of pure A

Xa = mole fraction of A

In our problem, we have that the vapor pressure of the solution is 241 torr.

Then, using Dalton´s law:

Pt = P(hexane) + P(Pentane)

Using Raoult´s law:

P(hexane) = P(pure hexane) * X(hexane)

P(pentane) = P(pure pentane) * X(pentane)

We also know that the sum of the molar fractions of each component in a solution equals 1:

X(hexane) + X(pentane) = 1

X(pentane) = 1 - X(hexane)

Replacing in the Dalton´s law in terms of X(hexane):

Pt = P(pure hexane) * X(hexane) + P(pure pentane) * (1 - X(hexane))

Solving for X(hexane):

Pt = P(pure hex) * X(hex) + P(pure pent) - P(pure pent) * X(hex)

Replacing with the data:

241 torr = 151 torr * X(hex) + 425 torr - 425 torr*X(hex)

-184 torr = -274 torr * X(hex)

-184torr/-274 torr = X(hex)

X(hex) = 0.67

The mole fraction of hexane in the solution is 1.59.

To find the mole fraction of hexane in the solution, we can use Raoult's law, which states that the partial pressure of a component in a solution is equal to the product of the mole fraction of that component and its vapor pressure.

Let's calculate the mole fraction of hexane:

Partial pressure of hexane = mole fraction of hexane * vapor pressure of hexane

241 torr = x * 151 torr

241 torr / 151 torr = x = 1.59

The mole fraction of hexane in the solution is 1.59.

1)

Answer:

[tex]a = 18.68 m/s^2[/tex]

Part b)

[tex]a = 1.9 g[/tex]

Explanation:

Rate of the spinning of the dancer is given as

[tex]f = 2.6 rev/s[/tex]

angular speed is given as

[tex]\omega = 2\pi f[/tex]

[tex]\omega = 2\pi(2.6) = 16.33 rad/s[/tex]

distance of the ear is given as

[tex]r = 7 cm = 0.07 m[/tex]

Part a)

Radial acceleration is given as

[tex]a = \omega^2 r[/tex]

[tex]a = (16.33)^2(0.07)[/tex]

[tex]a = 18.68 m/s^2[/tex]

Part b)

also we know that

[tex]g = 9.81 m/s[/tex]

so now we have

[tex]\frac{a}{g} = \frac{18.68}{9.81}[/tex]

[tex]a = 1.9 g[/tex]

2)

Answer:

Part a)

[tex]v = 226.2 m/s[/tex]

Part b)

[tex]a = 1.304 \times 10^3 g[/tex]

Explanation:

Length of the blades = 4.00 m

frequency of the blades = 540 rev/min

[tex]f = 540 \times \frac{1}{60} = 9 rev/s[/tex]

so angular speed is given as

[tex]\omega = 2\pi f[/tex]

[tex]\omega = 2\pi(9) = 56.5 rad/s[/tex]

Part a)

Linear speed of the tip of the blade is given as

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

[tex]v = (4.00)(56.5)[/tex]

[tex]v = 226.2 m/s[/tex]

Part b)

Radial acceleration of the tip of the blade

[tex]a = \frac{v^2}{r}[/tex]

[tex]a = \frac{226.2^2}{4}[/tex]

[tex]a = 1.28 \times 10^4 m/s^2[/tex]

also we know

[tex]\frac{a}{g} = \frac{1.28 \times 10^4}{9.81}[/tex]

[tex]a = 1.304 \times 10^3 g[/tex]

Answer:

Explanation:

(a)

Gold has an atomic mass of 197 g/mol means 197 g of gold contains 6.023×10^{-23} atoms.

therefore number of atoms in 17.4 g is

[tex]n= \frac{6.23\times10^{23}\times17.4}{197}[/tex]

[tex]5.32\times10^{22}[/tex]

Number of protons in each atom is 79.

therefore Total number of protons is 5.32×10^{22}×79 =4.2×10^24. protons

Their total positive charge is 4.2×10^{24}×1.6×10^{-19) =6.72×10^5 C.

(b)

equal number of protons should be there to neutralize the material therefore number of electrons is 4.2×10^(24.)

Answer:

Assuming the divers are not moving. The diving partner is at 1600 feet.

Explanation:

The sound wave have to  go to the diving partner and return, so its traveling the double of the total distance.

Been X= distance, V= speed, and t= time

[tex]x=v*t\\[/tex]

[tex]WaveTravelDistance=4000*0.8\\\\Distance to the partner=WaveTravelDistance/2\\\\WaveTravelDistance=3200feet\\\\Distance to the partner=1600feet[/tex]

Answer:

The International Astronomical Union (IAU)  has accepted 88 constellations in the sky.

Explanation:

Constellations has been used since the beginnings of civilizations and each one of them named them as they considered appropiate. It means Greeks' constellations were different than the ones described by Chinese, so it was necessary to gather all these constellations and make a great record with all of them, but there was a problem: Some constellations from different civilizations overlaped because they shared the same stars. There was necessary to put some order on this and that is when in 1922 the International Astronomical Union (IAU) defned a set of 88 moderm constellations  that would become the international standard to look at the night sky. Each one of them is unique and does not share stars with the other constellations.  

Answer:

the reason for the acceleration month that the coefficient of kinetic friction is less than the coefficient of satic frictionExplanation:

This exercise uses Newton's second law with the condition that the acceleration is zero, by the time the body begins to slide. At this point the balance of forces is

    fr- w || = 0

The expression for friction force is that it is proportional to the coefficient of friction by normal.

    fr = μ N

When the system is immobile, the coefficient of friction is called static coefficient and has a value, this is due to the union between  the surface, when the movement begins some joints are broken giving rise to coefficient of kinetic friction less than static.  

In consequence a lower friction force, which is why the system comes out of balance and begins to accelerate.

      μ kinetic <μ static

  In all this movement the normal with changed that the angle of the table remains fixed.

Consequently, the reason for the acceleration month that the coefficient of kinetic friction is less than the coefficient of satic friction

Answer:3.95 m/s

Explanation:

Given

Initial velocity(u)[tex]=20.1 m/s[/tex]

launch angle[tex]=47.8^{\circ}[/tex]

Player is 53 m away

Range of projectile

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

[tex]R=\frac{20.1^2\times sin95.6}{9.8}[/tex]

R=41.02 m

so he need to run 53-41.02=11.98 m

Time of flight of projectile [tex]=\frac{2usin\theta }{g}[/tex]

T=3.03 s

thus average speed of boy [tex]s_{avg}=\frac{11.98}{3.03}=3.95 m/s[/tex]

Answer:

11 kg of gas are emmited to the atmosphere

Explanation:

The problem consists in calculating the total mass of the gases produced in the combustion reaction of the carcoal.

In this case we consider the simple combustion reaction:

C+O2 -> CO2

In relation 1:1 molar with the oxygen.

In this case, the limitant reactant is the carbon, so we make the calculations based in the 3kg of carbon combusted.

First we convert the mass of carbon into mol of carbon

m,C=3kg=3000 g of C

Then we have the mol

n,C=3000g/(12 g/mol)=249.77 mol C

This number also corresponds to the number of oxygen mol reacting with the charcoal. Therefore we calculate the mass of oxygen.

m,O2=249.77 mol *(32 g/mol) = 7990 g =7.99 kg, aproximately 8 kg

Then the total mass of gas is the mass of charcoal plus the mass of oxygen

m,total=3kg + 8kg=11kg

Final answer:

The average angular velocity of the cylinder is calculated using the relationship between linear velocity and radius, with the formula ω = v / r. With a final linear velocity of 7.38 m/s and a radius of 5.08 cm, the average angular velocity is about 145.27 rad/s.

Explanation:

To calculate the average angular velocity of the cylinder, we must first understand the relationship between linear velocity and angular velocity. Angular velocity (ω) is related to linear velocity (v) through the radius of the object according to the equation ω = v / r, where r is the radius of the cylinder. Given that the final linear velocity is 7.38 m/s and the radius is 5.08 cm (or 0.0508 meters), we can calculate the average angular velocity.

Calculating the average angular velocity:

Therefore, the average angular velocity of the cylinder is approximately 145.27 rad/s.

Answer:

The number of electrons an atom has relative to a neutral atom of the same element

Explanation:

the oxidation state is a number that indicates the ability to give or receive electrons. This number usually has a sign, which can be positive or negative.

If the oxidation state is a positive number, it means that this atom has the ability to receive such an amount of electrons. If the oxidation state is negative, the element can give that amount of electrons.

Answer:

[tex]\frac{dA}{dt} = 188.5 m^2/s[/tex]

Explanation:

As we know that area of the circle at any instant of time is given as

[tex]A = \pi r^2[/tex]

now in order to find the rate of change in area we will have

[tex]\frac{dA}{dt} = 2\pi r\frac{dr}{dt}[/tex]

here we know that

rate of change of radius is given as

[tex]\frac{dr}{dt}= 1 m/s[/tex]

radius of the circle is given as

[tex]r = 30 m[/tex]

now we have

[tex]\frac{dA}{dt} = 2\pi (30)(1)[/tex]

[tex]\frac{dA}{dt} = 60\pi[/tex]

[tex]\frac{dA}{dt} = 188.5 m^2/s[/tex]

Answer:

K = 1.72 *10^{-11} N/m

Explanation:

given data:

L= 2.14*10^{-6} m

[tex]\Delta = 1.4%* of L[/tex]

           [tex]= \frac{1.4}{100} *2.14*10^{-6}[/tex]

           = 2.99*10^{-8} m

[tex]L' = L - \Delta L[/tex]

[tex]L ' = 2.14*10^{-6} -2.99*10^{-8}[/tex]

[tex]L = 2.11*10^{-6} m[/tex]

electrostatic force = spring force

[tex]\frac{ kq^2}{L'^2} = K \Delta L[/tex]

putting all equation to get spring constant K value

[tex]\frac{8.98*10^9 *1.6*10^{-19}}{2.11*10^{-6}} = K 2.99*10^{-8}[/tex]

K = 1.72 *10^{-11} N/m