This problem has been solved! See the answer A 6.0 kg object, initially at rest in free space, "explodes" into three segments of equal mass. Two of these segments are observed to be moving with equal speeds of 20 m/s with an angle of 60 degrees between their directions of motion. How much kinetic energy is released in this explosion?

Answer:

a) the new angular speed of the student is 0.9642 rad/s

b)

the kinetic energy of the student before the objects are pulled in is 1.9119 J

the kinetic energy of the student after the objects are pulled in is 2.4252 J

Explanation:

Given that;

mass of each object m = 1 kg

distance of objects from axis of rotation r = 0.9 m

Moment of inertia of each object initially [tex]I_{oi}[/tex]

[tex]I_{oi}[/tex] = mr² = 1kg ×(0.9m)² = 1 kg × 0.81 m²  = 0.81 kg.m²

moment of inertia of each object finally [tex]I_{of}[/tex]

[tex]I_{of}[/tex]  = mr² = 1kg × (0.33 m)² = 0.1089 kg.m²

Now

moment of inertia of student plus stool  [tex]I_{}[/tex] = 5 kg.m²

initial angular speed ω₀ = 0.76 rad/sec

final angular speed ω = ?

Now using conservation of angular momentum;

([tex]I_{}[/tex] + 2 [tex]I_{oi}[/tex] )ω₀ = ([tex]I_{}[/tex] + 2 [tex]I_{of}[/tex] )ω

so we substitute

(5 + 2 (0.81) )0.76 = (5 + 2 (0.1089) )ω

5.0312 = 5.2178 ω

ω =  5.0312 / 5.2178

ω  = 0.9642 rad/s

Therefore, the new angular speed of the student is 0.9642 rad/s

b)

K.E of student before = (0.5) ([tex]I_{}[/tex] + 2 [tex]I_{oi}[/tex] )ω₀²

= (0.5) (5 + 2 (0.81) )(0.76)²

= 0.5 × 6.62 × 0.5776

= 1.9119 J

Therefore, the kinetic energy of the student before the objects are pulled in is 1.9119 J

KE of student finally = (0.5) ([tex]I_{}[/tex] + 2 [tex]I_{of}[/tex] )ω²

= (0.5) (5 + 2 (0.1089) ) (0.9642)²

= 0.5 × 5.2178 × 0.9296

= 2.4252 J

Therefore, the kinetic energy of the student after the objects are pulled in is 2.4252 J

Answer:

the answer would be A. electricity don't specify the direction of any cardinal points the flow of charges moves.

Answer:

A

Explanation:

I did the test on ap3x

Answer:

a. [tex]\mathbf{F_{f_2} = 2 F_{f1}}[/tex], [tex]\mathbf { since \ the \ force \ o f \ friction \ is \ represented \ as \ the \ slope \ for \ each \ of \ the \ two \ curves.}[/tex]

Explanation:

From the information given;

Using the work-energy theorem

ΔKE = W = [tex]\mathbf{ F_f \times r}[/tex]

K = [tex]\mathbf{ F_f \times r}[/tex]

∴

[tex]\dfrac{K_1}{K_2} = \dfrac{F_{f1}}{F_{f2}} (\dfrac{r_1}{r_2})[/tex]

Since [tex]K_1 = K_2[/tex] and r_1 = 4, and r_2 = 2 (from the missing diagram which is attached below)

Then;

[tex]1 = \dfrac{F_{f1}}{F_{f2}} (\dfrac{4 \ m}{2 \ m})[/tex]

[tex]\mathbf{F_{f_2} = 2 F_{f1}}[/tex]

Answer:

the neutron is located in the nucleus of an atom

Answer:

nucleus

Explanation:

Atoms are made up of protons and neutrons located within the nucleus, with electrons in orbitals surrounding the nucleus.

Answer:

1399.2 ft

Explanation:

The initial velocity = 180 mph = [(180 * 5280) / (1 * 3600)] ft/s = 264 ft/s

[tex]In\ the \ horizontal\ direction(x)\\\\Initial\ velocity = v_{ox}=264\ ft/s\\\\distance\ travelled\ in\ x \ direction(x) =v_{ox}t\\\\\\For\ the\ vertical\ direction:\\\\initial\ velocity(y_{oy})=0\\\\vertical\ distance(y)=y_{oy}t+0.5gt^2\\\\but\ g\ =-32\ ft/s^2. Hence:\\\\y=0t+0.5(-32)t^2\\\\y=-16t^2\\\\At\ point\ B, y=-450, therefore:\\\\-450=-16t^2\\\\t^2=28.125\\\\t=5.3\ s\\\\The\ distance\ at\ which\ the\ pilot\ should\ release\ the\ water=x=v_{ox}t=264*5.3\\\\x=1399.2\ ft[/tex]

Answer:

a. Same � the electric fields point in opposite directions and therefore cancel at some midpoint.

b. bee 1 has a magnitude of charge less than bee 2

Explanation:

a. Do the bees have the same or opposite signs of charge?

They have the same charge. This is because since same charges would produce electric fields in opposite directions, that is the only way they can cancel out at some point. So, the charges are the same and the electric fields point in opposite directions and therefore cancel at some midpoint.

b. Suppose the net electric field is zero at a distance that is closer to bee 1. Does bee 1 have a magnitude of charge greater than or less than that of bee 2?

Let q be the charge on bee 1 and r its distance from the neutral electric field point. So, it electric field E = kq/r².

Also, let q' be the charge on bee 2 and d its distance from the neutral electric field point. So, it electric field E' = kq'/d².

Since E = E' at the neutral point.

kq/r² = kq'/d²

q/q' = r²/d² = (r/d)²

Given that r < d, so r/d < 1 and (r/d)² < 1

So, q/q' < 1

q < q'

So, the charge on bee 1 is less than that on bee 2  

Answer:

a)   f ’’ = f₀ [tex]\frac{1 + \frac{v}{c} }{1- \frac{v}{c} }[/tex] , b)   Δf = 2 f₀ [tex]\frac{v}{c}[/tex]

Explanation:

a) This is a Doppler effect exercise, which we must solve in two parts in the first the emitter is fixed and in the second when the sound is reflected the emitter is mobile.

Let's look for the frequency (f ’) that the mobile aorta receives, the blood is leaving the aorta or is moving towards the source

                    f ’= fo[tex]\frac{c+v}{c}[/tex]

This sound wave is reflected by the blood that becomes the emitter, mobile and the receiver is fixed.

                   f ’’ = f’ [tex]\frac{c}{ c-v}[/tex]

where c represents the sound velocity in stationary blood

therefore the received frequency is

                 f ’’ = f₀   [tex]\frac{c}{c-v}[/tex]

let's simplify the expression

                f ’’ = f₀ \frac{c+v}{c-v}

                f ’’ = f₀ [tex]\frac{1 + \frac{v}{c} }{1- \frac{v}{c} }[/tex]

b) At the low speed limit v <c, we can expand the quantity

                 (1 -x)ⁿ = 1 - x + n (n-1) x² + ...

                 [tex]( 1- \frac{v}{c} ) ^{-1} = 1 + \frac{v}{c}[/tex]

                f ’’ = fo [tex]( 1+ \frac{v}{c}) ( 1 + \frac{v}{c} )[/tex]

                f ’’ = fo [tex]( 1 + 2 \frac{v}{c} + \frac{v^2}{ c^2} )[/tex]

leave the linear term

               f ’’ = f₀ + f₀ 2[tex]\frac{v}{c}[/tex]

the sound difference

               f ’’ -f₀ = 2f₀ v/c

               Δf = 2 f₀ [tex]\frac{v}{c}[/tex]

Answer is C

Explanation:

Answer:

C. wavelength

good luck, i hope this helps :)

Answer:

The answer is below

Explanation:

The amplitude decreases by 2%  during each oscillation. Hence the decrease in amplitude can be represented by an exponential decay in the form:

y = abˣ; where x ad y are variables, a is the initial value and b is the factor.

Let y represent the amplitude after x oscillations. Since the initial amplitude is 10 cm, hence:

a = 10 cm, b = 2% = 0.02.

Therefore:

y = 10(0.02)ˣ

The amplitude after 25 oscillations is gotten by substituting x = 25 into the equation. Hence:

y = 10(0.02)²⁵

y= 3.355 * 10⁻⁴² cm

The amplitude after 25 oscillations is 3.355 * 10⁻⁴² cm

Answer:

Only the energy of the wave travels through the medium. In a transverse wave, particles of the medium vibrate up and down perpendicular to the direction of the wave. ... In a surface wave, particles of the medium vibrate both up and down and back and forth, so they end up moving in a circle.

Speed is the time rate of an object moving from one place to another, while velocity is the rate and direction of the object's movement. They are very similar but they don't mean the same thing.

Answer:

 rₙ = 1,325 10⁻⁹ m

Explanation:

To solve this problem we use the bohr atomic model

           Eₙ = -13.606 /n²     [eV]

the brackets indicate that the units are in electron volts.

let's reduce the photon energy to eV

        E = 4.5 10-19J (1 eV / 1.6 10⁻¹⁹ eV) = 2.8125 eV

This energy is in the visible range, so the transition must occur in this range, this is for the Balmer series whose initial number is n₀ = 2

for an atomic transition on two levels

         ΔE = Eₙ - E₀ = [tex]\frac{-13.606}{n^2} + \frac{13.606}{2^2}[/tex]

         2.8125 = [tex]\frac{-13.606}{n^2} + 3.4015[/tex]

         [tex]\frac{13.606}{n^2}[/tex] = 3.4015 - 2.8125 = 0.589

           n² = 13.606 / 0.589

           n² = 23.1

           n = 4.8

as n must be an integer

           n = 5

taking the quantum number as far as the electron goes, we substitute in the equation for the radius

           rn = n² a₀

where ao is the radius of the lowest level a₀ = 5.3 10⁻¹¹ m

            rₙ = 5 2 5.3 10⁻¹¹

            rₙ = 132.5 10⁻¹¹ m

            rₙ = 1,325 10⁻⁹ m

Answer:

Option B, Fix the piston in place so the volume of the pas remains constant

Explanation:

As we know

[tex]\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}[/tex]

The effect on variable due to another variable can be studied by keeping the third variable constant.

Hence, in order the study the variation of temperature with pressure or vice versa, the volume needs to fixed at a certain value.

Hence, option B is correct

Answer:

5760 J

Explanation:

From the question given above, the following data were obtained:

Mass of block = 48 kg

Height (h) = 12 m

Gravitational field strength (g) = 10 N/Kg

Gravitational potential energy (PE) =?

The gravitational potential energy stored by the block can simply be obtained as follow:

PE = mgh

PE = 48 × 10 × 12

PE = 5760 J

Therefore, the gravitational potential energy stored by the block is 5760 J

Answer:

       y =[tex]\frac{1.22L}{D}[/tex] [tex]\sqrt{\frac{h^2 m}{2eV} }[/tex]

Explanation:

Let's solve this exercise in parts. Let's start by finding the wavelength of the electrons accelerated to v = 10 103 V, let's use the DeBroglie relation

             λ= [tex]\frac{h}{p} = \frac{h}{mv}[/tex]

Let's use conservation of energy for speed

starting point

             Em₀ = U = e V

final point

             Em_f = K = ½ m v²

             Em₀ = Em_f

             eV = ½ m v²

             v =[tex]\sqrt{\frac{2eV}{m} }[/tex]

we substitute

             λ=  [tex]\sqrt{ \frac{h^2 m}{2eV}}[/tex]

the diffraction phenomenon determines the minimum resolution, for this we find the first zero of the spectrum

            a sin θ = m λ

first zero occurs at m = 1, also these experiments are performed at very small angles

            sin θ = θ

            θ = λ / a

This expression is valid for linear slits, in the microscope the slits are circular, when solving the polar coordinates we obtain

           θ = 1.22 λ / D

where D is the diameter of the opening

we substitute

          θ = [tex]\frac{1.22}{D}[/tex]   \sqrt{ \frac{h^2 m}{2eV}}  

this is the minimum angle that can be seen, if the distance is desired suppose that the distance of the microscope is L, as the angles are measured in radians

            θ = y / L

when substituting

where y is the minimum distance that can be resolved for this acceleration voltage

            y =[tex]\frac{1.22L}{D}[/tex] [tex]\sqrt{\frac{h^2 m}{2eV} }[/tex]

Answer:

meteor

Explanation:

A asteroid stays still and a meteor goes fast

Answer:

meteor

Explanation:

or some people call it a shooting star

Answer:

Explanation:

The only force acting on the cart is a component of its weight parallel to ramp downwards . No other force acts parallel to the ramp .

Even when the cart is moving up after the initial push , its weight is acting downwards so acceleration is acting downwards .

When the cart is stationary at the top position , its weight is acting downwards so acceleration is downwards at that moment also . When the cart is going downwards , still its weight is acting down so acceleration is acting downwards .

After this push, as the car moves up the ramp, the direction of the acceleration of the cart is _down _______ the ramp. After the reaches its highest point, turns around, and begins moving down the ramp, the direction of the acceleration of the cart is _down ________ the ramp. At the highest point the cart reaches on the ramp, when the cart momentarily comes to rest, the magnitude of the acceleration of the cart is _downwards ______.

Answer:

a) t = 0.25 s,  b)  x = 0.075 m

Explanation:

a) For this exercise we will use kinematic relationships in one dimension

         v = v₀ + a t

in the problem they indicate the initial velocity v₀ = 0.15 m / s, the final velocity v = 0.45 m / s and the acceleration of the squid a = 1.2 m / s²

          t = [tex]\frac{v -v_o}{a}[/tex]

we calculate

         t = [tex]\frac{0.45 - 0.15}{1.2}[/tex]

         t = 0.25 s

b) We can also find the distance traveled during this acceleration

         v² = v₀² + 2a x

         x = [tex]\frac{v^2 -v_o^2 }{2a}[/tex]

let's calculate

         x = [tex]\frac{0.45^2 - 0.15^2 }{2 \ 1.2}[/tex]

         x = 0.075 m

Answer:

V = 0.48 m/s

Explanation:

In this case, we need to analyze the given data by parts.

At first, we know that the woman is on a height of 84.5 m of a river. She drops a stone thinking that she may hit the raft that is traveling with a constant speed. When the raft is 6 m near the bridge, the woman drops the stone, and the stone hits the water when the raft is still 4 m far of the bridge.

With this given data, we can calculate the distance covered by the raft, because is traveling at a constant speed:

X = 6 - 4 = 2 m

And as it's traveling at constant speed then:

X = V.t

We have the distance of the raft, but not the time it took to cover that distance. This time will be the same time that the stone took to hit the water, therefore, if we can determine the time of the rock, well be determining the time of the raft to cover the distance, and then, we can determine it speed.

To determine the time of the rock, as the stone is going on a free fall, with an innitial speed of 0, the flight time of the rock will be:

y = gt²/2 ---> solving for t

2y/g = t²

t = √2y/g

If g = 9.8 m/s, and replacing the data we have that the flight time of the rock is:

t = √2*84.5 / 9.8

t = 4.15 s

So the rock took 4.15 s to hit the water, and it's also the time that the raft took to cover the distance of 2 m, then, it's speed:

V = X/t

V = 2 / 4.15

Hope this helps

Answer:

add them

100+150 = 250N same direction

that's the resultant

same direction

Answer:

The answer is below

Explanation:

a) Coulomb's law of electric force for charges at rest states that Like charges repel each other while unlike charges attract one another.

Therefore since the microbes has small positive charges, the microbe would be repelled by the positively charged electrodes and attracted by the negative charged electrodes.

Hence, the microbes would collect on the negatively charged electrodes.

b) The microbes can easily removed from the negative electrode for analysis by discharging the electrode from the source. Thereby making the electrode to be incapable of attracting the microbe.

Answer: Electrical energy is converted into other forms of energy, such as sound.

Explanation: Using the law of conservation of energy we know that energy can never be destryped it can be transferred or be transformed into from one form to another.

Electrical energy is converted into other forms of energy, such as sound.

The law of conservation of energy or matter, states that energy can neither be created nor destroyed but can be converted from one form to another.

When listening to music on radio, the electric energy supplied to the radio will be converted to mechanical energy of the moving parts of the radio which is then converted to sound energy.

Thus, we can conclude that electrical energy is converted into other forms of energy, such as sound.

Learn more here:https://brainly.com/question/2828402

Answer:a

Explanation:

Because its has to use tihs potential energy to overcome the atmosphere so the shuttle will not go back down

Answer:

41

Explanation:

Answer: 134 = 143 = 151 = 166 = 176

Hope this helps!!

Sorry if it's incorrect!!

:'(

Answer:

MGH=energy

3500*9.8*h=90000

h=90000/34300

h=2.62m

No. The object has to have motion for it to have kinetic energy.