A flywheel having constant angular acceleration requires 4.70 s to rotate through 164 rad . Its angular velocity at the end of this time is 101 rad/s . Find the angular velocity at the beginning of the 4.70 s interval. Find the angular acceleration of the flywheel.

Answer:

Energy consumed is 0.00033 Joules.

Explanation:

the formula of Energy is:

Energy = power/ time.

Answer:

Most importantly metals can pass an electric current without being affected and changed by the electricity. Electrical conductivity combined with ductility makes metals the most suitable materials for electrical transmission wires.

Answer:

23

Explanation:

3x-4

Answer:

jsbdhdndmlsusgsbkaksudgnslsosufhbf ffb

Explanation:

here the file has everything

This temperature would be 170° F on the Fahrenheit scale. Hence, option (C) is correct.

The physical concept of temperature indicates in numerical form how hot or cold something is. A thermometer is used to determine temperature. Thermometers are calibrated using a variety of temperature scales, which historically defined distinct reference points and thermometric substances.

The most popular scales are the Celsius scale, sometimes known as centigrade, with the unit symbol °C, the Fahrenheit scale (°F), and the Kelvin scale (K), with the latter being mostly used for scientific purposes.

the relation between Kelvin scale and Fahrenheit scale is given by:

(F - 32)/180 = (K - 273)/100

F - 32 = (350 - 273)(9/5)

F = 32 + (350 - 273)(9/5)

F = 170

Hence,  this temperature would be 170° F on the Fahrenheit scale.

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Answer:

6e-12

Explanation:

divide the time value by 1e+12

[tex]▪▪▪▪▪▪▪▪▪▪▪▪▪  {\huge\mathfrak{Answer}}▪▪▪▪▪▪▪▪▪▪▪▪▪▪[/tex]

Distance travelled by the truck is ~

And it's displacement is ~

[tex] \large \boxed{ \mathfrak{Step\:\: By\:\:Step\:\:Explanation}}[/tex]

See the diagram in attachment for reference ~

Let O be the initial point, It travels 60 km towards west till point B and then 80 km towards north till point P and returns to initial point O in a straight line, now as we can observe here, it forms a right angled Triangle.

The measure of two legs is 60 km and 80 km, let's find the hypotenuse ~

According to Pythagoras theorem ~

hypotenuse² = sum of squares of other two legs

that is ~

So, the distance between the point A and O is 100 km

Now, The total distance is equal to the distance covered through actual path that is ~

And displacement is the distance between the final point and initial point, but since the truck returns to the point from where it started the journey, so the final and initial point is same therefore displacement is equal to 0.

The average rate at which the neon atoms collide with one side of the container is [tex]3.31 \times 10^{26} \ atoms/s[/tex].

The given parameters;

The average kinetic energy of the gas molecules is calculated as follows;

[tex]K = \frac{3}{2} \frac{R}{N_a} T\\\\K = \frac{3 \times 8.314\times 298}{2\times 6.022 \times 10^{23}} \\\\K = 6.17\times 10^{-21} \ J/atoms[/tex]

The average speed of the gas molecules is calculated as follows;

[tex]K = \frac{1}{2}mv_{rms}^2\\\\v_{rms} = \sqrt{\frac{2K}{m} } \\\\v_{rms} = \sqrt{\frac{2\times 6.17 \times 10^{-21}}{3.35\times 10^{-26}} } \\\\v_{rms} = 607 \ m/s[/tex]

The time of collision of the gas molecules with the walls of the container is calculated as follows;

[tex]t = \frac{2d}{v} \\\\t = \frac{2\times 0.368}{607} \\\\t = 0.0012 \ s[/tex]

The average rate at which the gases collide with a single wall out of the 3 identical walls is calculated as follows;

[tex]rate =\frac{1}{3} \times \frac{n \times N_a}{t} \\\\rate = \frac{1.98 \times 6.02 \times 10^{23} \ atoms}{3 \times 0.0012 \ s} \\\\rate = 3.31 \times 10^{26} \ atoms/s[/tex]

Thus, the average rate at which the gases collide with one side of the container is [tex]3.31 \times 10^{26} \ atoms/s[/tex].

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Answer:

Option c.

Explanation:

The characteristics of the expression of the simple harmonic motion allows to find the results for the expression of the mass- block system are:

     A) The constant Ces: C = xinit

     B) The ocsntna S is: S = 0

     C) The equation of the system is: x = xinit cos wt

     D) If the reference system is at some extreme, the equation is:

              [tex]L - x_{init} = x_{init} \ cos \ wt[/tex]  

The simple harmonic movement is an oscillatory movement where the restoring force is proportional to the displacement, the general equation that describes this movement is indicated.

          x = C cos wt + S sin wt

Where x is the displacement C and S are constants. W the angular velocity and t the time.

A) The initial position of the body occurs when the time is zero, t = 0

We substitute.

           x = C cos 0 + S sin 0

           [tex]x_{init}[/tex] = C

B) The velocity of the particle is defined.

          [tex]v= \frac{dx}{dt} \\ v= C w \ sin \ wt - Sw \ cos \ wt[/tex]  

The initial velocity occurred for time zero t = 0

          v = - S w

It indicates that the initial velocity is zero, since the angular velocity must be different from zero, it implies that the constant is valid.

          S = 0

C) The equation for the block remains.

           x (t) = [tex]x_{init} \ cos \ wt[/tex]  

D) In ​​some cases it is measured with respect to another reference system, the most common are:

In these cases, the change that must be made is

          x =  [tex]L - x_{min}[/tex] t

we substitute

          [tex]L - x_{init} = x_{init} \ cos \ wt[/tex]  

          L = [tex]x_{init}[/tex]  (1 + cos wt)

In conclusion, using the characteristics of the expression of the simple harmonic motion we can find the results for the expression of the mass- block system are:

    A) The constant Ces: C = xinit

     B) The ocsntna S is: S = 0

     C) The equation of the system is: x = xinit cos wt

     D) If the reference system is at some extreme, the equation is:

              [tex]L - x_{init} = x_{init} \ cos \ wt[/tex]  

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Answer:

Explanation:

In the vertical analysis assuming launch from ground level.

0 = 0 + (41.3sinθ)(5.1) + ½(-9.8)5.1²

(41.3sinθ)(5.1) = ½(9.8)5.1²

(41.3sinθ) = ½(9.8)5.1

sinθ = ½(9.8)5.1/41.3

sinθ = 0.60508...

θ = 37.235°

vx = 41.3cos37.235

vx = 32.881452...

vx = 32.9 m/s

Answer:

Assuming there are no energy losses due to friction or drag, the gravitational potential energy will change into kinetic potential energy as the car reaches the bottom of the slope.

G.P.E = m*g*h

K.E = (m*v^2)/2

where

m = mass of toy car (kg)

g = gravity (m/s^2)

h = heigh of your car from the bottom (m)

v = velocity of the toy car as it reaches the bottom (m/s)

Equate K.E to G.P.E

G.P.E = K.E

m*g*h = (m*v^2)/2

make v the subject of the formula

v = (2*g*h)^(1/2)

Substitute g = 9.81 m/s^2 and h = 2m into the equation to get v

v = (2*9.81*2)^(1/2)

v = 6.264 m/s

Answer:

The box arrives first.

Explanation:

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According to the information, a solid sphere is an object that arrives at the bottom first. This is because it occupies less friction as compared to the box.

Friction may be defined as the resistance that is offered by the surfaces that are in contact when they move past each other. It is a type of force that opposes the motion of a solid object over another.

There are mainly four types of friction: static friction, sliding friction, rolling friction, and fluid friction. According to the context of this question, the sphere possesses less friction as compared to the box. This is because the box has an irregular surface that possesses high friction over the inclined surface.

Therefore, according to the information, a solid sphere is an object that arrives at the bottom first. This is because it occupies less friction as compared to the box.

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Answer:

Explanation:

Echo is caused by sound energy reflecting off of "hard" surfaces. It could be as simple as a change in density of the material the sound is traveling through.

In 0.8 s, the sound has traveled 0.8(1500) = 1200 m.

That means the object that reflected the sound is 600 m below the boat. The sound took 0.4 s to reach the object and another 0.4 s to return the echo.

-cant travel through space since there's no molecules to travel through

-sound travels 4.3 times faster in water than air

-sounds are waves that pass through our ears via vibrations and travel by vibrations of molecules

-different types of sound like audible, inaudible, infrasonic, ultrasonic,

-sounds waves are either longitudinal, mechanical and pressure waves

-sound travels at 767 miles per hour

Answer:

Explanation:

a) The spring force will equal the weight.

b) If up is positive

kx - mg = 0

mg = kx              kx = 25 N

c) m = kx/g = 25/10 = 2.5 kg

Answer:

Approximately [tex]2.5\; \rm m\cdot s^{-1}[/tex], assuming that no external force (e.g., gravitational pull) was acting on this rocket.

Explanation:

Assume that no external force is acting on this rocket. The system of the rocket and the fuel on the rocket would be isolated (an isolated system.) The momentum within this system would be conserved.

Let [tex]v_{0}\; \rm m\cdot s^{-1}[/tex] be the initial velocity of the rocket.

The velocity of the exhaust gas would be [tex](v_{0} - 5000)\; \rm m\cdot s^{-1}[/tex] since the gas is ejected away from the rocket.

Let [tex]\Delta v\; \rm m\cdot s^{-1}[/tex] denote the increase in the velocity of the rocket. The velocity of the rocket after ejecting the gas would be [tex](v + \Delta v)\; \rm m\cdot s^{-1}[/tex].

The momentum [tex]p[/tex] of an object of velocity [tex]v[/tex] and mass [tex]m[/tex] is [tex]p = m \cdot v[/tex].

The combined mass of the rocket and the fuel was [tex]10000\; \rm kg[/tex]. The initial momentum of this rocket-fuel system would be:

[tex]\begin{aligned}p_{0} &= m \cdot v\\ &= 10000\; {\rm kg} \times v_{0}\; {\rm m \cdot s^{-1}} \\ &= (10000\; v_{0})\; \rm {kg \cdot m\cdot s^{-1}}\end{aligned}[/tex].

The momentum of the [tex]5.0\; \rm kg[/tex] of fuel ejected at [tex](v_{0} - 5000)\; \rm m\cdot s^{-1}[/tex] would be:

[tex]\begin{aligned} & 5.0 \; {\rm kg} \times (v_{0} - 5000)\; {\rm m\cdot s^{-1}}\\ =\; & (5.0\, v_{0} - 25000)\; {\rm kg \cdot m \cdot s^{-1}}\end{aligned}[/tex].

After ejecting the [tex]5.0\; \rm kg[/tex] of the fuel, the mass of the rocket would be [tex]10000\; \rm kg - 5.0\; \rm kg = 9995\; \rm kg[/tex]. At a velocity of [tex](v + \Delta v)\; \rm m\cdot s^{-1}[/tex], the momentum of the rocket would be:

[tex]\begin{aligned} & 9995 \; {\rm kg} \times (v_{0} + \Delta v)\; {\rm m\cdot s^{-1}}\\ =\; & (9995\, v_{0} + 9995\, \Delta v)\; {\rm kg \cdot m \cdot s^{-1}}\end{aligned}[/tex].

Take the sum of these two quantities to find the momentum of the rocket-fuel system after the fuel was ejected:

[tex]\begin{aligned}p_{1} &= (5.0\, v_{0} - 25000)\; {\rm kg \cdot m\cdot s^{-1}\\ &\quad\quad + (9995\, v_{0} + 9995\, \Delta v)\; {\rm kg \cdot m \cdot s^{-1}} \\ &= (10000\, v_{0} + 9995\, \Delta v - 25000)\; {\rm kg \cdot m \cdot s^{-1}}\end{aligned}[/tex].

The momentum of the rocket-fuel system would be conserved. Thus [tex]p_{0} = p_{1}[/tex].

[tex](10000\, v_{0})\; {\rm kg \cdot m\cdot s^{-1}} = (10000\, v_{0} + 9995\, \Delta v - 25000)\; {\rm kg \cdot m \cdot s^{-1}}[/tex].

Solve this equation for [tex]\Delta v[/tex], the increase in the velocity of the rocket.

[tex]10000\, v_{0} = 10000\, v_{0} + 9995\, \Delta v - 25000[/tex].

[tex]9995\, \Delta v = 25000[/tex].

[tex]\begin{aligned}\Delta v &= \frac{25000}{9995} \approx 2.5\end{aligned}[/tex].

Thus, the velocity of the rocket would increase by approximately [tex]2.5\; \rm m\cdot s^{-1}[/tex] after ejecting the [tex]5.0\; \rm kg[/tex] of fuel.

The work done by the man against friction is 4,192.86 J.

The given parameters;

The work done by the man against friction is calculated as follows;

[tex]W = F \times d \times cos(75)\\\\W = 540 \times 30 \times cos(75)\\\\W = 4,192.86 \ J[/tex]

Thus, the work done by the man against friction is 4,192.86 J.

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Answer:

Explanation:

conservation of momentum during the collision

0.035(214) + 0.15(0) = 0.185v

v = 40.486 m/s

The kinetic energy after impact will convert to gravity potential energy

(ignoring air resistance)

mgh = ½mv²

     h = v²/2g

     h = 40.486² / (2(9.8))

     h = 83.6303...

     h = 84 m

Answer:

Explanation:

Plane 2 is moving north at

8.5sin20 = 2.9 m/s

Plane 2 is moving west at

8.5cos20 = 8.0 m/s

Part A

v = √((13 - 2.9)² + 8.0²) = 12.876... 13 m/s

Part B

θ = arctan((13 - 2.9) / 8.0) = 51.617... 52° N of E

Part C

13 m/s  52° S of W

relative velocity magnitude is independent of reference frame

Answer:

Answer: At its lowest point, the kinetic energy of a watermelon just as it touches the ground is zero if it does not touch anything on its way down.

Explanation:

This is because that upon having been dropped from a height, an object no longer has any kinetic energy at all. Kinetic energy transforms to gravitational potential energy during the fall and there's nothing left over for kinetic once you stop accelerating anymore. Fortunately, things don't stay still until they land very often! For example, if a person catches the fruit with his hands after some air resistance slows him down - making him more similar in speed to the lag of the trajectory - then he'll be able to share some of his saved up gravitational potential with that watermelon and do some

Answer:

[tex]20m/s^2[/tex]

Explanation:

Solution is attached. I apologize if it is a little messy.

We have that for the Question "find (a) the mass of CCl4 per second that passes point A and (b) the concentration of CCl4 at point A."

Answers:

From the question we are told

The concentration of [tex]CCl_4[/tex] at the left end of the tube is maintained at 1.71 x 10-2 kg/m3, and the diffusion constant is 21.9 x 10-10 m2/s. The CCl4 enters the tube at a mass rate of 5.86 x 10-13 kg/s

A) the mass flow rate of CCI_4 as it passes point A is the same as the mass flow rate at which CCI_4 enters the left end of the tube

Therefore, the mass flow rate of CCI_4 at point A

=  [tex]5.86*10^{-13} kg/s[/tex]

B) From Fick's law

[tex]\deltaC = \frac{mL}{DAt}\\\\ Assume L = 5*10^{-3}, A = 3*10^{-4}\\\\\deltaC = \frac{5.86*10^{-13} * 5*10^{-3}}{21.9*10^{-10} * 3*10^{-4}}\\\\\deltaC = 4.46*10^{-3}kg/m^3[/tex]

Then,

[tex]Concentration = 1.71*10^{-2} - 4.46*10^{-3}\\\\= 12.6*10^{-3}kg/m^3[/tex]

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Answer:

Its acceleration will increase.

Explanation:

For an object spinning around a central point, Its acceleration will increase if its distance from the centre is decreased.

The rate at which an item changes its velocity is known as acceleration, a vector quantity. If an object's velocity is changing, it is acceleration. The net acceleration that objects get as a result of the combined action of gravity and centrifugal force is known as the Earth's gravity

This is the acceleration of an object in a circle of radius r at a speed v. So, centripetal acceleration is greater at high speeds and in sharp curves smaller radii and for lager radii acceleration will be less.

acceleration, a = v²/r

For an object spinning around a central point, Its acceleration will increase if its distance from the centre is decreased.

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Answer:

Explanation:

The deflection of a charged particle by a magnetic field is proportional to its electric charge and to its velocity.   The deflection is also inversely proportional to its mass.   So given a proton and an electron going at the same velocity in a magnetic field and having equal (but opposite) electric charge the electron will deflect much more since the ratio of the masses is 1836.

The volume of a material is the total amount of matter that it can contain. The volume of the given coin has been determined to be 5.85 x [tex]10^{-6}[/tex] [tex]m^{3}[/tex]. Since the gold doubloon do not absorb water, then its volume remains constant at the ocean floor.

The volume of the gold doubloon can be determined by;

volume = [tex]\pi r^{2}[/tex] + h

where r is the radius of the coin and h is its thickness.

Such that; diameter = 6.1 cm (61 mm) and h = 2.0 mm

r = [tex]\frac{diameter}{2}[/tex]

 = [tex]\frac{61}{2}[/tex]

r = 30.5 mm

Thus,

volume of the coin = [tex]\frac{22}{7}[/tex] x [tex](30.5)^{2}[/tex] x 2

                               = 5847.2857

Therefore, the volume of the gold doubloon is 5847.3 [tex]mm^{3}[/tex]. This can also be expressed as 5.85 x [tex]10^{-6}[/tex] [tex]m^{3}[/tex].

Since the gold doubloon is  not miscible with water, thus its volume at a depth of 770 m at the ocean floor is the same as its initial volume.

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In 60 minutes or 3600 seconds, the tip of the minute hand traverses the circumference of a circle with radius 3.00 cm, so it moves with a tangential speed of

(3.00 cm)/(3600 s) ≈ 0.00083 cm/s = 8.3 μm/s