g Incandescent bulbs generate visible light by heating up a thin metal filament to a very high temperature so that the thermal radiation from the filament becomes visible. One bulb filament has a surface area of 30 mm2 and emits 60 W when operating. If the bulb filament has an emissivity of 0.8, what is the operating temperature of the filament

Answers

Answer 1

Answer:

2577 K

Explanation:

Power radiated , P = σεAT⁴ where σ = Stefan-Boltzmann constant = 5.6704 × 10⁻⁸ W/m²K⁴, ε = emissivity of bulb filament = 0.8, A = surface area of bulb = 30 mm² = 30 × 10⁻⁶ m² and T = operating temperature of filament.

So, T = ⁴√(P/σεA)

Since P = 60 W, we substitute the vales of the variables into T. So,

T = ⁴√(P/σεA)

= ⁴√(60 W/(5.6704 × 10⁻⁸ W/m²K⁴ × 0.8 × 30 × 10⁻⁶ m²)

= ⁴√(60 W/(136.0896 × 10⁻¹⁴ W/K⁴)

= ⁴√(60 W/(13608.96 × 10⁻¹⁶ W/K⁴)

= ⁴√(0.00441 × 10¹⁶K⁴)

= 0.2577 × 10⁴ K

= 2577 K


Related Questions

if you watch football let me know who you think is going to win super bowl 55 and what do you think the score going to be Kansas city chiefs or tampa bay buccaneers

Answers

Answer:

I think the bucs are gonna win because Tom Brady is on their team and it's rigged

but maybe I'm just thinking negatively lol

I don’t know much rlly but what this person said

Magnification of lens is 1. What does it mean?

Answers

Answer:

It means when you look into the lens your vision magnifies by x1

Explanation:

How could a change in straight line motion due to unbalanced forces be predicted from an understanding of inertia?

Answers

Answer:

If the force goes in the direction of movement, the speed must increase and if the net force goes in the opposite direction, the speed must decrease.

Explanation:

The principle of inertia or Newton's first law states that every body remains static or with constant velocity if there is no net force acting on it.

Based on this principle, if we have a net force, the velocity of the body changes by having an unbalanced force acting.

If the force goes in the direction of movement, the speed must increase and if the net force goes in the opposite direction, the speed must decrease.

A vertical piston-cylinder device contains a gas at a pressure of 100 kPa. The piston has a mass of 10 kg and a diameter for 14 cm. Pressure of the gas is to be increased by placing some weights on the piston. Determine the local atmospheric pressure and the mass of the weights that will doublethe pressure of the gas inside the cylinder.

Answers

Answer:

the local atmospheric pressure is  93.63 kPa

the mass of the weights is 156.9 kg

Explanation:

Given that;

Initial pressure of gas = 100 kPa

mass of piston = 10 kg and diameter = 14 cm = 0.14 m

g = 9.81 m/s²

Now,

P_gas = P_atm + P_piston

100 = P_atm + P_piston --------- let this equation 1

P_piston = M_piston × g / A = (10 × 9.81) / π/4×d²

P_piston = 98.1 / (π/4×( 0.14 )²)

P_piston = 98.1 / 0.01539 = 6374,269 Pa = 6.37 kPa

now, from equation 1

100 = P_atm + P_piston

we substitute

100 = P_atm + 6.37

P_atm = 100 - 6.37

P_atm = 93.63 kPa

Therefore, the local atmospheric pressure is  93.63 kPa

Now for pressure of the gas in the cylinder ⇒ 2×initial pressure

Pgas_2 = 2 × 100 = 200 kPa

Pgas_2 = P_atm + P_piston + P_weight

Pgas_2 =  P_gas  + P_weight

we substitute

200 kPa =  100 kPa  + P_weight

P_weight =  200 kPa -  100 kPa

P_weight = 100 kPa =  100,000 Pa

Also;

P_weight = M×g / A

100,000 Pa = ( M × 9.81 ) / (π/4 × (0.14)²)

100,000 × 0.01539 = M × 9.81

1539 = M × 9.81

M = 1539 / 9.81

M = 156.9 kg

Therefore, the mass of the weights is 156.9 kg

take a picture of an object in your house, describe the
energy stores and transfers that happen with it. You can be as imaginative as you wish
with the object (choose something unusual), but the stores you identify and transfers
that happen must be real.


pls give me ideas of what to take a photo of for this I'm really stuck :(​

Answers

A charger or a battery

One of the disadvantages of experimental research is that __________.
A.
it isn’t easily replicated
B.
it doesn’t often reflect reality
C.
the results aren’t generalizable
D.
conditions are not controllable


Please select the best answer from the choices provided

Answers

C I believe that would be it

Answer:

B

Explanation:

Two objects travel the same distance. The one that is moving faster will:


Take more time to go the distance

Take less time to go the same distance

Take the same time as the slower object

None of the above

Answers

Answer: take less time to go the same distance

Explanation:

Because if it is going faster let’s say mph 60 mph is 60 miles per hour if you are going 40 miles per hour it will take you longer to get to your destination.

a wooden block is cut into two pieces, one with three times the mass of the other. a depression is made in both faces of the cut so that a fire cracker can be placed in it and the block is reassembled. the reassembled block is set on rough surface and the fuse is lit. when the fire cracker explodes, the two blocks separate. what is the ratio of distances traveled by blocks?

Answers

Answer:

1/9

Explanation:

Let A denote the bigger piece and let B denote the smaller piece.

We are told that one with three times the mass of the other.

Therefore, we have;

M_a = 3M_b

Firecracker is placed in the block and it explodes and thus, momentum is conserved.

Thus;

V_ai = V_bi = 0

Where V_ai is initial velocity of piece A and V_bi is initial velocity of piece B.

Since initial momentum equals final momentum, we have;

P_i = P_f

Thus;

0 = (M_a × V_af) + (M_b × V_bf)

Since M_a = 3M_b, we have;

(3M_b × V_af) + (M_b × Vbf) = 0

Making V_af the subject, we have;

V_af = -⅓V_bf

The kinetic energy gained by each block during the explosion will later be lost due to the negative work done by friction. Thus;

W_f = -½M_b•(v_bf)²

Now, let's express the work is in terms of the force and the distance.

Thus;

W_f = F_f × Δx × cos 180°

Frictional force is also expressed as μmg

Thus;

W_f = -μM_b × g × Δx

Earlier, we saw that;

W_f = -½M_b•(v_bf)²

Thus;

-½M_b•(v_bf)²= -μM_b × g × Δx

Δx = (v_bf)²/2μg

Let the distance travelled by block A be Δx_a and that travelled by B be Δx_b

Thus;

Δx_a/Δx_b = ((v_ba)²/2μg)/((v_bf)²/2μg)

Δx_a/Δx_b = ((v_af)²/((v_bf)²)

Δx_a/Δx_b = (-⅓V_bf)²/(V_bf)²

Δx_a/Δx_b = 1/9

Car À moves at a speed of 8m/s for 43 seconds. Car B moves at a speed of 7 m/s for 50 seconds. Which car traveled a longer distance

Please show working

Answers

Distance = (speed) x (time)

Car A: Distance = (8 m/s) x (43 s)  =  344 meters

Car B: Distance = (7 m/s) x (50 s)  =  350 meters

350 meters is a longer distance than 344 meters.

Car-B traveled a longer distance than Car-A did.

Answer:

[tex]\boxed {\boxed {\sf Car \ B : 350 \ meters }}[/tex]

Explanation:

Distance is equal to the product of speed and time.

[tex]d=s*t[/tex]

1. Car A

Car A has a speed of 8 meters per second and travels for 43 seconds.

[tex]s= 8 \ m/s \\t= 43 \ s[/tex]

Substitute the values into the formula.

[tex]d= 8 \ m/s *43 \ s[/tex]

Multiply and note that the seconds will cancel out.

[tex]d= 8 \ m*43= 344 \ m[/tex]

2. Car B

Car B has a speed of 7 meters per second and travels for 50 seconds.

[tex]s= 7 \ m/s \\t= 50 \ s[/tex]

Substitute the values in and multiply.

[tex]d= 7 \ m/s * 50 \ s[/tex]

[tex]d= 7 \ m * 50 = 350 \ m[/tex]

350 meters is a longer distance than 344 meters, so Car B traveled the longer distance.

Consider a swimmer that swims a complete round-trip lap of a 50 m long pool in 100 seconds. The swimmer's... average speed is 0 m/s and average velocity is 0 m/s. average speed is 0.5 m/s and average velocity is 0.5 m/s. average speed is 1 m/s and average velocity is 0 m/s. average speed is 0 m/s and average velocity is 1 m/s.What is the swimmers average speed and average velocity?

Answers

Answer:

average speed is 1 m/s and average velocity is 0 m/s.

Explanation:

Given that :

Length of round trip = 50 m

Time taken = 100 seconds

The average speed :

Total distance / total time taken

Length of complete round trip :

(50 + 50) m, total. Distance = 100 m

100 / 100 = 1m/s

The average velocity :

Total Displacement / total time taken

Total Displacement of round trip = end point - start point = 0

0 / 100 = 0

Average speed is 1 m/s and average velocity is 0 m/s.

The average speed is defined as the ratio of distance to time. Speed is a scalar quantity hence it does not take direction into account while velocity is a vector quantity hence it takes direction into account.

The speed is obtained from;

Speed = Distance/time = 2(50 m)/100 s = 1 m/s.

The velocity is 0 m/s since it is complete round-trip lap.

Learn more about speed: https://brainly.com/question/7359669

Which of the following is a mixture?
a air
biron
Chydrogen
d nickel

Answers

The answer is to this is b

Answer:

it will option option A hope it helps

A group of 25 particles have the following speeds: two have speed 11 m/s, seven have 16 m/s , four have 19 m/s, three have 26 m/s, six have 31 m/s, one has 37 m/s, and two have 45 m/s.

Requiredd:
a. Determine the average speed.
b. Determine the rms speed.
c. Determine the most probable speed.

Answers

Answer:

a) Average speed is 24.04 m/s

b) the rms speed is 25.84 m/s

c) the most probable speed is 16 m/s

Explanation:

Given the data in the question;

a) Determine the average speed.

To determine the average speed, we simply divide total some of speed by number of particles;

Average speed =  [(2×11 m/s)+(7×16 m/s)+(4×19 m/s)+(3×26 m/s)+(6×31 m/s)+(1×37 m/s)+(2×45 m/s)] / 25    

= 601 / 25

= 24.04 m/s

Therefore, Average speed is 24.04 m/s

b) Determine the rms speed

we know that  (rms speed)² = sum of square speed / total number of particles

so

(rms speed)² =  [(2×11²)+(7×16²)+(4×19²)+(3×26²)+(6×31²)+(1×37²)+(2×45²)] / 25

(rms speed)² =  16691 / 25

(rms speed)² =  667.64

(rms speed) = √ 667.64

(rms speed) = 25.84 m/s

Therefore, the rms speed is 25.84 m/s

c) Determine the most probable speed.

Most particles (7) have velocity 16 m/s

i.e 7 is the maximum number of particle for a particular speed ,

Therefore, the most probable speed is 16 m/s

How much force is needed to accelerate a 65 kg rider AND her 215 kg motor scooter 8 m/s?? (treat
the masses as like terms)

Answers

Answer:

Force = 2240 Newton.

Explanation:

Given the following data;

Mass A= 65kg

Mass B = 215kg

Acceleration = 8m/s²

To find the force;

Force is given by the multiplication of mass and acceleration.

Mathematically, Force is;

[tex] F = ma[/tex]

Where;

F represents force.

m represents the mass of an object.

a represents acceleration.

First of all, we would have to find the total mass.

Total mass = Mass A + Mass B

Total mass = 65 + 215

Total mass = 280kg

Substituting into the equation, we have

[tex] Force = 280 * 8 [/tex]

Force = 2240 Newton.

Two children, each with a mass of 25.4 kg, are at fixed locations on a merry-go-round (a disk that spins about an axis perpendicular to the disk and through its center). One child is 0.78 m from the center of the merry-go-round, and the other is near the outer edge, 3.14 m from the center. With the merry-go-round rotating at a constant angular speed, the child near the edge is moving with translational speed of 11.5 m/s.

a. What is the angular speed of each child?
b. Through what angular distance does each child move in 5.0 s?
c. Through what distance in meters does each child move in 5.0 s?
d. What is the centripetal force experienced by each child as he or she holds on?
e. Which child has a more difficult time holding on?

Answers

Answer:

a) ω₁ = ω₂ = 3.7 rad/sec

b) Δθ₁ = Δθ₂ = 18.5 rad

c) d₁ = 14.5 m  d₂ = 57.5 m

d) Fc1 = 273.9 N Fc2 = 1069.8 N

e) The boy near the outer edge.

Explanation:

a)

Since the merry-go-round is a rigid body, any point on it rotates at the same angular speed.However, linear speeds of points at different distances from  the center, are different.Applying the definition of angular velocity, and the definition of angle, we can write the following relationship between the angular and linear speeds:

       [tex]v = \omega*r (1)[/tex]

Since we know the value of v for the child near the outer edge, and the value of r for this point, we can find the value of the angular speed, as follows:

       [tex]\omega = \frac{v_{out} }{r_{out} } = \frac{11.5m/s}{3.14m} = 3.7 rad/sec (2)[/tex]

As we have already said, ωout = ωin = 3.7 rad/sec

b)

Since the angular speed is the same for both childs, the angle rotated in the same time, will be the same for both also.Applying the definition of angular speed, as the rate of change of the angle rotated with respect to time, we can find the angle rotated (in radians) as follows:[tex]\Delta \theta = \omega * t = 3.7 rad/sec* 5.0 sec = 18.5 rad (3)[/tex]

        ⇒  Δθ₁ = Δθ₂ = 18.5 rad.

c)

The linear distance traveled by each child, will be related with the linear speed of them.Knowing the value of the angular speed, and the distance from each boy to the center, we can apply (1) in order to get the linear speeds, as follows:

       [tex]v_{inn} = \omega * r_{inn} = 3.7 rad/sec * 0.78 m = 2.9 m/s (4)[/tex]

      vout is a given of the problem ⇒ vout = 11. 5 m/s

Applying the definition of linear velocity, we can find the distance traveled by each child, as follows:

       [tex]d_{inn} = v_{inn} * t = 2.9m/s* 5.0 s = 14.5 m (5)[/tex]

      [tex]d_{out} = v_{out} * t = 11.5 m/s* 5.0 s = 57.5 m (6)[/tex]

d)

The centripetal force experienced by each child is the force that keeps them on a circular movement, and can be written as follows:

       [tex]F_{c} = m*\frac{v^{2}}{r} (7)[/tex]

Replacing by the values of vin and rin, since m is a given, we can find Fcin (the force on the boy closer to the center) as follows:

      [tex]F_{cin} = m*\frac{v_{in}^{2}}{r_{in}} = 25.4 kg* \frac{(2.9m/s)^{2} }{0.78m} = 273.9 N (8)[/tex]

In the same way, we get Fcout (the force on the boy near the outer edge):

      [tex]F_{cout} = m*\frac{v_{out}^{2}}{r_{out}} = 25.4 kg* \frac{(11.5m/s)^{2} }{3.14m} = 1069.8 N (9)[/tex]

e)

The centripetal force that keeps the boys in a circular movement, is not a different type of force, and in this case, is given by the static friction force.The maximum friction force is given by the product of the coefficient of static friction times the normal force.Since the boys are not accelerated in the vertical direction, the normal force is equal and opposite to the force due to gravity, which is the weight.As both boys have the same mass, the normal force is also equal.This means that for both childs, the maximum possible static friction force, is the same, and given by the following expression:[tex]F_{frs} = \mu_{s} * m* g (10)[/tex]If this force is greater than the centripetal force, the boy will be able to hold on.So, as the centripetal force is greater for the boy close to the outer edge, he will have a more difficult time holding on.

An RC car is carrying a tiny slingshot with a spring constant of 85 N/m at 0.2 m off the ground at 5.6 m/s. The sling shot is pulled back 3.5 cm from a relaxed state and shoots a 25 g steel pellet in the same direction the car is moving. What is the velocity of the steel pellet relative to the ground as it leaves the sling shot

Answers

Answer:

The velocity of the steel pellet relative to the ground when it leaves the sling shot is approximately 5.960 meters per second.

Explanation:

Let suppose that RC car-slingshot-steelpellet is a conservative system, that is, that non-conservative forces (i.e. friction, air viscosity) can be neglected. The velocity of the steel pellet can be found by means of the Principle of Energy Conservation and under the consideration that change in gravitational potential energy is negligible and that the RC car travels at constant velocity:

[tex]\frac{1}{2}\cdot (m_{C}+m_{P})\cdot v_{o}^{2} + \frac{1}{2}\cdot k\cdot x^{2} = \frac{1}{2}\cdot m_{C}\cdot v_{o}^{2} + \frac{1}{2}\cdot m_{P}\cdot v^{2}[/tex]

[tex]\frac{1}{2}\cdot m_{P}\cdot v_{o}^{2} + \frac{1}{2}\cdot k\cdot x^{2} = \frac{1}{2}\cdot m_{P}\cdot v^{2}[/tex]

[tex]m_{P}\cdot v_{o}^{2} + k\cdot x^{2} = m_{P}\cdot v^{2}[/tex]

[tex]v^{2} = v_{o}^{2} + \frac{k}{m_{P}}\cdot x^{2}[/tex]

[tex]v = \sqrt{v_{o}^{2}+\frac{k}{m_{P}}\cdot x^{2} }[/tex] (1)

Where:

[tex]v_{o}[/tex] - Initial velocity of the steel pellet, measured in meters per second.

[tex]v[/tex] - Final velocity of the steel pellet, measured in meters per second.

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

[tex]m_{P}[/tex] - Mass of the steel pellet, measured in kilograms.

[tex]m_{C}[/tex] - Mass of the RC car, measured in kilograms.

[tex]x[/tex] - Initial deformation of the spring, measured in meters.

If we know that [tex]v_{o} = 5.6\,\frac{m}{s}[/tex], [tex]k = 85\,\frac{N}{m}[/tex], [tex]m_{P} = 0.025\,kg[/tex] and [tex]x = 0.035\,m[/tex], then the velocity of the steel pellet relative to the ground when it leaves the sling shot is:

[tex]v = \sqrt{\left(5.6\,\frac{m}{s} \right)^{2}+\frac{\left(85\,\frac{N}{m} \right)\cdot (0.035\,m)^{2}}{0.025\,kg} }[/tex]

[tex]v \approx 5.960\,\frac{m}{s}[/tex]

The velocity of the steel pellet relative to the ground when it leaves the sling shot is approximately 5.960 meters per second.

what type of waves can only travel through a medium?

Answers

Answer:

Mechanical waves

Explanation:

Mechanical waves  are the waves that can travel only through a medium. Mechanical waves are disturbance of matter and require medium to transfer the energy. There are three types of mechanical waves that include  transverse wave, longitudinal wave and surface wave.

Some of the examples of mechanical waves are sound waves and seismic waves etcetera.

Hence, the correct answer is "Mechanical waves".

is 0.8 kilograms bigger then 80 grams

Answers

Answer:

Yes

Explanation:

0.8 kilograms is equal to 800 grams

Answer:

Yes, 0.8 kilograms is greater than 80 grams

Explanation:

0.8 kilograms is equal to 800 grams and 80 grams is equal to 0.08 kilogrmas.

Sorry if I'm wrong, correct me.

A sinusoidal wave is traveling on a string with speed 19.3 cm/s. The displacement of the particles of the string at x = 6.0 cm is found to vary with time according to the equation y = (2.6 cm) sin[1.8 - (5.8 s-1)t]. The linear density of the string is 5.0 g/cm. What are (a) the frequency and (b) the wavelength of the wave? If the wave equation is of the form

Answers

Answer:

Explanation:

equation of wave is given by the following equation

y = (2.6 cm) sin[1.8 - (5.8 s-1)t].

Comparing it with standard form of wave

y = A sin ( ωt - kx )

we get

ω = 5.8

2πn = 5.8

n = .92 per second

kx = 1.8

k x 6 = 1.8

k = 0.3

[tex]\frac{2\pi}{\lambda}[/tex] = 0.3

λ = 20.9 cm

As a person pushes a box across a floor, the energy from the person's moving arm is transferred to the box, and the box and the floor becomes warm. During the process, what happens to energy

Answers

Answer:

isnt heat transfer

Explanation:

sorry if im wrong

Which landform is produced at location E where the Mississippi River enters the Gulf of
Mexico?
a delta a drumlin an out wash an escarpment

Answers

Answer:

a delta

Explanation:

The landform produced at the location E where the Mississippi River enters the Gulf of Mexico is a delta.

A delta is a depositional landform where a smaller body of water enters into a larger one.

The Gulf of Mexico contains a larger body of water and as the Mississippi river enters into it, it splits up into many distributaries.

So, this feature is a delta.  

A physics student spends part of her day walking between classes or for recreation, during which time she expends energy at an average rate of 280 W. The remainder of the day she is sitting in class, studying or resting; during these activities, she expends energy at a rate of 100 W. If she expends a total of 1.1 x 10^7 J of energy in a 24 hour day, how much of the day did she spend walking

Answers

RESULT: Twalk= 3.64 hr

The time of the day she spent walking is equal to 3.70 hrs.

What is power?

Power can be explained as the rate of doing work in unit time. The SI unit of measurement of power is J/s or Watt (W). Power can be described as a time based quantity. The mathematical expression for power can be represented as mentioned below.

Power = work/time

P = W/t

Given, the energy spends part of her day walking, Ew = 280 W

The energy is spent by sitting in the class, Es = 100 W

The total energy spends, Et = 1.1 × 10⁷J

[tex]E_w \times t + E_s(24\times 60\times 60-t)= 1.1 \times 10^7J[/tex]

[tex]280 \times t + 100(24\times 60\times 60-t)= 1.1 \times 10^7[/tex]

280 t + 0.86 × 10⁷ - 100 t = 1.1 × 10⁷

180 t = 0.24 × 10⁷

t =  0.24 × 10⁷/180 × 3600

t = 3.70 hr

Learn more about power, here:

brainly.com/question/14070854

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A mass m is gently placed on the end of a freely hanging spring. The mass then falls 33 cm before it stops and begins to rise. What is the frequency of the oscillation

Answers

Answer:

Explanation:

The mass falls by .33 m before it begins to rise . At that point loss of potential energy is equal to gain of elastic energy .

1/2 k x² = mgx

.5 x k x .33² = m x 9.8 x .33

k / m = 59.4

frequency of oscillation =  [tex]\frac{1}{2\pi} \times\sqrt{\frac{k}{m} }[/tex]

= [tex]\frac{1}{2\pi} \times\sqrt{59.4}[/tex]

= 1.22 per second .

what is the direction of the third force that would cause the box to remain stationary on the ramp ?

Answers

An arrow pointing from the bottom of the ramp to the top, I assume it would be friction.

The arrow on the bottom pointing down due to friction the bow would not be able to go down the ramp

What kind of scattering (Rayleigh, Mie, or non-selective) would you expect to be most important when radiation of the specified wavelength encounters the following natural or anthropogenic particles?
Slides 16-31, Lecture 2 ought to help - slides 19, 24, and 31 are key.
Wavelength O2 molecules Smoke particles Cloud droplets Rain droplets
(size 10^-10 m) (size 0.3 (μm) (20 μm) (size 3 mm)
550 nm
11 μm
1600 nm
1 cm

Answers

Solution :

1. Rayleigh scattering takes place when the particle size is smaller than the wavelength (λ).

2. Mie scattering takes place when particle size is nearly equal to the wavelength (λ).

3. Non-selective scatter takes place when particle size in greater than the wavelength  (λ).

We have the sizes of different particles :

[tex]$O_2 \rightarrow 10^{10} \ m $[/tex]

Smoke particles [tex]$\rightarrow 3 \times 10^{-7} \ m$[/tex]

Cloud droplets [tex]$\rightarrow 2 \times 10^{-5} \ m$[/tex]

Rain droplets [tex]$\rightarrow 3 \times 10^{-3} \ m$[/tex]

Wavelength           [tex]$ O_2 $[/tex]         Smoke particles    Cloud droplets     Rain droplets

                            [tex]$10^{-10} \ m$[/tex]        [tex]$ 3 \times 10^{-7} \ m$[/tex]           [tex]$ 2 \times 10^{-5} \ m$[/tex]              [tex]$ 3 \times 10^{-3} \ m$[/tex]

[tex]$5500 \times 10^{-4} \ m$[/tex]      Rayleigh  Non-selective      Non-selective     Non-selective

[tex]$11 \times 10^{-6} \ m $[/tex]         Rayleigh    Rayleigh            Non-selective      Non-selective

[tex]$1600 \times 10^{-10} \ m $[/tex]    Rayleigh  Non-selective      Non-selective     Non-selective

[tex]$10^{-2} \ m $[/tex]                 Rayleigh      Rayleigh               Rayleigh          Mie

It has been argued that power plants should make use of off-peak hours (such as late at night) to generate mechanical energy and store it until it is needed during peak load times, such as the middle of the day. One suggestion has been to store the energy in large flywheels spinning on nearly frictionless ball-bearings. Consider a flywheel made of iron, with a density of 7800 kg/m^3 , in the shape of a uniform disk with a thickness of 11.3 cm.

Required:
a. What would the diameter of such a disk need to be if it is to store an amount of kinetic energy of 14.1 MJ when spinning at an angular velocity of 93.0 rpm about an axis perpendicular to the disk at its center?
b. What would be the centripetal acceleration of a point on its rim when spinning at this rate?

Answers

Answer:

Explanation:

kinetic energy = 14.1 MJ = 14.1 x 10⁶ J

Let radius of flywheel be r .

volume of flywheel = π r² x t where t is thickness

= 3.14 x r² x .113 m³

= .04 r² m³

mass = volume x density

= .04 r² x 7800 = 312.73 r²kg

moment of inertia I = 1 / 2 mass x radius²

= .5 x 312.73 r² x r²

= 156.37 r⁴ kg m²

angular velocity ω = 2π x 93/60

= 9.734 rad /s

kinetic energy = 1/2 Iω² where ω is angular velocity

= .5 x 156.37 r⁴ x 9.734²

= 7408.08 r⁴

Given

7408.08 r⁴ =  14.1 x 10⁶

r⁴ = .19 x 10⁴

r = .66 x 10

= 6.60 m .

Diameter = 13.2 m

b )

centripetal acceleration of a point on its rim = ω² r

= 9.734² x 6.6

= 625.35 m /s²

An 8.00 kg mass moving east at 15.4 m/s on a frictionless horizontal surface collides with a 10.0 kg object that is initially at rest. After the collision, the 8.00 kg object moves south at 3.90 m/s. (a) What is the velocity of the 10.0 kg object after the collision

Answers

Answer:

9.3m/s

Explanation:

Based on the law of conservation of momentum

Sum of momentum before collision = sum of momentum after collision

m1u1 +m2u2 = m1v1+m2v2

m1 = 8kg

u1 = 15.4m/s

m2 = 10kg

u2 = 0m/s(at rest)

v1 = 3.9m/s

Required

v2.

Substitute

8(15.4)+10(0) = 8(3.9)+10v2

123.2=31.2+10v2

123.2-31.2 = 10v2

92 = 10v2

v2 = 92/10

v2 = 9.2m/s

Hence the velocity of the 10.0 kg object after the collision is 9.2m/s

A student is driving through a mountainous region where the road is at some times flat, at some times inclined upward, and at some time inclined downward. The student maintains a speed of 20 m/s on the roadway, but is required to make an emergency stop on the three sepearte occasions. On levels roadway, it takes 25 m to stop. On a downward-sloping roadway, it takes 40 m to stop. On an upward-sloping roadway, it takes 18 m to stop. Explain why the stopping distances are different. (Focus answer using work and energy, other concepts may be used as well but be sure work and energy are included.)

Answers

Answer:

Explanation:

It is frictional force of the ground that helps in bringing the vehicle to stop . In the process of stopping , negative work is done on the car by friction force to overcome its kinetic energy .

At levelled road , for stoppage

Kinetic energy of vehicle = Work done by frictional force . = friction force x displacement .

At upward slopping road , gravitational force acting downward also helps the vehicle to stop do friction has to do less work .

At upward inclined  road , for stoppage

Kinetic energy of vehicle = Work done by frictional force + work done by gravitational force  = (friction force + gravitational force ) x displacement .

Hence displacement is less .

At downward slopping road ,  friction has to do more work because friction has to do work against gravitational force acting downwards wards and kinetic energy of the vehicle  also .

At downward inclined  road , for stoppage

Kinetic energy of vehicle + work done by gravitational force  = Work done by frictional force = friction force  x displacement .

Hence displacement is more .

Hence displacement is more in the downward slopping.

What is Displacement?

Displacement is defined as the change in position of an object. It is a vector quantity and has a direction and magnitude.

It is frictional force of the ground that helps in bringing the vehicle to stop . In the process of stopping , negative work is done on the car by friction force to overcome its kinetic energy .

At levelled road , for stoppage

Kinetic energy of vehicle = Work done by frictional force . = friction force x displacement .

At upward slopping road , gravitational force acting downward also helps the vehicle to stop do friction has to do less work .

At upward inclined  road , for stoppage

Kinetic energy of vehicle = Work done by frictional force + work done by gravitational force  = (friction force + gravitational force ) x displacement .

Hence displacement is less .

At downward slopping road ,  friction has to do more work because friction has to do work against gravitational force acting downwards wards and kinetic energy of the vehicle  also .

At downward inclined  road , for stoppage

Kinetic energy of vehicle + work done by gravitational force  = Work done by frictional force = friction force  x displacement .

Hence displacement is more in the downward slopping.

To learn more about displacement refer to the link:

brainly.com/question/11934397

#SPJ5

"45 meters north" is an example of

Answers

Answer:

Displacement

Explanation:

The quantity 45m north is a typical example of displacement.

Displacement is the distance traveled by a body in a specific direction. Displacement is a vector quantity with both magnitude and direction.

When we are specifying the displacement of a body, the direction must be indicated accurately. Therefore, the quantity given is displacement

A truck travelling down the street suddenly brakes, applying a 14 N force over 3.5 seconds. What was the impulse over the given time.

Answers

Answer:

49 Ns

Explanation:

Given data

Force= 14N

time = 3.5seconds

Applying the expression for impulse

P= Ft

substitute

P=14*3.5

P=49 Ns

Hence the impulse is 49 Ns

A ratio is another name for a decimal true or false

Answers

True....................
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