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kotykmax [81]
4 years ago
5

Application: Economics. A vector ~p = h150, 225, 375i represents the price in dollars of certain models of bicycles sold by a bi

cycle shop at costs given by the vector ~c = h125, 200, 360i. (a) What is the vector ~r representing the revenue made from selling these bikes? (b) Suppose the shop owner offers a store-wide discount of 10% on the price of the bicycles. How much revenue does he make if he sells 10 o
Physics
1 answer:
DerKrebs [107]4 years ago
6 0

Answer:

a. r=25,25, 15

b. Total revenue =225,225, 135

Explanation:

Application: Economics. A vector ~p = 150, 225, 375

represents the price in dollars of certain models of bicycles sold by a bicycle shop at costs given by the vector ~c = 125, 200, 360.

(a) What is the vector ~r representing the revenue made from selling these bikes? (b) Suppose the shop owner offers a store-wide discount of 10% on the price of the bicycles. How much revenue does he make if he sells 10 o

Lets's simplify the values step by step

A vector p = 150, 225, 375 is the price

vector c = 125, 200, 360.  represent the cost of the bicycle

therefore

vector r=the revenue from selling the bicycle will be

p-c:    

150, 225, 375-(125, 200, 360)

r=(25,25, 15)

b.

if there is a discount of 10% for 10 bicycles, i presume

then the total revenue will be

revenue /one bicycle*10% discount*10 bicycles

to get the

r=(100%(25,25, 15)-10%(25,25, 15))*10

if there is a 10% slash in the price, it means that the item will now be 90% of its price

r=225,225,135

Total revenue =225,225, 135

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Answer

given,

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Force on the stick on one side = F =  9 N

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magnitude of acceleration

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F = m a

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Learning Goal:
Rudiy27

Answer:

The questions are not complete so this is the complete questions

1. How much work W does the motor do on the platform during this process?

2. What is the angular velocity ωf of the platform at the end of this process?

3. What is the rotational kinetic energy, Ek, of the platform at the end of the process described above?

4. How long does it take for the motor to do the work done on the platform calculated in Part 1?

5. What is the average power delivered by the motor in the situation above?

6. . Note that the instantaneous power P delivered by the motor is directly proportional to ω, so P increases as the platform spins faster and faster. How does the instantaneous power P•f being delivered by the motor at the time t•f compare to the average power

P(average) calculated in Part e?

Explanation:

Given that,

The torque τ=25Nm

Moment of inertia I =50kgm²

The platform is initially at rest,

ω•i=0 rad/sec

Revolution the torque produce is 12

Then, θ=12 revolution

1 revolution=2πrad

So, θ=24πrad

1. Work done in a rotational motion is give as

W=τ•Δθ

Given that the τ=25Nm and the initial angular displacement is 0rad

The final angular displacement is 24πrad

Δθ =(θ2-θ1)

Δθ=24π-0

Δθ=24πrad

Then,

W=τ•Δθ

W=25(24π)

W=25×24π

W=1884.96J

To 4s.f, W=1885J

2. Final angular velocity ωf

Using the angular equation

ω•f²=ω•i²+2•α•Δθ

We need to get angular acceleration

The torque is given as

τ=I•α

Given that,

I is moment if inertia =50kgm²

τ=25Nm

α=τ/I

α= 25/50

α=0.5rad/s²

Now, using the angular acceleration

ω•f²=ω•i²+2•α•Δθ

ω•f²=0²+2×0.5×24π

ω•f²=0+75.398

ω•f²=75.398

ω•f=√75.398

ω•f=8.68 rad/sec.

3. We need to find rotational Kinetic energy and it is given as

K.E, = ½I•ω²

Given that, I=50kgm² and ω•f=8.68rad/sec

Then,

K.E, =½I•ω²

K.E, =½×50×8.68²

K.E, =1884.96J

To 4s.f,

K.E, =1885J

Which is the same as the work done by the motor.

4. Time taken to complete part 1,

Using the rotational equation

ω•f=ω•i+α•t

Since, ω•f=8.68 rad/sec and ω•i=0

And α=0.5rad/s²

Then,

ω•f=ω•i+α•t

8.68=0+0.5t

8.68=0.5t

Then, t=8.68/0.5

t=17.36secs

5. The average power of rotational motion is given as

P(average) =Workdone/timetaken

Since,

Work done =1884.96J

Time taken =17.36sec

P(average) =Workdone/timetaken

P(average)=1884.96/17.36

P(average)= 108.58Watts

To 4s.f

P(average)=108.6Watts

6. We need to find •, it is given as

• =τ•ωf

Given that, ω•f=8.68rad/sec, τ=25Nm

•=25×8.68

•=217Watts

Then, the ratio of • to P(average) is

Ratio = •/ P(average)

Ratio= 217/108.58

Ratio=1.9985

Then, the ratio is approximately 2

Ratio=2

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