PHYSICS COLLEGE

A boy flies a kite with the string at a 30∘ angle to the horizontal. The tension in the string is 4.5 N. Part A Part complete How much work does the string do on the boy if the boy stands still?

Answers

Answer 1

Answer:

Answer:

Work done is zero

Explanation:

given data

Angle of kite with horizontal = 30 degree

tension in the string = 4.5 N

WE KNOW THAT

Work = force * distance

horizontal force = $Tcos\theta = 4.5*cos30 = 3.89 N$

DISTANCE = 0 as boy stands still. therefore

work done = 3.89 *0 = 0

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Hotel rooms in Smalltown go for $100, and 1,000 rooms are rented on atypical day.a. To raise revenue, the mayor decides to charge hotels a tax of $10 perrented room. After the tax is imposed, the going rate for hotel roomsrises to $108, and the number of rooms rented falls to 900. Calculate theamount of revenue this tax raises for Smalltown and the deadweightloss of the tax. (Hint: The area of a triangle is l/2Xbase X height.)h. The mayor now doubles the tax to $20. The price rises to $116, andthe number of rooms rented falls to 800. Calculate tax revenue anddeadweight loss with this larger tax. Are they double, more thandouble, or less than double? Explain.

Answers

Answer:

doubling the size of the tax more than doubles the deadweight loss while less than doubles the revenue generated

Explanation:

(a)

The quantity of rooms rented before tax, Q1 = 1000 rooms.

The quantity of rooms rented after the imposition of tax Q2 = 900 rooms.

Size of the tax = $10

Price paid by buyer = $108

Price received by seller = $98

Deadweight loss = 1/2 x (Q2 — Q1) x (size of the tax)

Deadweight loss = 1/2 x (1000 — 900) x ($10) = $500

Tax revenue generated = size of tax * (Q2) = $10 x (900) = $9000

The quantity of rooms rented before tax, Q1 = 1000 rooms

The quantity of rooms rented after the imposition of tax, Q2 = 800 rooms Size of the tax = $20

Price paid by buyer = $116

Price received by seller = $96

Deadweight loss = 1/2 x (Q2 — Q1) x (size of the tax)

New Deadweight loss = 1/2 x (1000 — 800) x ($20) = $2000

Thus, dead weight loss quadruples post doubling the size of tax. New Tax revenue generated = size of tax x (Q2) = $20 x (800) = $16000 Thus, revenue generated less than doubles post doubling the size of tax.

A virtual image produced by a lens is always

Answers

The correct answer is:

"located in front of lens"

just took PF test and this was right answer

The distance in the x direction between two control points on a vertical aerial photograph is 4.5". If the distance between these same two points is 3.6" on another photograph having a scale of 1:24,000, determine the scale of the first vertical aerial photograph. Of the focal length of the camera is 6"and the average elevation at these points is 100 ft, determine the flying height from which each photograph was taken

Answers

Answer:

Use proportions to find the scale of the first photo, then use that scale and other given information to fill in the equation

S=f/(H-h)

Where:

S = scale of the photo

f = focal length of the camera (in feet)

H = flying height

h = average elevation

A block-spring system consists of a spring with constant k = 475 N/m attached to a 2.50 kg block on a frictionless surface. The block is pulled 5.50 cm from equilibrium and released from rest. For the resulting oscillation, find the amplitude, angular frequency, frequency, and period. What is the maximum value of the block's velocity and acceleration?

Answers

Explanation:

It is given that,

Spring constant of the spring, k = 475 N/m

Mass of the block, m = 2.5 kg

Elongation in the spring from equilibrium, x = 5.5 cm

(a) We know that the maximum elongation in the spring is called its amplitude. So, the amplitude for the resulting oscillation is 5.5 cm.

(b) Let $\omega$ is the angular frequency. It is given by :

$\omega=\sqrt{(k)/(m)}$

$\omega=\sqrt{(475)/(2.5)}$

$\omega=13.78\ rad/s$

(c) Let T is the period. It is given by :

$T=(2\pi)/(\omega)$

$T=(2\pi)/(13.78)$

T = 0.45 s

(d) Frequency,

$f=(1)/(T)$

$f=(1)/(0.45)$

f = 2.23 Hz

(e) Let v is the maximum value of the block's velocity. It is given by :

$v_(max)=\omega* A$

$v_(max)=13.78* 5.5* 10^(-2)$

$v_(max)=0.757\ m/s$

The value of acceleration is given by :

$a=\omega^2A$

$a=(13.78)^2* 5.5* 10^(-2)$

$a=10.44\ m/s^2$

Hence, this is the required solution.

Final answer:

The amplitude of the block-spring system is 0.055 m, with an angular frequency of 13.77 rad/s, and a frequency of approximately 2.19 Hz. The system has a period of approximately 0.46 s, with the maximum velocity being 0.76 m/s, and the maximum acceleration being 189 m/s².

Explanation:

In this block-spring system, we can determine the oscillation properties with the given parameters: mass (m = 2.50 kg), spring constant (k = 475 N/m), and displacement (x = 5.50 cm = 0.055 m).

Amplitude (A): This is the maximum displacement from the equilibrium position, in this case, it is 0.055 m, the distance from which the block is pulled and released.
Angular Frequency (ω): This can be calculated by the formula ω = √(k/m) which is equal to √(475/2.50) = 13.77 rad/s.
Frequency (f): It is given by the formula f = ω / 2π ≈ 2.19 Hz.
Period (T): The time for one complete cycle of the motion, calculated as T = 1/f ≈ 0.46 s.
Maximum value of velocity (v max): Calculated by the formula v = ω*A = 13.77 * 0.055 = 0.76 m/s.
Maximum value of acceleration (a max): Maximum acceleration occurs when the block is at the maximum displacement (amplitude), found with formula a = ω²*A = 189 m/s².

Learn more about Oscillations here:

brainly.com/question/30111348

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Hoover Dam on the Colorado River is the highest dam in the United States at 221m, with a power output of 680 MW. The dam generates electricity by flowing water down to a point 150 m below the stop, at an average flow rate of 650 m3/s.

Answers

Answer:

The power in this flow is $9.56*10^(8)\ W$

Explanation:

Given that,

Distance = 221 m

Power output = 680 MW

Height =150 m

Average flow rate = 650 m³/s

Suppose we need to calculate the power in this flow in watt

We need to calculate the pressure

Using formula of pressure

$Pressure=\rho g h$

Where, $\rho$ = density

h = height

g = acceleration due to gravity

Put the value into the formula

$Pressure=1000*9.8*150$

$Pressure=1470000\ Nm^2$

We need to calculate the power

Using formula of power

$P=Pressure* flow\ rate$

Put the value into the formula

$P=1470000*650$

$P=9.56*10^(8)\ W$

Hence, The power in this flow is $9.56*10^(8)\ W$

A disgruntled autoworker pushes a small foreign import offacliff with a height of h. the vehicle lands a distance away
fromthe cliff. Determine how fast the vehicle was pushed off
thecliff.