Ask question

Ask question

All categories

2 years ago

5

Originally 21 acts were tentatively scheduled to perform at the arena; only 10 have been booked. What is the shortfall if the ot

her acts do not perform

1 answer:

Levart [38]2 years ago

5 0

Answer:

Here is the complete question.

Originally 21 acts were tentatively scheduled to perform at the arena; only 10 have been booked. What is the shortfall if the other acts do not perform, assuming an average profit margin of $175,000 for each concert?

The shortfall is of $\$1925000$

Explanation:

If all the acts were performed then the earnings would be $21\times 175,000=\$3,675,000$

Now as mentioned in the question that only $10$ have been performed so far.

Now the earning from these $10$ acts $=10\times 175,000=1,750,000$

To find the shortfall.

We have to subtract the expected earnings with the actual earning from the concert.

So shortfall $=(3,675,000-1,750,000)\$=\$1,925,000$

So the shortfall is of $\$1,925,000$ if the other acts were not performed.

You might be interested in

Explain why the brakes of a car get much hotter than the brakes of a bicycle?

ValentinkaMS [17]

Answer: Car brakes produces more energy then the bicycle because the cars wheels produces a much bigger force that makes the car go and to stop that force the car uses greater amount of energy that transfers to heat but in a bicycle the wheels do not turn that fast so when you press the break there is less energy that transfer to heat.

Explanation:

8 0

2 years ago

Now suppose the initial velocity of the train is 4 m/s and the hill is 4 meters tall. If the train has a mass of 30000 kg, what

hoa [83]

Answer:

<h2>187,500N/m</h2>

Explanation:

From the question, the kinectic energy of the train will be equal to the energy stored in the spring.

Kinetic energy = 1/2 mv² and energy stored in a spring E = 1/2 ke².

Equating both we will have;

1/2 mv² = 1/2ke²

mv² = ke²

m is the mass of the train

v is the velocity of then train

k is the spring constant

e is the extension caused by the spring.

Given m = 30000kg, v = 4 m/s, e = 4 - 2.4 = 1.6m

Substituting this values into the formula will give;

30000*4² = k*1.6²

$k = \frac{30,000*16}{1.6^2}\\ \\k = \frac{480,000}{2.56}\\ \\k = 187,500Nm^{-1}$

The value of the spring constant is 187,500N/m

7 0

2 years ago

A single crystal of aluminum is oriented for a tensile test such that its slip plane normal makes angle of 28.1 with the tensil

NISA [10]

Answer:

we have to find out the critical resolved shear stress. As it it given in the question

Ф = 28.1°and the possible values for λ are 62.4°, 72.0° and 81.1°.

a) Slip will occur in the direction where cosФ cosλ are maximum. Cosine for all possible λ values are given as follows.

cos(62.4°) = 0.46

cos(72.0°) = 0.31

cos(81.1°) = 0.15

Thus, the slip direction is at the angle of 62.4° along the tensile axis.

b) now the critical resolved shear stress can be find out by the following equation.

τ $_{crss}$ = σ $_{Y}$ ( cosФ cosλ) $_{max}$

now by putting values,

= (1.95MPa)[ cos(28.1) cos(62.4)] = 0.80 MPa (114 Psi) 7.23

3 0

2 years ago

You are provided with three polarizers with filters making angles of (A) 90 ∘ , (B) 180 ∘ and (C) −45 ∘ with respect

irinina [24]

Answer:

Order of maximum transmission of the polarizer is A, C and B.

Solution:

As per the question:

For the first polarizer, the angle is quite insignificant:

(A) $90^{\circ}$ :

The light intensity after passing through the first polarizer is $I_{o}$ and this intensity does not depend on the angle of the polarizer.

Consider $90^{\circ}$ with the vertical, the intensity is given by:

$I = I_{o}cos^{2}90^{\circ}$

$I = I_{o}cos(2(45^{\circ})) = I_{o}(\frac{1+cos90^{\circ})}{2} = \frac{I_{o}}{2}$

(B) $180^{\circ}$ :

Suppose the second polarizer is $45^{\circ}$ with the vertical.

Now, intensity through the second polarizer:

$I' = Icos^{2}(\theta_{2} - \theta_{1}) = \frac{I_{o}}{2}cos^{2}(- 45 - 90)$

$I' = \frac{I_{o}}{2}cos^{2}135^{\circ} = \frac{I_{o}}{4}$

Now, if we consider the second polarizer to be $180^{\circ}$ ,

$I' = \frac{I_{o}}{2}cos^{2}180^{\circ} = \frac{I_{o}}{2}cos^{2}(180^{\circ} - 90^{\circ}) = 0$

(C) $- 45^{\circ}$ :

Now,

Intensity through the third polarizer, if it is $180^{\circ}$ with the vertical:

$I' = Icos^{2}(\theta_{2} - \theta_{1}) = \frac{I_{o}}{2}cos^{2}(180 - (- 45))$

$I' = \frac{I_{o}}{8}$

5 0

2 years ago

Which statement about work and power correctly describes an automobile race?

miv72 [106K]

Well this question looks like it makes some assumptions. So assuming that both cars have the same mass and experience the same wind resistance regardless of speed and same internal frictions, then we could say "The car that finishes last has the lowest power". The reason is that for a given race the cars must overcome losses associated with motion. Since they all travel the same distance, the amount of work will be the same for both. This is because work is force times distance. If the force applied is the same in both cases (identical cars with constant wind resistance) and the distance is the same for both (a fair race track) then W=F·d will be the same.
Power, however, is the work done divided by the time over which it is done. So for a slower car, time t will be larger. The power ratio W/t will be smaller for the longer time (slower car).

7 0

2 years ago

Read 2 more answers