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ki77a [65]
3 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
Physics
1 answer:
Levart [38]3 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.

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Answer:

v=3.564\ m.s^{-1}

\Delta v =2.16\ m.s^{-1}

Explanation:

Given:

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  • length of slide, l=3\ m

(A)

height between John and William, h=1.8\ m

<u>Using the equation of motion:</u>

v_J^2=u_J^2+2 (g.sin\theta).l

where:

v_J = final velocity of John at the end of the slide

u_J = initial velocity of John at the top of the slide = 0

Now putting respective :

v_J^2=0^2+2\times (9.8\times \frac{1.8}{3})\times 3

v_J=5.94\ m.s^{-1}

<u>Now using the law of conservation of momentum at the bottom of the slide:</u>

<em>Sum of initial momentum of kids before & after collision must be equal.</em>

m_J.v_J+m_w.v_w=(m_J+m_w).v

where: v = velocity with which they move together after collision

30\times 5.94+0=(30+20)v

v=3.564\ m.s^{-1} is the velocity with which they leave the slide.

(B)

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<u>Now we find the force along the slide due to the body weight:</u>

F=m_J.g.sin\theta

F=30\times 9.8\times \frac{1.8}{3}

F=176.4\ N

<em><u>Hence the net force along the slide:</u></em>

F_R=71.4\ N

<em>Now the acceleration of John:</em>

a_j=\frac{F_R}{m_J}

a_j=\frac{71.4}{30}

a_j=2.38\ m.s^{-2}

<u>Now the new velocity:</u>

v_J_n^2=u_J^2+2.(a_j).l

v_J_n^2=0^2+2\times 2.38\times 3

v_J_n=3.78\ m.s^{-1}

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\Delta v =5.94-3.78= 2.16\ m.s^{-1}

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If the mass of the object is doubled and the speed is halved then kinetic energy will change by a factor of:
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A certain moving electron has a kinetic energy of 0.991 × 10−19 J. Calculate the speed necessary for the electron to have this e
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Putting in the values in the equation:

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v=466462m/s

The speed necessary for the electron to have this energy is 466462 m/s

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