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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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Two skaters stand facing each other. One skater's mass is 60 kg, and the other's
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Answer:

v' = 2.4 m/s

Explanation:

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Mass of one skater, m = 60 kg

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3 0
2 years ago
A bowler throws a bowling ball of radius R = 11.0 cm down the lane with initial speed = 8.50 m/s. The ball is thrown in such a w
Ad libitum [116K]

Answer:

a) 1.18 seconds

b) 8.6 m

c) 5.19 revolutions

d) 6.07 m/s

Explanation:

<u>Step 1: </u>Data given

radius of the ball = 11.0 cm

Initial speed of the ball = 8.50 m/s

The coefficient of kinetic friction between the ball and the lane is 0.210.

<em></em>

<em>(a) For what length of time does the ball skid?</em>

The velocity at time t can be written as v(t) = v0 + at

 ⇒ with v(t) = the velocity at time t

⇒ with v0 : the initial velocity = 8.50 m/s

⇒ with a = the acceleration (in m/s²)

   ⇒The acceleration (negative) due to friction: a = -µg

           ⇒ with µ = 0.210

          ⇒ with g = 9.81 m/s²

v(t) =8.5m/s - 0.21*9.81m/s² * t = 8.5 - 2.06t

Torque τ = Iα = (2m(0.11m)²/5)α = 0.00484m*α

τ = F * r = µm*g*R = 0.21 * M * 9.81m/s² * 0.11m = 0.227m

so α = 0.227m / 0.00484m = 46.9 rad/s²

angular velocity ω(t) = ωo + αt = 0 + 46.9 rad/s² * t

The ball stops sliding when v(t) = ω(t) * r

8.5 - 2.06t  = 46.9*0.11*t = 5.159t

7.219t = 8.5

<u>t = 1.18 seconds</u>

<em>b) How far down the lane does it skid?</em>

s = Vo*t + ½at² = 8.5m/s * 1.18s - ½* 2.06 m/s² * (1.18s)² = <u>8.6 m</u>

<em>c) How many revolutions does it make before it starts to roll?</em>

The angular acceleration of the ball is:

α =  τ/I

 ⇒ with  τ = the torque experienced by the ball due the frictional force

   ⇒  τ = fk*R

α = fk*R /I

 ⇒ I = 2/5 m*R²

 ⇒ fk = µk*m*g

α = (µk*m*g*R)/(2/5mR²)

α = 5µk*g /2R

The angular displacement of the ball is:

∅ = 1/2αt²

⇒ The ball does not have an initial angular velocity

∅ =1/2*(5µk*g/2)*t²

∅ = 5µkgt²/4R

∅ = (5*0.21*9.81*1.18²)/(4*11.0 *10^-2)

∅ = 32.6 rad

Number of revolutions = 32.6 rad /2π

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<em>(d) How fast is it moving when it starts to roll?</em>

v = Vo + at = 8.5m/s - 2.06m/s² * 1.18s = <u>6.07 m/s</u>

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