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kobusy [5.1K]
3 years ago
10

What separates musical theater from other dance styles?

Physics
1 answer:
Rus_ich [418]3 years ago
7 0

Answer: Musical theater is like acting in a musical.

Explanation: Other dance styles don't involve singing.

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The experimentation step of scientific inquiry involves _______.
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<span>The answer is letter D.

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Read 2 more answers
The melting of polar ice is one effect of
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The melting of polar ice is one effect of the greenhouse effect, or also global warming.

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5 0
3 years ago
A DJ starts up her phonograph player. The turntable accelerates uniformly from rest, and takes t1 = 11.9 seconds to get up to it
Degger [83]

Answer:

a)\omega_1=8.168\,rad.s^{-1}

b)n_1=7.735 \,rev

c)\alpha_1 =0.6864\,rad.s^{-2}

d)\alpha_2=4.1454\,rad.s^{-2}

e)t_2=1.061\,s

Explanation:

Given that:

  • initial speed of turntable, N_0=0\,rpm\Rightarrow \omega_0=0\,rad.s^{-1}
  • full speed of rotation, N_1=78 \,rpm\Rightarrow \omega_1=\frac{78\times 2\pi}{60}=8.168\,rad.s^{-1}
  • time taken to reach full speed from rest, t_1=11.9\,s
  • final speed after the change,  N_2=120\,rpm\Rightarrow \omega_2=\frac{120\times 2\pi}{60}=12.5664\,rad.s^{-1}
  • no. of revolutions made to reach the new final speed,  n_2=11\,rev

(a)

∵ 1 rev = 2π radians

∴ angular speed ω:

\omega=\frac{2\pi.N}{60}\, rad.s^{-1}

where N = angular speed in rpm.

putting the respective values from case 1 we've

\omega_1=\frac{2\pi\times 78}{60}\, rad.s^{-1}

\omega_1=8.168\,rad.s^{-1}

(c)

using the equation of motion:

\omega_1=\omega_0+\alpha . t_1

here α is the angular acceleration

78=0+\alpha_1\times 11.9

\alpha_1 = \frac{8.168 }{11.9}

\alpha_1 =0.6864\,rad.s^{-2}

(b)

using the equation of motion:

\omega_1\,^2=\omega_0\,^2+2.\alpha_1 .n_1

8.168^2=0^2+2\times 0.6864\times n_1

n_1=48.6003\,rad

n_1=\frac{48.6003}{2\pi}

n_1=7.735\, rev

(d)

using equation of motion:

\omega_2\,^2=\omega_1\,^2+2.\alpha_2 .n_2

12.5664^2=8.168^2+2\alpha_2\times 11

\alpha_2=4.1454\,rad.s^{-2}

(e)

using the equation of motion:

\omega_2=\omega_1+\alpha_2 . t_2

12.5664=8.168+4.1454\times t_2

t_2=1.061\,s

4 0
3 years ago
During a braking test, a car is brought to rest beginning from an initial speed of 60 mi/hr in a distance of 120 ft. With the sa
maw [93]

Answer:

Explanation:

Given

Initial speed u=60\ mi/hr\approx 88\ ft/s

distance traveled before coming to rest d_1=120\ ft

using equation of motion

v^2-u^2=2as

where v=final velocity

u=initial velocity

a=acceleration

s=displacement

0-(88)^2=2\times a\times 120---1

for u_2=80\ mi/hr\approx 117.33\ ft/s

using same relation we get

0-(117.33)^2=2\times a\times (d_2)----2

divide 1 and 2 we get

(\frac{88}{117.33})^2=\frac{120}{d_2}

d_2=213.32\ ft

So a distance if 213.32 ft is required to stop the vehicle with 80 mph speed

8 0
3 years ago
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