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kobusy [5.1K]
2 years ago
10

What separates musical theater from other dance styles?

Physics
1 answer:
Rus_ich [418]2 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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A 95-kg astronaut is stranded from his space shuttle. He throws a 2-kg hammer away from the shuttle with a velocity of 19 m/s .
geniusboy [140]

Answer:

The astronaut will be propelled towards the shuttle at the rate of 0.4 m/s.

Explanation:

Given;

mass of  the astronaut, m₁ = 95 kg

mass of the hammer thrown, m₂ = 2 kg

velocity of the hammer, v₂ = 19 m/s

let the recoil velocity of the shuttle = v₁

Apply the principle of conservation of linear momentum;

m₁v₁ = m₂v₂

v₁ = m₂v₂/m₁

v₁ = (2 x 19) / 95

v₁ = 0.4 m/s

Therefore, the astronaut will be propelled towards the shuttle at the rate of 0.4 m/s.

5 0
2 years ago
Earl is using his hands to hold a metal pan 10 centimeters above a hot burner. How can this scenario be changed to demonstrate c
Olenka [21]

Answer: touch the pan to the burner

Explanation:

There are three modes of heat transfer:

conduction, convection and radiation.  

For conduction, the heat transfers from a hot object to a cold object when the two are in contact.

For convection there is bulk motion of fluid occurs which transfers the heat.

For heat transfer by radiation, medium is not required.

Thus, to demonstrate conduction between pan and burner, the pan must touch the burner.

8 0
3 years ago
Read 2 more answers
Electromagnetic waves are commonly referred to as _________
mart [117]
It’s going to be both answer A and B but if you can only answer one then it’s going to be B
3 0
2 years ago
A uniform disk with mass 35.2 kg and radius 0.200 m is pivoted at its center about a horizontal, frictionless axle that is stati
Sergio [31]

Answer:

a) v = 1.01 m/s

b) a = 5.6 m/s²

Explanation:

a)

  • If the disk is initially at rest, and it is applied a constant force tangential to the rim, we can apply the following expression (that resembles Newton's 2nd law, applying to rigid bodies instead of point masses) as follows:

       \tau = I * \alpha  (1)

  • Where τ is the external torque applied to the body, I is the rotational inertia of the body regarding the axis of rotation, and α is the angular acceleration as a consequence of the torque.
  • Since the force is applied tangentially to the rim of the disk, it's perpendicular to the radius, so the torque can be calculated simply as follows:
  • τ = F*r (2)
  • For a solid uniform disk, the rotational inertia regarding an axle passing through its center  is just I = m*r²/2 (3).
  • Replacing (2) and (3) in (1), we can solve for α, as follows:

       \alpha = \frac{2*F}{m*r} = \frac{2*34.5N}{35.2kg*0.2m} = 9.8 rad/s2 (4)

  • Since the angular acceleration is constant, we can use the following kinematic equation:

        \omega_{f}^{2}  - \omega_{o}^{2} = 2*\Delta \theta * \alpha (5)

  • Prior to solve it, we need to convert the angle rotated from revs to radians, as follows:

       0.2 rev*\frac{2*\pi rad}{1 rev} = 1.3 rad (6)

  • Replacing (6) in (5), taking into account that ω₀ = 0 (due to the disk starts from rest), we can solve for ωf, as follows:

       \omega_{f} = \sqrt{2*\alpha *\Delta\theta} = \sqrt{2*1.3rad*9.8rad/s2} = 5.1 rad/sec (7)

  • Now, we know that there exists a fixed relationship the tangential speed and the angular speed, as follows:

        v = \omega * r (8)

  • where r is the radius of the circular movement. If we want to know the tangential speed of a point located on the rim of  the disk, r becomes the radius of the disk, 0.200 m.
  • Replacing this value and (7) in (8), we get:

       v= 5.1 rad/sec* 0.2 m = 1.01 m/s (9)

b)    

  • There exists a fixed relationship between the tangential and the angular acceleration in a circular movement, as follows:

       a_{t} = \alpha * r (9)

  • where r is the radius of the circular movement. In this case the point is located on the rim of the disk, so r becomes the radius of the disk.
  • Replacing this value and (4), in (9), we get:

       a_{t}  = 9.8 rad/s2 * 0.200 m = 1.96 m/s2 (10)

  • Now, the resultant acceleration of a point of the rim, in magnitude, is the vector sum of the tangential acceleration and the radial acceleration.
  • The radial acceleration is just the centripetal acceleration, that can be expressed as follows:

       a_{c} = \omega^{2} * r  (11)

  • Since we are asked to get the acceleration after the disk has rotated 0.2 rev, and we have just got the value of the angular speed after rotating this same angle, we can replace (7) in (11).
  • Since the point is located on the rim of the disk, r becomes simply the radius of the disk,, 0.200 m.
  • Replacing this value and (7) in (11) we get:

       a_{c} = \omega^{2} * r   = (5.1 rad/sec)^{2} * 0.200 m = 5.2 m/s2 (12)

  • The magnitude of the resultant acceleration will be simply the vector sum of the tangential and the radial acceleration.
  • Since both are perpendicular each other, we can find the resultant acceleration applying the Pythagorean Theorem to both perpendicular components, as follows:

       a = \sqrt{a_{t} ^{2} + a_{c} ^{2} } = \sqrt{(1.96m/s2)^{2} +(5.2m/s2)^{2} } = 5.6 m/s2 (13)

6 0
3 years ago
Opal adds 25 grams of salt to a one-liter glass beaker filled up to its volume mark with pure water. She stirs the water until t
Mars2501 [29]

By what i know i think that the answer would be A a homogeneous mixture.

5 0
3 years ago
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