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Bumek [7]
3 years ago
15

A heavy flywheel rotating on its central axis is slowing down because of friction in its bearings. At the end of the first minut

e of slowing, its angular speed is 0.68 of its initial angular speed of 260 rev/min. Assuming a constant angular acceleration, find its angular speed (rev/min) at the end of the second minute.
Physics
1 answer:
Soloha48 [4]3 years ago
6 0

Answer:

ω₂ = 93.6 rev / min

Explanation:

ω₀ = 260 rev / min

ω₁ = 0.68*ω₀ = 0.68*(260 rev / min) = 176.8 rev / min

ω₂ = ?

t₁ = 1 min

t₂ = 2 min

We can apply the equation:

ω₁ = ω₀ + α*t₁     ⇒    α = (ω₁ - ω₀) / t₁  

⇒   α = (176.8 rev / min - 260 rev / min) / 1 min = - 83.2 rev / min²

then we can use the same formula, knowing the angular acceleration:

ω₂ = ω₀ + α*t₂   ⇒   ω₂ = (260 rev / min) + (- 83.2 rev / min²)*(2 min)

⇒   ω₂ = 93.6 rev / min

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What two kinds of crust are involved in a subduction zone
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Answer:

Oceanic crust and continental crust

Explanation:

A subduction zone is normally between oceanic crust which is made of basalt and continental crust which is made of granite. Oceanic crust is denser than continental crust. So when oceanic crust collides with continental crusts, it subsducts underneath the continental crust since it is denser.

5 0
3 years ago
A small truck has a mass of 2145 kg. How much work is required to decrease the speed of the vehicle from 25.0 m/s to 12.0 m/s on
MAXImum [283]

Answer:

The work required is -515,872.5 J

Explanation:

Work is defined in physics as the force that is applied to a body to move it from one point to another.

The total work W done on an object to move from one position A to another B is equal to the change in the kinetic energy of the object. That is, work is also defined as the change in the kinetic energy of an object.

Kinetic energy (Ec) depends on the mass and speed of the body. This energy is calculated by the expression:

Ec=\frac{1}{2} *m*v^{2}

where kinetic energy is measured in Joules (J), mass in kilograms (kg), and velocity in meters per second (m/s).

The work (W) of this force is equal to the difference between the final value and the initial value of the kinetic energy of the particle:

W=\frac{1}{2} *m*v2^{2}-\frac{1}{2} *m*v1^{2}

W=\frac{1}{2} *m*(v2^{2}-v1^{2})

In this case:

  • W=?
  • m= 2,145 kg
  • v2= 12 \frac{m}{s}
  • v1= 25 \frac{m}{s}

Replacing:

W=\frac{1}{2} *2145 kg*((12\frac{m}{s} )^{2}-(25\frac{m}{s} )^{2})

W= -515,872.5 J

<u><em>The work required is -515,872.5 J</em></u>

3 0
3 years ago
6 A test of a driver's perception/reaction time is being conducted on a special testing track with level, wet pavement and a dri
mylen [45]

Answer:

a. 10.5 s b. 6.6 s

Explanation:

a. The driver's perception/reaction time before drinking.

To find the driver's perception time before drinking, we first find his deceleration from

v² = u² + 2as where u = initial speed of driver = 50 mi/h = 50 × 1609 m/3600 s = 22.35 m/s, v = final speed of driver = 0 m/s (since he stops), a = deceleration of driver and s = distance moved by driver = 385 ft = 385 × 0.3048 m = 117.35 m

So, a = v² - u²/2s

substituting the values of the variables into the equation, we have

a = v² - u²/2s

a = (0 m/s)² - (22.35 m/s)²/2(117.35 m)

a =  - 499.52 m²/s²/234.7 m

a = -2.13 m/s²

Using a = (v - u)/t where u = initial speed of driver = 50 mi/h = 50 × 1609 m/3600 s = 22.35 m/s, v = final speed of driver = 0 m/s (since he stops), a = deceleration of driver = -2.13 m/s² and t = reaction time

So, t = (v - u)/a

Substituting the values of the variables into the equation, we have

t = (0 m/s - 22.35 m/s)/-2.13 m/s²

t = - 22.35 m/s/-2.13 m/s²

t = 10.5 s

b. The driver's perception/reaction time after drinking.

To find the driver's perception time after drinking, we first find his deceleration from

v² = u² + 2as where u = initial speed of driver = 50 mi/h = 50 × 1609 m/3600 s = 22.35 m/s, v = final speed of driver = 30 mi/h = 30 × 1609 m/3600 s = 13.41 m/s, a = deceleration of driver and s = distance moved by driver = 385 ft = 385 × 0.3048 m = 117.35 m

So, a = v² - u²/2s

substituting the values of the variables into the equation, we have

a = v² - u²/2s

a = (13.41 m/s)² - (22.35 m/s)²/2(117.35 m)

a = 179.83 m²/s² - 499.52 m²/s²/234.7 m

a = -319.69 m²/s² ÷ 234.7 m

a = -1.36 m/s²

Using a = (v - u)/t where u = initial speed of driver = 50 mi/h = 50 × 1609 m/3600 s = 22.35 m/s, v = final speed of driver = 30 mi/h = 30 × 1609 m/3600 s = 13.41 m/s, a = deceleration of driver = -1.36 m/s² and t = reaction time

So, t = (v - u)/a

Substituting the values of the variables into the equation, we have

t = (13.41 m/s - 22.35 m/s)/-1.36 m/s²

t = - 8.94 m/s/-1.36 m/s²

t = 6.6 s

4 0
3 years ago
Match the technology that uses radio waves to the field in which it is used
Elden [556K]

Answer:

Naturally occurring radio waves are made by lightning or by astronomical objects. Artificially generated radio waves are used for fixed and mobile radio communication, broadcasting, radar and other navigation systems, communications satellites, computer networks and innumerable other applications.

Explanation:

5 0
2 years ago
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