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AleksAgata [21]
2 years ago
8

A horizontal 810-N merry-go-round of radius 1.60 m is started from rest by a constant horizontal force of 55 N applied tangentia

lly to the merry-go-round. Find the kinetic energy of the merry-go-round after 4.0 s. (Assume it is a solid cylinder. Also assume the force is applied at the outside edge.)
Physics
1 answer:
Sloan [31]2 years ago
6 0

Answer:

576 joules

Explanation:

From the question we are given the following:

weight = 810 N

radius (r) = 1.6 m

horizontal force (F) = 55 N

time (t) = 4 s

acceleration due to gravity (g) = 9.8 m/s^{2}

K.E = 0.5 x MI x ω^{2}

where MI is the moment of inertia and ω is the angular velocity

MI = 0.5 x m x r^2

mass = weight ÷ g = 810 ÷ 9.8 = 82.65 kg

MI = 0.5 x 82.65 x 1.6^{2}

MI = 105.8 kg.m^{2}

angular velocity (ω) = a x t

angular acceleration (a) = torque ÷ MI

where torque = F x r = 55 x 1.6 = 88 N.m

a= 88 ÷ 105.8 = 0.83 rad /s^{2}

therefore

angular velocity (ω) = a x t = 0.83 x 4 = 3.33 rad/s

K.E = 0.5 x MI x ω^{2}

K.E = 0.5 x 105.8 x 3.33^{2} = 576 joules

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The sun's energy begins as what form of energy?
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The idea that earth was the center of the universe is an example of ---that has been long dicarded
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Geocentric model of the universe

Explanation:

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2 years ago
A high-speed flywheel in a motor is spinning at 450 rpm when a power failure suddenly occurs. The flywheel has mass 40.0 kg and
alexira [117]

Answer:

A) \omega_f=17.503\ rad.s^{-1}

B) t=55.6822\ s

C) \theta=1312\ rad

Explanation:

Given:

  • mass of flywheel, m=40\ kg
  • diameter of flywheel, d=0.72\ m
  • rotational speed of flywheel, N_i=450\ rpm \Rightarrow \omega_i=\frac{450\times 2\pi}{60} =15\pi\ rad.s^{-1}
  • duration for which the power is off, t_0=35\ s
  • no. of revolutions made during the power is off, \theta=180\times 2\pi=360\pi\ rad

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

\theta=\omega_i.t+\frac{1}{2} \alpha.t^2

360\pi=15\pi\times 35+\frac{1}{2} \times \alpha\times35^2

\alpha=-0.8463\ rad.s^{-2}

Negative sign denotes deceleration.

A)

Now using the equation:

\omega_f=\omega_i+\alpha.t

\omega_f=15\pi-0.8463\times 35

\omega_f=17.503\ rad.s^{-1} is the angular velocity of the flywheel when the power comes back.

B)

Here:

\omega_f=0\ rad.s^{-1}

Now using the equation:

\omega_f=\omega_i+\alpha.t

0=15\pi-0.8463\times t

t=55.6822\ s is the time after which the flywheel stops.

C)

Using the equation of motion:

\theta=\omega_i.t+\frac{1}{2} \alpha.t^2

\theta=15\pi\times 55.68225-0.5\times 0.8463\times 55.68225^2

\theta=1312\ rad revolutions are made before stopping.

3 0
3 years ago
A 500-N weight sits on the small piston of a hydraulic machine. The small piston has area 2.0 cm2. If the large piston has area
weqwewe [10]

Answer:

W₂= 10000 N

Explanation:

Pascal´s Principle can be applied in the hydraulic press:

If we apply a small force (F1) on a small area piston A1, then, a pressure (P) is generated that is transmitted equally to all the particles of the liquid until it reaches a larger area piston and therefore a force (F2) can be exerted that is proportional to the area (A2) of the piston:

Pressure is defined as the force (F) applied per unit area (A)

P=F/A   (N/m²)

P1=P2

\frac{F_{1} }{A_{1} } = \frac{F_{2} }{A_{2} }

F_{2} = \frac{F_{1}*A_{2}  }{A_{1}}  Equation (1)

Data

W₁ = weight sits on the small piston

F₁ = W₁= 500 N

A₁ = 2.0 cm²

A₂ = 40 cm²

Calculation of the weight  (W₂) can the large piston support

We replace data in the equation (1)

F_{2} = \frac{(500)*(40) }{2}

F₂ = 10000 N

W₂= F₂= 10000 N

6 0
3 years ago
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