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icang [17]
1 year ago
12

A roundabout in a fairground requires an input power of 2.5 kW when operating at a constant angular velocity of 0.47 rad s–1 . (

a) Show that the frictional torque in the system is about 5 kN m. (3) 1. (b) When the power is switched off, the roundabout decelerates uniformly because the
Physics
1 answer:
natita [175]1 year ago
5 0

Answer:

Explanation:

Since the roundabout is rotating with uniform velocity ,

input power = frictional power

frictional power = 2.5 kW

frictional torque x angular velocity = 2.5 kW

frictional torque x .47 = 2.5 kW

frictional torque = 2.5 / .47 kN .m

= 5.32 kN . m

= 5 kN.m

b )

When power is switched off , it will decelerate because of frictional torque .

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The Vehicle Assembly Building at the Kennedy Space Center in Florida has a volume of 3,666,500 m3. Convert this volume to liters
maksim [4K]
<h2>Volume of vehicle Assembly Building at the Kennedy Space Center in Florida = 3.67 x 10⁹ L</h2>

Explanation:

The Vehicle Assembly Building at the Kennedy Space Center in Florida has a volume of 3,666,500 m³.

Volume = 3,666,500 m³

1 m³ = 1000 L

So volume = 3,666,500 x 1000 = 3666500000 L

Volume of vehicle Assembly Building at the Kennedy Space Center in Florida = 3666500000 L

Volume of vehicle Assembly Building at the Kennedy Space Center in Florida = 3.67 x 10⁹ L

3 0
2 years ago
A solid steel cylinder is standing (on one of its ends) vertically on the floor. The length of the cylinder is 3.2 m and its rad
maksim [4K]

To solve this problem it is necessary to apply the concepts related to Young's Module and its respective mathematical and modular definitions. In other words, Young's Module can be expressed as

\Upsilon = \frac{F/A}{\Delta L/L_0}

Where,

F = Force/Weight

A = Area

\Delta L= Compression

L_0= Original Length

According to the values given we have to

\Upsilon_{steel} = 200*10^9Pa

\Delta L = 5.6*10^{-7}m

L_0 = 3.2m

r= 0.59m \rightarrow A = \pi r^2 = \pi *0.59^2 = 1.0935m^2

Replacing this values at our previous equation we have,

\Upsilon = \frac{F/A}{\Delta L/L_0}

200*10^9 = \frac{F/1.0935}{5.6*10^{-7}/3.2}

F = 38272.5N

Therefore the Weight of the object is 3.82kN

4 0
1 year ago
How much energy is needed to change the speed of a 1600 kg sport utility vehicle from 15.0 m/s to 40.0 m/s?
podryga [215]

Answer:

Energy needed = 1100 kJ

Explanation:

Energy needed = Change in kinetic energy

Initial velocity = 15 m/s

Mass, m = 1600 kg

\texttt{Initial kinetic energy = }\frac{1}{2}mv^2=0.5\times 1600\times 15^2=180000J

Final velocity = 40 m/s

\texttt{Final kinetic energy = }\frac{1}{2}mv^2=0.5\times 1600\times 40^2=1280000J

Energy needed = Change in kinetic energy = 1280000-180000 = 1100000J

Energy needed = 1100 kJ

4 0
2 years ago
If a helicopter's mass is 4,500kg and the net force on it is 18,000 N upward, what is its acceleration?
Vinvika [58]
The acceleration is0.25m/s^2
4 0
1 year ago
Read 2 more answers
A box sliding on a horizontal frictionless surface runs into a fixed spring, compressing it a distance x1 from its relaxed posit
inn [45]

Answer:twice of initial value

Explanation:

Given

spring compresses x_1 distance for some initial speed

Suppose v is the initial speed and k be the spring constant

Applying conservation of energy

kinetic energy converted into spring Elastic potential energy

\dfrac{1}{2}mv^2=\dfrac{1}{2}kx_1^2----1

When speed doubles

\dfrac{1}{2}m(2v)^2=\dfrac{1}{2}kx_2^2----2

divide 1 and 2

\dfrac{1}{4}=\dfrac{x_1^2}{x_2^2}

x_2=2x_1

Therefore spring compresses twice the initial value

   

7 0
1 year ago
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