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RSB [31]
2 years ago
15

A wheel starts from rest and has an angular acceleration that is given by α (t) = (6.0 rad/s4)t2. After it has turned through 10

rev its angular velocity is:
Physics
1 answer:
marissa [1.9K]2 years ago
7 0

Answer:

75 rad/s

Explanation:

The angular acceleration is the time rate of change of angular velocity. It is given by the formula:

α(t) = d/dt[ω(t)]

Hence: ω(t) = ∫a(t) dt

Also, angular velocity is the time rate of change of displacement. It is given by:

ω(t) = d/dt[θ(t)]

θ(t) = ∫w(t) dt

θ(t) = ∫∫α(t) dtdt

Given that: α (t) = (6.0 rad/s4)t² = 6t² rad/s⁴. Hence:

θ(t) = ∫∫α(t) dtdt

θ(t) = ∫∫6t² dtdt =∫[∫6t² dt]dt

θ(t) = ∫[2t³]dt = t⁴/2 rad

θ(t) = t⁴/2 rad

At θ(t) = 10 rev = (10 *  2π) rad = 20π rad, we can find t:

20π = t⁴/2

40π = t⁴

t = ⁴√40π

t = 3.348 s

ω(t) = ∫α(t) dt = ∫6t² dt = 2t³

ω(t) = 2t³

ω(3.348) = 2(3.348)³ = 75 rad/s

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A glider moving with a speed of 200 kilometers/hour experiences a cross wind of 30 kilometers/hour. What is the resultant speed
liraira [26]
You didn't say so, but we must assume that the "200 km/hr" is
the glider's air-speed, that is, speed relative to the air. 

If the air itself is moving at 30 km/hr relative to the ground and
across the glider's direction, then the glider's speed relative to
the ground is

                 √(200² + 30²)

             =  √(40,000 + 900)

             =  √(40,900)  =  202.24... km/hr   (rounded) 
5 0
2 years ago
A uniform Rectangular Parallelepiped of mass m and edges a, b, and c is rotating with the constant angular velocity ω around an
Sonbull [250]

Answer:

(a) k = \frac{Mw^{2} }{6} (a^{2} +b^{2} )

(b)  τ = \frac{M}{3} (a^{2} +b^{2} ) ∝

Explanation:

The moment of parallel pipe rotating about it's axis is given by the formula;

I = \frac{M}{3} (a^{2} +b^{2} )   ---------------------------------1

(a) The kinetic energy of a parallel pipe is also given as;

k =\frac{1}{2} Iw^{2} --------------------------------2

Putting equation 1 into equation 2, we have;

k = \frac{M}{6} (a^{2} +b^{2} )w^{2}

k = \frac{Mw^{2} }{6} (a^{2} +b^{2} )

(b) The angular momentum is given by the formula;

τ = Iw -----------------------3

Putting equation 1 into equation 3, we have

τ = \frac{Mw}{3} (a^{2} +b^{2} )

But

τ = dτ/dt = \frac{M}{3} (a^{2} +b^{2} )\frac{dw}{dt}   ------------------4

where

dw/dt = angular acceleration =∝

Equation 4 becomes;

τ = \frac{M}{3} (a^{2} +b^{2} ) ∝

8 0
2 years ago
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
The diagram below shows a person using a device called a jetpack. Water is forced downwards from the jetpack and produces an upw
klasskru [66]

Answer:

The condition necessary for the person to be able to remain stationary in mid-air is that there must be an upward force also referred to as THRUST

Explanation:

Thrust in Physics is defined as the force that propels, forces or pushes the mass of a certain object in a specified or particular direction.

Thrust can defined in the terms of a Jet pack can be defined as the force that is required to propel the mass of a person in an upward direction.

Based on the diagram that we have been shown the question, we can see that water is the fuel used to proper the user or person upwards hence, the jet pack used in the question is an HydroJet pack

The condition necessary for the person to be able to remain stationary in mid-air is that the propelling force called the thrust must be upwards such that it can overcome these 4 things:

a) Gravitational force

b) Mass of the jet pack user

c) Mass of the jet pack itself

d) The water which serves as fuel for the jet pack.

5 0
2 years ago
The speed of sound in air is 320 ms-1 and in water it is 1600 ms-1. It takes 2.5 s for sound to reach a certain distance from th
Nonamiya [84]

Answer:

Distance covered by the sound in air is 800 meter and the time taken by the sound in water for the same distance is 0.5 seconds.

Explanation:

Given:

Speed of sound in air = 320 m/s

Speed of sound in water = 1600 m/s

Time taken to reach certain distance in air = 2.5 sec

a.

We have to find the distance traveled by sound in air.

Distance = Product of speed and time.

⇒ Distance = Speed\times time\ taken

⇒ Distance = 320\times 2.5

⇒ Distance = 800 meters.

b.

Now we have to find how much time the sound will take to travel in water.

⇒ Time = Ratio of distance and speed

⇒ Time =\frac{distance}{speed}

⇒ Time =\frac{800}{1600}   <em>   ...distance = 800 m and speed = 1600 m/s</em>

⇒ Time =\frac{1}{2}

⇒ Time =0.5 seconds.

Distance covered by the sound in air is 800 meter and the time taken by the sound in water for the same distance is 0.5 seconds.

7 0
2 years ago
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