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IrinaK [193]
3 years ago
10

A 7300 N elevator is to be given an acceleration of 0.150g by connecting it to a cable of negligible weight wrapped around a tur

ning cylindrical shaft. If the shaft’s diameter can be no larger than 16.0 cm due to space limitations, what must be its minimum angular acceleration to provide the required acceleration of the elevator?
Physics
1 answer:
Firlakuza [10]3 years ago
6 0

Answer:

α =18.75  rad/s²

Explanation:

Given that

Acceleration a = 0.15 g

We know that   g =10 m/s²

a= 0.15 x 10 = 1.5  m/s²

d= 16 cm

Radius r= 8 cm

Lets take angular acceleration =α rad/s²

As we know that

a= α r

Now by putting the values

1.5 = α  x 0.08

α =18.75  rad/s²

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See the explanation for the answer.

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slader A transcontinental flight of 4890 km is scheduled to take 40 min longer westward than eastward. The airspeed of the airpl
11111nata11111 [884]

Answer:

v_{s}=65.2km/h

Explanation:

Given data

Flight distance S=4890 km

Time difference Δt=t₂-t₁=40 min

Air speed of plane=980 km/h

To find

Speed of jet stream

Solution

When moving in the same direction as the jet stream time taken as t₁=d/(v+vs),v is velocity of plane and vs is velocity of plane

While moving in opposite direction t₂=d/(v+vs)

So

t_{2}-t_{1}=\frac{d}{(v-v_{s}) } - \frac{d}{(v+v_{s}) }\\t_{2}-t_{1}=\frac{d(v+v_{s})-d(v-v_{s})}{(v-v_{s})(v+v_{s})} \\t_{2}-t_{1}=\frac{2dv_{s}}{(v)^{2} -(v_{s})^{2} }\\0.666667h=\frac{2(4890km)v_{s}}{(980km/h)^{2} -(v_{s})^{2} }\\0.666667((980km/h)^{2} -(v_{s})^{2})=9780v_{s}\\640267-0.666667(v_{s})^{2}-9780v_{s}=0\\0.666667(v_{s})^{2}+9780v_{s}-640267=0

Apply quadratic formula to solve for vs

So

v_{s}=65.2km/h

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