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kirill [66]
3 years ago
10

An electric fan is running on HIGH. After fan has been running for 1.3 minutes, the LOW button is pushed. The fan slows down to

109 rad/sec in 1.87 seconds. The blades of the fan have a diameter of 6.7 meters and their deceleration rate is 4.7 rad/sec2.
What was the initial angular speed of the fan in rev/sec?
ωi =
Physics
1 answer:
mafiozo [28]3 years ago
8 0

Answer:

    wo = 18.75 rev / s

Explanation:

This is an exercise in endowment kinematics, it indicates that the final angular velocity is w_f = 109 rad / s, the time to reach this velocity is t = 1.87 s and the deceleration a = 4.7 rad / s²

         w_f = w₀ - a t

         w₀ = w_f + a t

         w₀ = 109 + 4.7 1.87

         w₀ = 117.8 rad / s

let's reduce to revolutions / s

         w₀ = 117.8 rad / s (1 rev / 2pi rad)

         w₀ = 18.75 rev / s

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Answer:

8 N North.

Explanation:

Given that,

One force has a magnitude of 10 N directed north, and the other force has a magnitude of 2 N directed south.

We need to find the magnitude of net force acting on the object.

Let North is positive and South is negative.

Net force,

F = 10 N +(-2 N)

= 8 N

So, the magnitude of net force on the object is 8 N and it is in North direction (as it is positive). Hence, the correct option is (d) "8N north".

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Explain how thermal energy (temperature) affects chemical changes.
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Explanation:

hope this helps!

5 0
3 years ago
The first PCS technology used a form of time division multiplexing called ____.​ a. ​Time Division Multiple Access (TDMA) b. ​Gl
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Answer:

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3 0
3 years ago
n deep space, sphere A of mass 47 kg is located at the origin of an x axis and sphere B of mass 110 kg is located on the axis at
Volgvan

Answer:

a)-1.014x 10^{-7J

b)3.296 x  10^{-7J

Explanation:

For Sphere A:

mass 'Ma'= 47kg

xa= 0

For sphere B:

mass 'Mb'= 110kg

xb=3.4m

a)the gravitational potential energy is given by

U_{i = -GMaMb/ d

U_{i= - 6.67 x 10^{-11} x 47 x 110/ 3.4 => -1.014x 10^{-7J

b) at d= 0.8m (3.4-2.6) and U_{i=-1.014x 10^{-7J

The sum of potential and kinetic energies must be conserved as the energy is conserved.

K_{i + U_{i= K_{f + U_{f

As sphere starts from rest and sphere A is fixed at its place, therefore K_{i is zero

U_{i= K_{f + U_{f

The final potential energy is

U_{f= - GMaMb/d

Solving for 'K_{f '

K_{f = U_{i + GMaMb/d => -1.014x 10^{-7 + 6.67 x 10^{-11} x 47 x 110/ 0.8

K_{f = 3.296 x  10^{-7J

6 0
3 years ago
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