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stealth61 [152]
3 years ago
15

A firefighter with a weight of 756 N slides down a vertical pole with an acceleration of 2.96 m/s2, directed downward. What are

the magnitudes and directions (choose the positive direction up) of the vertical forces
(a) on the firefighter from the pole and
(b) on the pole from the firefighter?
Physics
1 answer:
Katen [24]3 years ago
3 0

Answer:

a) F = 527.65 N, Force applied is upwards.

b)F = - 527.65 N, where, negative sign depicts Force is applied downwards.

Explanation:

Data provided:

Weight of the firefighter = 756 N

Mass of the firefighter = 756/9.8 = 77.14 Kg

Acceleration, a = 2.96 m/s²

a) In the absence of the pole the firefighter would have been moving down with an acceleration of 9.8 m/s (i.e the acceleration due to the gravity), but due to the presence of the pole the acceleration of the firefighter has been reduced. thus, a force is applied by the pole on the firefighter to reduce the acceleration.

therefore, we have

F = ma(net) = 77.14 × (9.8-2.96) = 527.65 N, Force applied is upwards.

B) According to the Newton's third law, the force will be equal and opposite to the force in the part a)

thus, we have

F = - 527.65 N

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A phonograph record 0.15 m in its radius rotates 18 times per 90 seconds what is the frequency?
ioda

Answer:

The frequency of the phonograph record is 0.2 Hz

Explanation:

The frequency of an object moving in uniform circular motion is the number of completed cycles the object makes in a specified time period

The given parameters of the phonograph record are;

The radius of the record = 0.15 m

The number of times the phonograph record rotates, n = 18 times

The time it takes the phonograph record to rotate the 18 times, t = 90 seconds

The frequency of the phonograph record, f = (The number of times the phonograph record rotates) ÷ (The time it takes the phonograph record to rotate the 18 times)

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The frequency of the phonograph record = 0.2 Hz.

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3 years ago
A space station, in the form of a wheel 119 m in diameter, rotates to provide an "artificial gravity" of 2.20 m/s2 for persons w
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3 years ago
g The electric power needs of a community are to be met by windmills with 40-m-diameter rotors. The windmills are to be located
Ksenya-84 [330]

Answer:

Explanation:

Given Data

The diameter of the wind mills is d = 40m

Velocity of the air is V = 6 m / s

Required power output is:  P ₀ = 2100 k W

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E = V ² / 2

Substitute the value in above expression

E = ( 6 m / s ) ² / 2

E = 18 m ² / s ² x (1kJ/kg / 1000m²/s²)

E = 0.018 k J / k g

Expression to calculate the density of the air is

P v =m R T

m /v = P  /RT ⋯ ⋯( I )

Here  

m  is the mass of the air,  

v  is the volume of the air,  

P  is the atmospheric pressure,  

T  is the standard temperature at the atmospheric pressure and  

R  is the gas constant

As the density is

ρ = m /V

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ρ = 101  kP a /( 0.287 k J / k g ⋅ K ) ( 298 K )

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m = ρ A V ⋯ ⋯ ( I I )

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Substitute the value in above expression

A = π /4 ( 40 m ²)

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Substitute the value in expression (II)

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P w = m ( V ² /2 )

Substitute the value in above expression

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n = P o /P w

Substitute the value in above expression

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Answer:

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