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inysia [295]
3 years ago
9

The alarm at a fire station rings and a 87.5-kg fireman, starting from rest, slides down a pole to the floor below (a distance o

f 4.10 m). Just before landing, his speed is 1.75 m/s. What is the magnitude of the kinetic frictional force exerted on the fireman as he slides down the pole?
Physics
1 answer:
blsea [12.9K]3 years ago
6 0

Answer:

F_f=840N

Explanation:

From the question we are told that

Weight of fireman W_f= 87.5kg

Pole distance D=4.10m

Final speed is V_f 1.75m/s

Generally the equation for velocity is mathematically represented as

v^2 = v_0^2 + 2 a d

Therefore Acceleration a

a'= v^2 / 2 d

a'= 0.21m/s^2

Generally the equation for Frictional force F_f is mathematically given as

F_f=m*a

F_f=m*(g-0.21)

F_f=87.5*(9.81-0.21)

Therefore

F_f=840N

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3 years ago
A rock is dropped off a cliff and falls the first half of the distance to the ground in 2.0 seconds. how long will it take to fa
son4ous [18]

Answer:

The answer is 0.83 seconds.

Explanation:

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h=1/2*g*t^2

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h=1/2*9.8*2^2=19.6

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39.2=1/2*9.8*t^2\\t^2=8\\t=2.83

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Read 2 more answers
Each corner of a right-angled triangle is occupied by identical point charges "A", "B", and "C" respectively. Draw a sketch of t
NISA [10]

Answer:

Fnet = F√2

Fnet = kq²/r² √2

Explanation:

A exerts a force F on B, and C exerts an equal force F on B perpendicular to that.  The net force can be found with Pythagorean theorem:

Fnet = √(F² + F²)

Fnet = F√2

The force between two charges particles is:

F = k q₁ q₂ / r²

where

k is Coulomb's constant, q₁ and q₂ are the charges, and r is the distance between the charges.

If we say the charge of each particle is q, then:

F = kq²/r²

Substituting:

Fnet = kq²/r² √2

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If BHALA AHMAD KHAN applied the 20N force is applied on an object moving with the velocity 30 m/s. calculate the power in KW.
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