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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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A pogo stick has a spring with a force constant of 2.50 × 104 N/m , which can be compressed 11.2 cm. what maximum height, in met
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Answer:

h = 36.4 cm

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by energy of conservation

K.E_i +P.E_i= K.E_f + P.E_f

\dfrac{1}{2}kx^2 = mgh

\dfrac{1}{2}\times 2.5 \times 10^4 \times 0.112^2 = 44\times 9.8 \times h

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7 0
3 years ago
Compute the velocity of an electron that has been accelerated through a difference of potential of 100 volts. express your answe
Elodia [21]

The velocity of an electron that has been accelerated through a difference of potential of 100 volts will be 5.93 * 10^{6} m/s

Electrons move because they get pushed by some external force. There are several energy sources that can force electrons to move. Voltage is the amount of push or pressure that is being applied to the electrons.

By conservation of energy, the kinetic energy has to equal the change in potential energy, so KE=q*V. The energy of the electron in electron-volts is numerically the same as the voltage between the plates.

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charge of electron = 1.6 × 10^{-19} C

mass of electron  = 9.1 × 10^{-31} kg

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potential energy is stored in the form of work done

potential energy = work done = Force * displacement

                                                   = q * (E * d)  

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kinetic energy = 1/2 * m * v^{2}

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equation both the equations

1/2 *  9.1 × 10^{-31} *  v^{2} = 1.6 × 10^{-17}

v^{2} = 0.352 * 10^{14} m/s

v^{2} = 35.2 * 10^{12}

    = 5.93 * 10^{6} m/s

To learn more about  kinetic energy in electric field  here

brainly.com/question/8666051

#SPJ4

3 0
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