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Sauron [17]
3 years ago
11

Larry drops a 5kg ball off of a building. The ball hits the ground 4.7s later. How tall is the building?

Physics
1 answer:
musickatia [10]3 years ago
8 0

Explanation:

Given:

v₀ = 0 m/s

a = 9.8 m/s²

t = 4.7 s

Find: Δy

Δy = v₀ t + ½ at²

Δy = (0 m/s) (4.7 s) + ½ (9.8 m/s²) (4.7 s)²

Δy ≈ 110 m

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Answer:

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Explanation:

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3 years ago
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A circuit supplied with 110V carries 5 Amps current. Calculate the Resistance value of the circuit?
IgorC [24]

Answer:

The value is R =  22 \  \Omega

Explanation:

From the question we are told that

     The voltage is  V  =  110 \  V

    The current is  I  =  5 \  A

Generally the resistance value is mathematically represented as

         R =  \frac{V}{I}

=>      R =  \frac{110}{5}

=>      R =  22 \  \Omega

6 0
3 years ago
A railroad car of mass 2.50*10^4 kg is moving with a speed of 4.00 m/s. It collides and couples with three other coupled railroa
NemiM [27]

Answer:

(a) 2.5 m/s

(b) 37.5 KJ

Explanation:

(a)

From the law of conservation of momentum, Initial momentum=Final momentum

mV_1+3mV_2=(m+3m)V_f=4mV_f

V_1+3V_2=4V_f and making V_f the subject then

V_f=\frac {V_1+3V_2}{4} and since V_1 is initial velocity of car, value given as 4 m/s, V_2 is the initial velocity of the three cars stuck together, value given as 2 m/s and v_f is the final velocity which is unknown. By substitution

V_f=\frac {4+(3\times2)}{4}=2.5 m/s

(b)

Initial kinetic energy is given by

\frac {mV_1^{2}}{2}+\frac {3mV_2^{2}}{2}=\frac {m(V_1^{2}+3V_2^{2}}{2}=\frac {2.5\times 10^{4}(4^{2}+3(2^{2}))}{2}=350\times10^{3} J= 350 KJ

Final kinetic energy is given by

\frac {4mV_f^{2}}{2}=\frac {4\times 2.5\times 10^{4}\times 2.5^{2}}{2}=312.5\times 10^{3} J=312.5 KJ

The energy lost is given by subtracting the final kinetic energy from the initial kinetic energy hence

Energy lost=350-312.5=37.5 KJ

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3 years ago
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Volgvan

Answer:

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Explanation:

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A projectile of mass 0.2 kg and an initial velocity of 50 m/s collides with the end of a blade attached to a turbine. The rotati
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Answer:

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