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2 years ago

6

A 60-watt light bulb carries a current of 0. 5 ampere. the total charge passing through it in one hour is________.

1 answer:

tino4ka555 [31]2 years ago

8 0

<u>Answer:</u>

charge passed = 1800 C

<u>Explanation:</u>

• We can use the following formula to calculate charge:

$\boxed{Q = It}$ ,

where:

Q = charge in Coulombs

I = current in amperes (0.5 A)

t = time in seconds (1 h = 60 min = 3600 s).

• Substituting the values into the formula:

charge = 0.5 × 3600

= 1800 C

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From the question we are told that

Mass $m=6kg$

Velocity of mass $V_m=16$

Force of Tunnel $F_t=8N$

Length of Tunnel $L_t=1.6$

Height of frictional incline $H_i=2.9$

Angle of inclination $\angle =16 \textdegree$

Acceleration due to gravity $g=9.8m/s^2$

First Frictional surface has a coeﬃcient $\alpha_1 =0.21\ for\ d_c=1$

Second Frictional surface has a coeﬃcient $\alpha _2=0.1$

Generally the initial Kinetic energy is mathematically given by

$K.E=\frac{1}{2}mv^2$

$K.E=\frac{1}{2}(6)(16)^2$

$K.E=768$

Generally the work done by the Tunnel is mathematically given as

$w_t=F_t*d_t$

$w_t=8*1.6$

$w_t=12.8J$

Therefore

$Total energy\ E_t=Initial\ kinetic energy\ K.E*Work done\ by\ tunnel\ W_t$

$E_t=K.E+E_t\\E_t=768J+12.8J$

$E_t=780.8J$

Generally the energy lost while climbing is mathematically given as

$E_c=mgh$

$E_c=(6)(9.8)(2.9)$

$E_c=170.52J$

Generally the energy lost to friction is mathematically given as

$E_f=\alpha *m*g*cos\textdegree*d_c$

$E_f=0.21*6*9.8*cos16*1$

$E_f=11.86965942 \approx 12J$

Generally the energy left in the form of mass $Em$ is mathematically given as

$E_m=E_t+E_c+E_f$

$E_m=(768J)-(170.52)-(12)$

$E_m=585.48J$

Since

$E_m=\alpha_2*g*m*d$

Therefore

It slide along the second frictional region

$D=\frac{585.46}{0.1*9.81*6}$

$D=99.4665307m \approx 99.5m$

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