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3 years ago

A charge of 24 C pass a certain point in a conductor in 10 s. What is the size of current in the conductor?

Physics

2 answers:

arsen [322]3 years ago

8 0

Answer:

2.4A

Explanation:

I = Q/t

I = 24/10

= 2.4A

allsm [11]3 years ago

6 0

Answer:

write the name of any five districts of nepal

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Which statement is correct about how the temperature of an object changes?

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The molecules making up object collide with the molecules of the other and some of the kinetic energy from a warmer object is transferred to the cooler one.

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3 years ago

Two teams are playing tug of war. Team A pulls to the right with a force of 450 N. Team B pulls to the left with a force of 415

melamori03 [73]

Answer:

Two teams are playing tug of war. Team A pulls to the right with a force of 450 N. Team B pulls to the left with a force of 415 N

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3 years ago

A 25.0 kg bag of peat moss sits in the back of a flatbed truck, driving up a hill. The bag experiences a 225N normal force. The

malfutka [58]

Answer:

$\theta = 23.32^o$

$\mu_s = 0.27$

$s = 0.948 \ m$

Explanation:

From the question we are told that

The mass of the bag is $m_b = 25.0 \ kg$

The normal force experienced is $F_n = 225 \ N$

The maximum acceleration of the bag is $a = 2.40 \ m/s^2$

Generally this normal force experience by the bag is mathematically represented as

$F_n = mg cos \theta$

=> $225 = (25 * 9.8) cos \theta$

=> $0.9183 = cos \theta$

=> $\theta = cos^{-1}[0.9183]$

=> $\theta = 23.32^o$

Generally for the bag not to slip , it means that the frictional force is equal to the sliding force

$F_f = F_s$

Hence $F_f$ is mathematically represented as

$F_f = \mu_s * F_n$

While $F_s$ is mathematically represented as

$F_s = m * a$

$\mu_s * F_n = m * a$

=> $\mu_s * 225 = 25 * 2.40$

=> $\mu_s = 0.27$

Generally from the workdone equation we have that

$KE_f - KE_i = W_f$

Here $W_f$ is the work done by friction which is mathematically represented as

$W_f = m * g * \mu_k * s$

Here s is the distance covered by the bag

$KE_f$ is zero given that velocity at rest is zero

and

$KE_i = \frac{1}{2} * m* v_i^2$

$\frac{1}{2} * m* v_i^2 = m * g * \mu_k * s$

=> $\frac{1}{2} * v_i^2 = g * \mu_k * s$

substituting 2.55 m/s for v_i and 0.350 for \mu_k we have that

$\frac{1}{2} * 2.55^2 = 9.8 * 0.350 * s$

=> $s = 0.948 \ m$

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4 years ago

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Answer:

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