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NNADVOKAT [17]
3 years ago
5

Calculate the current in a circuit if 500 C of charge passes through it in 10 minutes. *

Physics
1 answer:
Sloan [31]3 years ago
7 0

Answer:

Choice a. Approximately 0.83\; \rm A on average.

Explanation:

The electric current through a wire is the rate at which electric charge flows through a cross-section of this wire.

Assume that electric charge of size Q flowed through a wire cross-section over a period of time t. The average current in that wire would be:

\displaystyle I = \frac{Q}{t}.

For this question:

  • Q = 500\; \rm C, whereas
  • t = 10\; \rm \text{minutes}.

Therefore, the average current in this circuit would be:

\displaystyle I = \frac{Q}{t} = \frac{500\; \rm C}{10\; \text{minutes}} = 50\; \rm C /\text{minute}.

However, the units in the choices are all in \rm A (for Amperes.) One Ampere is equal to one \rm C / \text{second}. It will take some unit conversations to change the unit of I = 50\; \rm C/ \text{minute} (coulombs-per-minute) to coulombs-per-second.

\begin{aligned}I &= 50\; \rm C/ \text{minute} \\ &= \frac{50\; \rm C}{1\; \rm \text{minute}} \times \frac{1 \; \text{minute}}{60\; \rm \text{seconds}} \approx 0.83\; \rm C/ \text{second} = 0.83 \; \rm A\end{aligned}.

Hence, the most accurate choice here would be choice a.

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Calculate the work done if a boy lifts a bag of cement 500N to the floor of a lorry 2.5m above the ground
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W = 1250 J = 1.25 KJ

Explanation:

The work done by the boy is due to the change in the position of the cement vertically. Hence, the work done in this case will be equal to the potential energy of the cement:

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W = Work done = ?

mg = W = weight of cement = 500 N

h = height covered = 2.5 m

Therefore,

W = (500\ N)(2.5\ m)

<u>W = 1250 J = 1.25 KJ</u>

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A small block of mass m1 = 0.4 kg is placed on a long slab of mass m2 = 2.8 kg. Initially, the slab is stationary and the block
mel-nik [20]

Answer:

v₁ = 0.375 m / s ,   x = 0.335 m

Explanation:

Let's analyze this interesting exercise, the block moves and has a friction force with the tile, we assume that the speed of the block is constant, so the friction force opposes the block movement. For the only force that acts (action and reaction) this friction force exerted by the block that is in the direction of movement of the tile.

We can also see that the isolated system formed by the block and the tile will reach a stable speed where friction cannot give the system more energy, this speed can be found by treating the system with the conservation of linear momentum.

initial moment. Right at the start of the movement

       p₀ = m v₀ + 0

final moment. Just when it comes to equilibrium

      p_{f} = (m + M) v₁

how the forces are internal

       p₀ =p_{f}

       m v₀ = (m + M) v₁

       v₁ = m /m+M    v₀

let's calculate

       v₁ = 0.4 /(0.4 + 2.8)  3

       v₁ = 0.375 m / s

 

Let's apply Newton's second law to the Block, to find the friction force

Y axis

       N - W = 0

       N = W

       N = m g

where m is the mass of the block

the friction force has the formula

      fr = μ N

      fr = μ m g

We apply Newton's second law to slab    

X axis

       fr = M a

where M is the mass of the slab

       μ m g = M a

       a = μ g m / M

let's calculate

       a = 0.15  9.8  0.4 / 2.8

       a = 0.21 m / s²

With kinematics we can find the position

       v²= v₀²+2 a x

as the slab is initially at rest, its initial velocity is zero

       v² = 2 a x

       x = v2 / 2a

let's calculate

        x = 0.375²/2 0.21

        x = 0.335 m

4 0
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