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2 years ago

Which statement is the correct representation of these electric field lines?

Physics

2 answers:

irinina [24]2 years ago

6 0

C . plate a is negatively charged and plate b is positively charged

Taya2010 [7]2 years ago

5 0

Answer:

C. Plate A is negatively charged and plate B is positively charged.

Explanation:

The direction of the lines of an electric field represents the direction to which a positive test charge placed in that field would move. Therefore, since positive charges are repelled by positive charges and attracted by negative charges, this means that:

- Electric field lines point away from positive charges, and

- Electric field lines point towards negative charges

In the figure in the problem, we see that electric field lines go away from plate B and towards plate A. This means that:

- Plate B is positively charged

- Plate A is negatively charged

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A 55-kg block, starting from rest, is pushed a distance of 5.0 m across a floor by a horizontal force Fp whose magnitude is 140

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Answer:

The answer is "151.25 J and -547.64 J".

Explanation:

$u = 0\\\\v = 2.35\ \frac{m}{sec}\\\\d = 5.0 \ m\\\\$

Using formula:

$v^2 = u^2 + 2 \times a \times d\\\\2.35^2 = 0^2 + 2 \times a \times 5\\\\a = \frac{2.35^2}{10} \\\\$

$= 0.55 \ \frac{m}{sec^2}\\\\$

$F_{net} = m \times a\\\\F_{net} = 55 \times 0.55 = 30.25\ N\\\\$

Calculating the Work by net force

$W = F_{net}\times d\\\\W = 30.25 \times 5 = 151.25 \ J\\\\$

The above work is converted into thermal energy.

Now,

$F_{net} = F_p - F_f\\\\F_p = 140 \ N\\\\F_f = u_k\times m \times g = u_k \times 55 \times 9.81\\\\F_f = 539.55 \times u_k\\\\30.25 = 140 - u_k \times 55 \times 9.81\\\\u_k = \frac{(140 - 30.25)}{(55\times 9.81)}\\\\uk = 0.203 = \text{Coefficient of friction}\\\\W_f = -F_f \times d\\\\W_f = -0.203 \times 55 \times 9.81 \times 5\\\\Work\ done\ by\ friction = -547.64 \ J$

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