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Tamiku [17]
3 years ago
7

What is the correct displacement for the following vectors: 4 km south, 2 km north, 5 km south, and 5 km north? 2 km south 16 km

total 11 km west 6 km south
Physics
1 answer:
lesantik [10]3 years ago
6 0

<u>Answer:</u>

The correct answer option is: 2 km South.

<u>Explanation:</u>

We are to find out the displacement for the given vectors:

- 4 km south,

- 2 km north,

- 5 km south; and

- 5 km north

So starting off with the vector going 4 km south we have a displacement of 4 km south. Then it goes 2 km so we are left with 4 - 2 = 2 km south.

Next, it goes 5 km south so we will have a displacement of 2 + 5 = 7 km south. Then again it goes 5 km north so we are left with a displacement of 7 - 5 = 2 km South.

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

Explanation:

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

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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} \\\\

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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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