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atroni [7]
3 years ago
8

Alan starts from his home and walks 1.3 km east to the library. He walks an additional 0.68 km east to a music store. From there

, he walks 1.1 km north to a friend’s house and an additional 0.42 km north to a grocery store before he finally returns home along the same path. What is his final displacement and total distance traveled?
Physics
1 answer:
zepelin [54]3 years ago
3 0

Answer: final Displacement = 0 km, total distance covered =7 km

Explanation:

Given the following :

From home to library = 1.3 km East

Library to music store = 0.68km East

Music store to friend's house = 1.1km North

Friend's house to grocery store = 0.42 km North

Displacement is the net change in distance traveled.

Eastward distance :

To = (1.3 + 0.68)km = 1.98km East

Fro = (0.68 + 1.3)km = 1.98 km East

Δ distance = (1.98 - 1.98) = 0

Northward direction:

To = (1.1 + 0.42)km = 1.52km north

Fro = (0.42 + 1.1)km = 1.52km North

Δ distance = (1.98 - 1.98) = 0

Hence final Displacement = 0

Total distance covered = 2 × (1.3 + 0.68 + 1.1 + 0.42) = 2 × 3.5

= 7km

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2 years ago
A 2.4-kg ball falling vertically hits the floor with a speed of 2.5 m/s and rebounds with a speed of 1.5 m/s. What is the magnit
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Answer:

9.6 Ns

Explanation:

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2 years ago
How do I calculate the tension in the horizontal string?
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ANSWER

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EXPLANATION

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• The mass of the ball, m = 1.8kg

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In this position, the ball is at rest, so, by Newton's second law of motion, for each direction we have,

\begin{gathered} T_{1y}-F_g=0_{}_{}_{} \\ T_2-T_{1x}=0 \end{gathered}

The components of the tension of the first string can be found considering that they form a right triangle, where the vector of the tension is the hypotenuse,

\begin{gathered} T_{1y}=T_1\cdot\cos 30\degree \\ T_{1x}=T_1\cdot\sin 30\degree \end{gathered}

We have to find the tension in the horizontal string, T₂, but first, we have to find the tension 1 using the first equation,

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Solve for T₁,

T_1=\frac{m\cdot g}{\cos30\degree}=\frac{1.8kg\cdot9.8m/s^2}{\cos 30\degree}\approx20.37N

Now, we use the second equation to find the tension in the horizontal string,

T_2-T_1\sin 30\degree=0

Solve for T₂,

T_2=T_1\sin 30\degree=20.37N\cdot\sin 30\degree\approx10.19N

Hence, the tension in the horizontal string is 10.19N, rounded to the nearest hundredth.

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