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Aloiza [94]
3 years ago
13

Consider a swimmer that swims a complete round-trip lap of a 50 m long pool in 100 seconds. The swimmer's... average speed is 0

m/s and average velocity is 0 m/s. average speed is 0.5 m/s and average velocity is 0.5 m/s. average speed is 1 m/s and average velocity is 0 m/s. average speed is 0 m/s and average velocity is 1 m/s.What is the swimmers average speed and average velocity?
Physics
1 answer:
KonstantinChe [14]3 years ago
6 0

Answer:

average speed is 1 m/s and average velocity is 0 m/s.

Explanation:

Given that :

Length of round trip = 50 m

Time taken = 100 seconds

The average speed :

Total distance / total time taken

Length of complete round trip :

(50 + 50) m, total. Distance = 100 m

100 / 100 = 1m/s

The average velocity :

Total Displacement / total time taken

Total Displacement of round trip = end point - start point = 0

0 / 100 = 0

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What is the maximum speed (in units of m/s) with which a car can round a
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Answer:

B)  15.5 m/s

Explanation:

r = 60m

μs = 0.4

using the formula max V = √r*g*μs (flat roadway)

v = sqrt(60 * 10 * 0.4)

v = 15.5 m/s

5 0
3 years ago
Suppose you increase your walking speed from 4 m/s to 12 m/s in a period of 2 s. What is your acceleration?
Karo-lina-s [1.5K]

You should write ' 4 ' in the first box,
and ' m/s² ' in the second box.

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3 years ago
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9. The acceleration due to gravity on the
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W=mg
W=75(1.6)
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3 years ago
What happens when a ray of light is directed at a mirror, a glass block and a prism?
zimovet [89]
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5 0
3 years ago
If the moment acting on the cross section is M=630N⋅m, determine the maximum bending stress in the beam. Express your answer to
-BARSIC- [3]

Answer:

2.17 Mpa

Explanation:

The location of neutral axis from the top will be

\bar y=\frac {(240\times 25)\times \frac {25}{2}+2\times (20\times 150)\times (25+(\frac {150}{2}))}{(240\times 25)+2\times (20\times 150)}=56.25 mm

Moment of inertia from neutral axis will be given by \frac {bd^{3}}{12}+ ay^{2}

Therefore, moment of inertia will be

\frac {240\times 25^{3}}{12}+(240\times 25)\times (56.25-25/2)^{2}+2\times [\frac {20\times 150^{3}}{12}+(20\times 150)\times ((25+150/2)-56.25)^{2}]=34.5313\times 10^{6} mm^{4}}

Bending stress at top= \frac {630\times 10^{3}\times (175-56.25)}{34.5313\times 10^{6}}=2.1665127\approx 2.17 Mpa

Bending stress at bottom=\frac {630\times 10^{3}\times 56.25}{34.5313\times 10^{6}}=1.026242858\approx 1.03 Mpa

Comparing the two stresses, the maximum stress occurs at the bottom and is 2.17 Mpa

8 0
4 years ago
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