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LenKa [72]
3 years ago
6

scott and jafar are ready to work a problem. in this simulation, assume the coordinates of the points are as follows. a (0 s, 30

m); b (10 s, 50 m); c (20 s, 38 m); d (30 s, 0 m); e (40 s, −38 m); and f (50 s, −50 m) recall the definitions of average speed, vavg ≡ d δt , and average velocity in the x direction, vx ,avg ≡ δx δt . find the average velocity from circled a to circled b.
Physics
1 answer:
max2010maxim [7]3 years ago
7 0

Answer:

Explanation:

point a represents time 0 and position coordinate 30

point b represents time 10 s , and position coordinate 50 m .

time elapsed = 10 - 0 = 10 s .

displacement = 50 m - 30 m

= 20 m

average velocity = displacement / time elapsed

= 20 / 10

= 2 m /s .

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Assume that the position vector of A is r=i+j+k . Determine the moment about the origin O if the force F=(1)i+(0)j+(5)k . The mo
ddd [48]

Answer:

M₀ = 5i - 4j - k

Explanation:

Using the cross product method, the moment vector(M₀) of a force (F) is about a given point is equal to cross product of the vector A from the point (r) to anywhere on the line of action of the force itself. i.e

M₀ = r x F

From the question,

r = i + j + k

F = 1i + 0j +  5k

Therefore,

M₀ = (i + j + k) x (1i + 0j +  5k)

M₀ = \left[\begin{array}{ccc}i&j&k\\1&1&1\\1&0&5\end{array}\right]

M₀ = i(5 - 0) -j(5 - 1) + k(0 - 1)

M₀ = i(5) - j(4) + k(-1)

M₀ = 5i - 4j - k

Therefore, the moment about the origin O of the force F is

M₀ = 5i - 4j - k

3 0
2 years ago
How long will it take to go 150000m traveling at 50km/hr
Tom [10]

Answer:

3000 hurs

Explanation:  just divide 150000 by 50 and get 3000

4 0
2 years ago
A beam of light has a wavelength of 4.5 x10^-7 meter in a vacuum. the frequency of this light is
valkas [14]
The basic relationship between frequency and wavelength for light (which is an electromagnetic wave) is
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f= \frac{c}{\lambda}= \frac{3 \cdot 10^8 m/s}{4.5 \cdot 10^{-7} m}=6.7 \cdot 10^{14} Hz
3 0
3 years ago
4 what is the difference between an array's size declarator and a subscript?
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7 0
3 years ago
You see lightning and 30 seconds later you hear thunder. how far away is the thunderstorm? take the speed of sound to be 339 m/s
Jobisdone [24]
Let the observer be 'd' distance away from the thunderstorm and let light take 't' time to reach the observer
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For sound,
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Putting value from 1 in 2.
               d = 10^4 m(approx)
3 0
3 years ago
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