Answer:a b c
Explanation: I’m not sure tho
Odpowiedź:
0,049 m / s
Wyjaśnienie:
Biorąc pod uwagę, że:
Dystans biegu = 900m
Czas trwania = 205 minut
Długość przejścia = 300 m
Zajęty czas = 205 minut
Średnia prędkość :
(Przebieg + pokonany dystans) / całkowity czas
Średnia prędkość :
(900 m +. 300 m) / 205 + 205
1200 m / 410 minut
Minuty do sekund
1200 / (410 * 60)
1200/24600
= 0,0487804
= 0,049 m / s
Answer:
B. NET force: 2 resultant motion: left
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C. Net force: 3 Resultant motion: Left
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D. Net Force: 7 Resultant motion: right
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E. Net Force:0 resultant motion: NO MOTION
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F. NET Force: 3 resultant motion: Down
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G. NET FORCE: 10 resultant motion: up
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H. Net force: 3 Resultant motion: left
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I. Net force: 50 Resultant motion: right
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J. NET FORCE: 75 Resultant motion: down
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K. Net force :200 Resultant motion: Right
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L. Net force: 0 resultant motion:No motion
Explanation:
Answer:
4.245s
Explanation:
Given that,
Hypothetical value of speed of light in a vacuum is 18 m/s
Speed of the car, 14 m/s
Time given is 6.76 s, and we're asked to find the observed time, T
The relationship between the two times can be given as
T = t / √[1 - (v²/c²)]
The missing variable were looking for is t, and we can find it if we rearrange the formula and make t the subject
t = T / √[1 - (v²/c²)]
And now, we substitute the values and insert into the equation
t = 6.76 * √[1 - (14²/18²)]
t = 6.76 * √[1 - (196/324)]
t = 6.76 * √(1 - 0.605)
t = 6.76 * √0.395
t = 6.76 * 0.628
t = 4.245 s
Therefore, the time the driver measures for the trip is 4.245s
