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
If the measurement is in joules then you can push something or pull something as long as you are moving the object. Formula: f*n force times newtons
In this question force is measured in g cm/s2 so we know that to get the answer we times g by cm/s2
50 × 20 = 1000
Answer: 30.34m/s
Explanation:
The sum of forces in the y direction 0 = N cos 28 - μN sin28 - mg
Sum of forces in the x direction
mv²/r = N sin 28 + μN cos 28
mv²/r = N(sin 28 + μcos 28)
Thus,
mv²/r = mg [(sin 28 + μ cos 28)/(cos 28 - μ sin 28)]
v²/r = g [(sin 28 + μ cos 28)/(cos 28 - μ sin 28)]
v²/36 = 9.8 [(0.4695 + 0.87*0.8829) - (0.8829 - 0.87*0.4695)]
v²/36 = 9.8 [(0.4695 + 0.7681) / (0.8829 - 0.4085)]
v²/36 = 9.8 (1.2376/0.4744)
v²/36 = 9.8 * 2.6088
v²/36 = 25.57
v² = 920.52
v = 30.34m/s
