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
Answer:
As light from a star races through our atmosphere, it bounces and bumps through the different layers, bending the light before you see it. Since the hot and cold layers of air keep moving, the bending of the light changes too, which causes the star's appearance to wobble or twinkle.
The two displacement functions are
x₁ = 4t
x₂ = -161 + 48t - 4t²
where
x₁, x₂ are in meters
t is time, s
The distance between the two objects is
x = x₁ - x₂
= 4t + 161 - 48t + 4t²
x = 4t² - 44t + 161
Write this equation in the standard form for a parabola.
x = 4[t² - 11t] + 161
= 4[ (t - 5.5)² - 5.5² ] + 161
x = 4(t-5)² + 40
Ths is a parabola that faces up and has its vertex (lowest point) at (5, 40).
Therefore the closest approach of the two objects is 40 m.
The graph of x versus t confirms the result.
Answer: The distance of the closest approach is 40 m.
Answer:
The equation says that due to variation in temperature is
delt T = .59 m/s / C = 16 C * .59 m/s = 9.44 m/s
So v = 332 m/s + 9.44 m/s = 341 m/s (to three significant figures)
Answer:
-0.01 mm
Explanation:
We are given that
The value of one division of vernier scale =0.5 mm
The value of one main scale division=0.49 mm
We have to find the value of least count of the instrument in mm.
We know that
Leas count of vernier caliper=1 main scale division-1 vernier scale division
Least count of vernier caliper=0.49-0.50=-0.01 mm
Hence, the least count of the instrument=-0.01 mm
Answer: -0.01 mm