speed of tortoise is given as v1 = 0.14 m/s
speed of hare is given as v2 = 20*0.14 = 2.8 m/s
now let say the total length of the path is "d"
so the total time taken by the tortoise to cover this
now given that hare took rest for 1 min
so total time of run for hare is (t - 60)s
so the distance that hare covered is given by
now by above two equations
and the time t is given by
so part a)
t = 63 s
part b)
d = 8.82 m
Answer:
Option B) Speed is the answer.
Answer:
These are Diffraction Grating Questions.
Q1. To determine the width of the slit in micrometers (μm), we will need to use the expression for distance along the screen from the center maximum to the nth minimum on one side:
Given as
y = nDλ/w Eqn 1
where
w = width of slit
D = distance to screen
λ = wavelength of light
n = order number
Making x the subject of the formula gives,
w = nDλ/y
Given
y = 0.0149 m
D = 0.555 m
λ = 588 x 10-9 m
and n = 3
w = 6.6x10⁻⁵m
Hence, the width of the slit w, in micrometers (μm) = 66μm
Q2. To determine the linear distance Δx, between the ninth order maximum and the fifth order maximum on the screen
i.e we have to find the difference between distance along the screen (y₉-y₅) = Δx
Recall Eqn 1, y = nDλ/w
given, D = 27cm = 0.27m
λ = 632 x 10-9 m
w = 0.1mm = 1.0x10⁻⁴m
For the 9th order, n = 9,
y₉ = 9 x 0.27 x 632 x 10-9/ 1.0x10⁻⁴m = 0.015m
Similarly, for n = 5,
y₅ = 5x 0.27 x 632 x 10-9/ 1.0x10⁻⁴m = 0.0085m
Recall, Δx = (y₉-y₅) = 0.015 - 0.0085 = 0.0065m
Hence, the linear distance Δx between the ninth order maximum and the fifth order maximum on the screen = 6.5mm
Answer:
Line the numbers from smallest to largest the subtract the smallest from the largest numbers.
Explanation:
The average speed is 20.8 m/s
Explanation:
The average speed for the trip is given by:
where
d is the distance covered
t is the time elapsed
For the trip in this problem, we have:
d = 187 km = 187,000 m is the distance travelled
The initial time is 10:00 pm while the arriving time is 12:30 am: this means that the time elapsed is 2.5 hours. Converting into seconds,
Therefore, the average speed for the trip is
Learn more about average speed:
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