Chameleon tongue reaches 23 cm.
Train's final speed is 32 m/s.
The distance the tongue travels is divided into 2 phases.
1. The acceleration phase.
2. The coasting phase.
For the acceleration phase, the formula d = 0.5AT^2 determines how far the tongue travels while accelerating. The during the coasting phase, the tongue continues onward without changing its velocity, so it's formula is d = VT. The peak velocity of the tongue will be reached after 20 ms. So let's calculate it.
d = 0.5AT^2
d = 0.5*290 m/s^2 * (0.020 s)^2
d = 145 m/s^2 * 0.0004 s^2
d = 0.058 m
Now to handle coasting
d = 0.058 m + 0.030 s * 290 m/s^2 * 0.20 s
d = 0.058 m + 0.174 m
d = 0.232 m
Rounding to 2 significant digits gives 0.23 meters, or 23 cm.
For the train, we need to determine the acceleration. We know the velocity changed from 5.0 m/s to 14.0 m/s over a period of 8.0 seconds. So the acceleration is:
(14.0 m/s - 5.0 m/s)/8.0 s = (9.0 m/s)/8.0 s = 1.125 m/s^2
At the edge of town, the train is traveling at 14 m/s and accelerates at 1.125 m/s^2 for 16 seconds, so:
14 m/s + 1.125 m/s^2 * 16 s = 14 m/s + 18 m/s = 32 m/s
So the final speed of the train is 32 m/s
Answer:
The height of the cliff is, h = 78.4 m
Explanation:
Given,
The horizontal velocity of the projectile, Vx = 20 m/s
The range of the projectile, s = 80 m
The projectile projected from a height is given by the formula
<em> S = Vx [Vy + √(Vy² + 2gh)] / g
</em>
Therefore,
h = S²g/2Vx²
Substituting the values
h = 80² x 9.8/ (2 x 20²)
= 78.4 m
Hence, the height of the cliff is, h = 78.4 m
Answer:
1.44*10^-3m
Explanation:
Given that distance BTW two bright fringes is
DetaY = lambda* L/d
So for second wavelength
Deta Y2= Lambda 2* L/d
=lambda 2 x deta y1/ lambda1
So substituting
= 360 x 10^-9 x (1.6*10^-3/640*10^-9)
1.44*10^ -3m
45✖️77 and then you will get your answer