The bodies of arthropods are supported, not by internal bones, but by a hardened exoskeleton<span> made of </span>chitin<span>, a substance produced by many non-arthropods as well. In arthropods, the nonliving exoskeleton is like a form-fitting suit of armor. It is produced by the "skin" and then hardens into a protective outer-covering.</span>
λ=v/f
λ-wavelength
v-speed
f-frequency
we have the wavelength(6.2 x 10^-6meters) and we use the speed of light which is equal to 3*10^8m/s
6.2*10^-6m=3*10^8m/s/f
f=(3*10^8m/s)/(6.2*10^-6)≈0.48*10^14Hz
Answer:
PE=0.92414J and KE=0.28175J
Explanation:
Gravitational potential energy=mass*gravity*height
PE=mgh
Data,
M=0.046kg
H=2.05m
g=9.8m/s^2
PE=0.046kg * 9.8m/s^2 * 2.05m
PE =0.92414J
KE=1/2mv^2
M=0.046kg
V=3.5m/s
KE=[(0.046kg)*(3.5m/s)^2]\2
KE=0.28175J
Answer:
The solar radiation is first intercepted by Earth's atmosphere, just a small part of the radiation is absorbed by gases such as water vapor. Some of the radiation is reflected back to space by the clouds and Earth's surface.
Answer:
a) v = 54.7m/s
b) v = (58 - 1.66a) m/s
c) t = 69.9 s
d) v = -58.0 m/s
Explanation:
Given;
The height equation of the arrow;
H = 58t - 0.83t^2
(a) Find the velocity of the arrow after two seconds. m/s;
The velocity of the arrow v can be given as dH/dt, the change in height per unit time.
v = dH/dt = 58 - 2(0.83t) ......1
At t = 2 seconds
v = dH/dt = 58 - 2(0.83×2)
v = 54.7m/s
(b) Find the velocity of the arrow when t = a. m/s
Substituting t = a into equation 1
v = 58 - 2(0.83×a)
v = (58 - 1.66a) m/s
(c) When will the arrow hit the surface? (Round your answer to one decimal place.) t = s
the time when H = 0
Substituting H = 0, we have;
H = 58t - 0.83t^2 = 0
0.83t^2 = 58t
0.83t = 58
t = 58/0.83
t = 69.9 s
(d) With what velocity will the arrow hit the surface? m/s
from equation 1;
v = dH/dt = 58 - 2(0.83t)
Substituting t = 69.9s
v = 58 - 2(0.83×69.9)
v = -58.0 m/s
