Answer:
(a) θ1 = 942.5rad, (b) θ2 = 13195 rad
Explanation:
(a) Given
ωo = 0 rad/s
ω = 3600rev/min = 3600×2(pi)/60 rad/s
ω = 377rad/s
t1 = 5s
θ1 = (ω + ωo)t/2
θ1 = (377 +0)×5/2
θ1 = 942.5 rads
(b) ωo = 377rad/s
ω = 0 rad/s
t2 = 70s
θ2 = (ω + ωo)t/2
θ2 = (377 +0)×70/2
θ2 = 13195 rad
Answer:
0.00278 km/sec2 or 36,000 km/hr2 .
Explanation:
Answer:
C 350W
Explanation:
Given power output to walk on a flat ground to be 300W, h = 0.05x, v = 1.4m/s
m = 70kg and g =9.8m/s².
x = horizontal distance covered
Total energy used = potential energy used in climbing and the energy used in a walking the horizontal distance.
E = mgh + 300t
Where t is the time taken to cover the distance
x = vt and h = 0.05vt
So
E = mg×0.05×vt + 300t
Substituting respective values
E = 70×9.8×0.05×1.4t +300t = 348t
P = E/t = 348W ≈ 350W.
Answer: The HUMAN EYE
Explanation:
The human eye is made up of different parts which ranges from controlling the amount of light that enters the eye to the focusing of the image that is formed. The camera is a device which is both mechanically and electronically operated which shares a number of similarities with the eye.
In the human eye, the IRIS helps to regulate the amount of rays passing through the pupil to the lens by either contracting or dilating in light or dark environment respectively. While in the camera, the DIAPHRAGM controls the amount of light entering the camera.
The PUPIL serves as the passage for light into the eye while in the camera, the APERTURE does the same.
The photosensitive surface in the eye is the YELLOW SPOT while in the camera, the photosensitive surface is the PHOTOGRAPHIC FILM.
