Answer:
14523.55J
Explanation:
The work done by the jogger against gravity is given by the following equation;
where m is the mass, g is acceleration due to gravity taken as and h is the height of the hill.
Since the length of the hill is 132m and it is inclined at 12 degrees to the horizontal, the height is thus given as follows;
Substituting this into equation (1) with all other necessary parameters, we obtain the following;
<u>Answer:</u>
950-kg subcompact car be moving to have the same kinetic energy of truck at 116.08 km/hr
<u>Explanation:</u>
Kinetic energy of body with mass m and moving at velocity v is given by
KE of truck
KE of subcompact car
Equating
So 950-kg subcompact car be moving to have the same kinetic energy of truck at 116.08 km/hr
Answer:
Explanation:
a ) If the image of this object is viewed with the eyepiece adjusted for minimum eyestrain (image at the far point of the eye) , the image from object lens must have been formed at the focus of eye lens . So the objective image must have been formed at 19.5 - 2.75 = 16.75 cm from the object lens.
b ) Let the object distance be u
For object lens
v = 16.75 cm , f = .35 cm
1/v - 1/u = 1/f
1/16.75 - 1/u = 1/ .35
.0597 - 1/u = 2.857
1/u = - 2.7973
u = .3575 cm
c ) Angular magnification
=
v₀ and u₀ are image and object distance for object lens , D = 25 cm and f_e is focal length of eye lens
= (16.75 / .3575) x( 25 / 2.75)
= 46.85 x 9.09
= 426
Answer:
the approximate size of the smallest object on the Earth that astronauts can resolve by eye when they are orbiting 250 km above the Earth is y = 31.495 m
Explanation:
Using Rayleigh criterion for the limiting angle of resolution of an eye
rad
Thus; the separation between the two sources is expressed as:
Thus; the approximate size of the smallest object on the Earth that astronauts can resolve by eye when they are orbiting 250 km above the Earth is y = 31.495 m