<span>B. because they cannot make light rays converge</span>
Answer:
Radiation
Explanation:
The sun radiates energy to the earth to make it warmer near the equator.
Answer:
a) In order to catch the ball at the level at which it is thrown in the direction of motion.
b)Speed of the receiver will be 7.52m/s
Explanation:
Calculating range,R= Vo^2Sin2theta/g
R= (20^2×Sin(2×30)/9.8 = 35.35m
Let receiver be(R-20) = 35.35-20= 15.35m
The horizontal component of the ball is:
Vox= Vocostheta= 20× cos30°
Vox= 17.32m/s
Time taken to coverR=35.35m with 17.32m/s will be:
t=R/Vox= 35.35/17.32
t= 2.04seconds
b)Speed required to cover 15.35m at 2.04seconds
Vxreciever= d/t = 15.35/2.04 = 7.52m/s
2.0y i think this is guess but if it right then thats good
-- The string is 1 m long. That's the radius of the circle that the mass is
traveling in. The circumference of the circle is (π) x (2R) = 2π meters .
-- The speed of the mass is (2π meters) / (0.25 sec) = 8π m/s .
-- Centripetal acceleration is V²/R = (8π m/s)² / (1 m) = 64π^2 m/s²
-- Force = (mass) x (acceleration) = (1kg) x (64π^2 m/s²) =
64π^2 kg-m/s² = 64π^2 N = about <span>631.7 N .
</span>That's it. It takes roughly a 142-pound pull on the string to keep
1 kilogram revolving at a 1-meter radius 4 times a second !<span>
</span>If you eased up on the string, the kilogram could keep revolving
in the same circle, but not as fast.
You also need to be very careful with this experiment, and use a string
that can hold up to a couple hundred pounds of tension without snapping.
If you've got that thing spinning at 4 times per second and the string breaks,
you've suddenly got a wild kilogram flying away from the circle in a straight
line, at 8π meters per second ... about 56 miles per hour ! This could definitely
be hazardous to the health of anybody who's been watching you and wondering
what you're doing.
