Answer:
Explanation:
Let the distance from spotlight to wall be 15m, and distance from the man to the building be 
.
#Therefore the height of the shadow as a function of the above is 
Hence, height of the shadow is expressed as s=(15-x)m
#See attached photo for illustration
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A. The closet point in the Moon's orbit to Earth . . . . . perigee
B. The farthest point in the Moon's orbit to Earth . . . . . apogee
C. The Sun's orbit that is closest to the Moon . . . . . a meaningless description
D. The closest point in Earth's orbit of the Sun . . . . . perihelion
-- The farthest point in Earth's orbit of the Sun . . . . . aphelion
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Answer:
The answer to your question is letter A. r = 1.07 x 10⁻¹⁴ m
Explanation:
Data
F = 2 N
d = ?
q = 1.6 x 10 ⁻¹⁹ C
k = 8.987 Nm²/C²
Formula
Solve for r
Substitution
Simplification
r = 
r = 
Result
r = 1.07 x 10⁻¹⁴ m
Answer:
answer is option 4
Explanation:
you have to use option 4 because u need to find out initial velocity (Vi)
Answer:
Explanation:
a )
change in the gravitational potential energy of the bear-Earth system during the slide = mgh
= 45 x 9.8 x 11
= 4851 J
b )
kinetic energy of bear just before hitting the ground
= 1/2 m v²
= .5 x 45 x 5.8²
= 756.9 J
c ) If the average frictional force that acts on the sliding bear be F
negative work done by friction
= F x 11 J
then ,
4851 J - F x 11 = 756.9 J
F x 11 = 4851 J - 756.9 J
= 4094.1 J
F = 4094.1 / 11
= 372.2 N
