Light travels in straight lines. Once a light has been produced, it will keep moving in a straight line until it hits something else. Shadows are evidence of light traveling in straight lines. An object blocks light so that it can’t reach the surface where we see the shadow.
Answer:
Explanation:
Let the first height be h . second height .75h
third height .75h . fourth height .75²h
fifth height .75²h , sixthth height .75³ and so on
Total distance consists of two geometric series as follows
1 ) first series
h + .75h + .75²h + .75³h......
2 ) second series
.75h +.75²h +.75³h + .75⁴h .......
Sum of first series :
first term a = h , commom ratio r = .75
sum = a / (1 - r )
= h / 1 - .75
= h / .25
4h
sum of second series :--
first term a = .75 h , commom ratio r = .75
sum = a / (1 - r )
= .75h / 1 - .75
= .75h / .25
3h
Total of both the series
= 4h + 3h
= 7h .
h = 1 m
Total distance = 7 m
The answer to number 1 is D and the answer for the second one is 2
^-^
Answer:
the balls reached a height of 4.9985 m
Explanation:
Given the data in the question;
mass one m = 3.8 kg
mass two M = 2.1 kg
Initial velocities
u = 22 m/s
U = { moving downward} = 12 m/s
Now, using the law conservation of linear moment;
mu + MU = v( m + M )
we solve for "v" which is the velocity of the ball s after collision;
v = (mu + MU) / ( m + M )
so we substitute our given values into the equation
v = ( ( 3.8 × 22 ) + ( 2.1 × -12) ) / ( 3.8 + 2.1 )
v = ( 83.6 - 25.2 ) / 5.9
v = 58.4 / 5.9
v = 9.898 m/s
Now, we determine required height using the following relation;
v"² - v² = 2gh
where v" is the velocity at the top which is 0 m/s and g = -9.8 m/s²
0 - v² = 2gh
v² = -2gh
so we substitute
( 9.898 )² = -2 × -9.8 × h
97.97 = 19.6 × h
h = 97.97 / 19.6
h = 4.9985 m
Therefore, the balls reached a height of 4.9985 m
