To solve this problem it is necessary to apply the rules and concepts related to logarithmic operations.
From the definition of logarithm we know that,
In this way for the given example we have that a logarithm with base 10 expressed in the problem can be represented as,
We can express this also as,
By properties of the logarithms we know that the logarithm of a power of a number is equal to the product between the exponent of the power and the logarithm of the number.
So this can be expressed as
Since the definition of the base logarithm 10 of 10 is equal to 1 then
The value of the given logarithm is equal to 6
B is the answer for your question the answer is b an elevator slowing down
Answer:
It does both. Once they get close enough the air does start to get charged, but then they eventually discharge when they touch.
Explanation:
Answer:
a)Distance traveled during the first second = 4.905 m.
b)Final velocity at which the object hits the ground = 38.36 m/s
c)Distance traveled during the last second of motion before hitting the ground = 33.45 m
Explanation:
a) We have equation of motion
S = ut + 0.5at²
Here u = 0, and a = g
S = 0.5gt²
Distance traveled during the first second ( t =1 )
S = 0.5 x 9.81 x 1² = 4.905 m
Distance traveled during the first second = 4.905 m.
b) We have equation of motion
v² = u² + 2as
Here u = 0, s= 75 m and a = g
v² = 0² + 2 x g x 75 = 150 x 9.81
v = 38.36 m/s
Final velocity at which the object hits the ground = 38.36 m/s
c) We have S = 0.5gt²
75 = 0.5 x 9.81 x t²
t = 3.91 s
We need to find distance traveled last second
That is
S = 0.5 x 9.81 x 3.91² - 0.5 x 9.81 x 2.91² = 33.45 m
Distance traveled during the last second of motion before hitting the ground = 33.45 m
There is no image!?...was there meant to be something attached?
