To find out scientific notation, you want to make sure that number is less than 10. So do 5.000000, you don't rally need the zeros but I just want to make my point. So use 10^x meaning ten the whatever power adds zeros like 5.000000x10^6 meaning it is increasing it by six zeros moving it out of the decimals and letting become 5,000,000.
To solve this problem we will use the concepts related to Magnification. Magnification is the process of enlarging the apparent size, not physical size, of something. This enlargement is quantified by a calculated number also called "magnification".
The overall magnification of microscope is
Where
N = Near point
l = distance between the object lens and eye lens
= Focal length
= Focal of eyepiece
Given that the minimum distance at which the eye is able to focus is about 25cm we have that N = 25cm
Replacing,
Therefore the correct answer is C.
Juan lives a hundred miles away from Bill then the average speed that he reaches Bill's Home in 50 seconds means that Juan lives 50 seconds away from bill because 50 + 50 seconds equals 100 seconds so Juansorry my bad Juan lives 50 miles away from Belle but he probably how I must have run to get to bill in 50 seconds .
Answer:
6200 J
Explanation:
Momentum is conserved.
m₁ u₁ + m₂ u₂ = m₁ v₁ + m₂ v₂
The car is initially stationary. The truck and car stick together after the collision, so they have the same final velocity. Therefore:
m₁ u₁ = (m₁ + m₂) v
Solving for the truck's initial velocity:
(2700 kg) u = (2700 kg + 1000 kg) (3 m/s)
u = 4.11 m/s
The change in kinetic energy is therefore:
ΔKE = ½ (m₁ + m₂) v² − ½ m₁ u²
ΔKE = ½ (2700 kg + 1000 kg) (3 m/s)² − ½ (2700 kg) (4.11 m/s)²
ΔKE = -6200 J
6200 J of kinetic energy is "lost".
Answer:
-54.12 V
Explanation:
The work done by this force is equal to the difference between the final value and the initial value of the energy. Since the charge starts from the rest its initial kinetic energy is zero.
The change in electrostatic potential energy 
, of one point charge q is defined as the product of the charge and the potential difference.
