The atom must have gained 1 or more electrons or must have lost 1 or more electrons.
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
Answer : B) The cow pulls back on the girl.
From newton’s third law we know that every action has a reaction force pushing back. So when the girl pulls on a cow, the cow is pulling back on her.
Answer:
Option B. 6.25 J/S
Explanation:
Data obtained from the question include:
t (time) = 2secs
F (force) = 50N
d (distance) = 0.25m
P (power) =?
The power can be obtained by using the formula P = workdone/time.
P = workdone / time
P = (50 x 0.25)/ 2
P = 6.25J/s
