Heya!!
For calculate aceleration, lets applicate second law of Newton:
<u>Δ Being Δ</u>
F = Force = 78,3 N
m = Mass = 24,5 kg
a = Aceleration = ?
⇒ Let's replace according the formula and clear "a":
⇒ Resolving
Result:
The aceleration is <u>3,19 meters per second squared (m/s²)</u>
Good Luck!!
Answer:
DMs are not accessible anymore. I assume Zuka is a staff member? the only way to talk to a staff member anymore is to report something, but even then, the probably won't even look at what they're deleting :/
May I have brainliest please? :)
F=dP/dt. So you want the momentum to change as slowly as possible in time to minimize the force. So as you catch the egg, let your hand move backward with it for awhile, slowly bringing it to a stop. If you hold your hand steady when you catch it the force due to the impact could break it.
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
