Answer: hello options related to your question is missing attached below is the missing part of your question
answer: No charge of the length of the bonds expected because the rod did not touch the charge source ( option A )
Explanation:
When the Charge is first, Furthest away and second and closest to the source charge. <em>The spring like bonds can be said to have No charge of the length of the bonds expected because the rod did not touch the charge source </em><em>when Furthest away the bond with charge will be less effective </em>
Answer:
B
Explanation:
OOf we are doing this stuff atm
So if its faster at the front and slow at the back you can tell that its not slowing down because less of a force is there however at the front there is more of a force. Friction is low which means that its not makimg much contact so no sudden change of forces thats also why its B
The energy carried by a single photon of frequency f is given by:
where
is the Planck constant. In our problem, the frequency of the photon is
, and by using these numbers we can find the energy of the photon:
Answer:
a) a = 4.57 m/s², b) a = 6.48 m / s²
, c) a = 1.42 m / s²,d) r = 82.3 m
Explanation:
The centripetal acceleration is the acceleration responsible for the change of direction of the acceleration vector and occurs in circular movements, the expression is
a = v² / r
let's apply this precaution to our cases
a) let's calculate
a = 8²/14
a = 4.57 m/s²
b) an automobile at v = 65 km / h (1000 m / 1km) (1 h / 3600 s) =18,055 m/s
let's reduce feet to meters
1 ft = 0.3048 m
r = 165 ft (0.3048 m / 1 ft) = 50.292 m
a = 18,055 2 / 50,292
a = 6.48 m / s²
c) we calculate
a = 1.25²2 / 1.1
a = 1.42 m / s²
d) we look for the radius
a = v² / r
r = v² / a
we reduce
v = 80 km / h (1000 m / 1km) (1h / 3600s) = 22.22 ms
r = 22.22²/6
r = 82.3 m
e) the cenripeta acceleration is used to take the curves on the highway,
Used in centrifuges to separate compounds
It is used in the games of the park of atraccio
Used in CD players and computer hard drives
