Answer:
i think its the third one or second one
Answer:
Final velocity (v) = 36 m/s
Distance traveled (s) = 2,160 m
Explanation:
Given:
Initial velocity (u) = 0
Acceleration (a) = 0.3 m/s
Time travel (t) = 2 minutes = 120 seconds
Find:
Final velocity (v) = ?
Distance traveled (s) = ?
Computation:
v = u + at
v = 0 + 0.3(120)
v = 0.3(120)
v = 36 m/s
Final velocity (v) = 36 m/s
Distance traveled (s) = ut + (1/2)at²
Distance traveled (s) = (0.5)(0.3 × 120 × 120)
Distance traveled (s) = 2,160 m
The type of current produced by a battery is direct current. C.
Answer:
The smallest distance the student that the student could be possibly be from the starting point is 6.5 meters.
Explanation:
For 2 quantities A and B represented as
and 
The sum is represented as
For the the values given to us the sum is calculated as

Now the since the uncertainity inthe sum is 
The closest possible distance at which the student can be is obtained by taking the negative sign in the uncertainity
Thus closest distance equals
meters