Answer:
t_pass = 2.34 m
t_stop = 4.68 s
Thus, for the car passing at constant speed the pedestrian will have to wait less.
Explanation:
If the car is moving with constant speed, then the time taken by it will be given as:
where,
t_pass = time taken = ?
D = Distance covered = 23 m
v = constant speed = (22 mi/h)(1609.34 m/1 mi)(1 h/3600 s) = 9.84 m/s
Therefore,
<u>t_pass = 2.34 m</u>
<u></u>
Now, for the time to stop the car, we will use third equation of motion to get the acceleration first:
Now, for the passing time we use first equation of motion:
<u>t_stop = 4.68 s</u>
It will float because the density is less than 1
Answer:
(a). The electric potential at 1.650 cm is .
(b). The electric potential at 2.81 cm is .
Explanation:
Given that,
Radius of sphere R=2.81 cm
Charge = +2.35 fC
Potential at center of sphere
(a). We need to calculate the potential at a distance r = 1.60 cm
Using formula of potential difference
The electric potential at 1.650 cm is .
(b). We need to calculate the potential at a distance r = R
Using formula of potential difference
The electric potential at 2.81 cm is .
Hence, This is the required solution.
Answer:
0.0083km or 8.3m
Explanation:
Given parameters:
Distance = 5km
Time = 10min
Unknown:
Distance moved in 1sec = ?
Solution:
To solve this problem, let us first find the speed of the motorbike.
Speed =
Speed = = 0.5km/min
So;
In 1sec:
60 sec = 1 min
1 sec = min
Distance moved in 1sec = speed x time
Distance moved = 0.5km/min x min = 0.0083km or 8.3m
Answer:group polarization
Explanation: