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>
Answer:
a) Deceleration = 201.76 m/s^2
b) Distance traveled = 197.68 m
Explanation:
Initial Speed = 632 mi/h
Initial Speed (in meters/second) = (632 * 1.609 * 1000) / (60 * 60) = 282.4 m/s
Time to decelerate = 1.4 seconds
a) Change in speed = Acceleration * time
-282.4 = Acceleration * 1.4
Acceleration = -201.76 m/s^2
Deceleration = 201.76 m/s^2
b) Distance traveled = average speed * time
average speed = 282.4 / 2 = 141.2 m/s
Distance traveled = 141.2 * 1.4
Distance traveled = 197.68 m
Answer:
that jet will fly 1200 km in 2 hours
Answer:
2 charges of electron (2C)
Explanation:
I = Q/t
2 = Q/1
Q = 2×1= 2C
Q = 2 charge of electron
Answer:
4.84615 m
Explanation:
Here the linear momentum is conserved and it is assumed that they will meet at the center of gravity of the system.
The center of gravity is given by
where
i = Particle
m = Mass
r = Position
n = Number of particles
The father's location is taken as reference
The father will be 4.84615 m from the origin.