With the switch open, there's no current in the circuit, and therefore
no voltage drop across any of the dissipative elements (the resistor
or the battery's internal impedance). So the entire battery voltage
appears across the switch, and the voltmeter reads 12.0V .
Answer:
20.4m/s²
Explanation:
Given parameters:
Initial velocity = 0m/s
Distance = 53m
Time = 5.2s
Unknown:
Acceleration = ?
Solution:
This is a linear motion and we use the right motion equation;
S = ut +
at²
S is the distance
u is the initial velocity
a is the acceleration
t is the time
Insert the parameters and solve;
53 = (0x 5.2) +
x a x 5.2
53 = 2.6a
a =
= 20.4m/s²