Operant conditioning, sometimes called <em>instrumental learning</em>, was first extensively studied by Edward L. Thorndike, who observed the behavior of cats trying to escape from home-made puzzle boxes.
Hope this helps!
Answer:
u = - 38.85 m/s^-1
Explanation:
given data:
acceleration = 2.10*10^4 m/s^2
time = 1.85*10^{-3} s
final velocity = 0 m/s
from equation of motion we have following relation
v = u +at
0 = u + 2.10*10^4 *1.85*10^{-3}
0 = u + (21 *1.85)
0 = u + 38.85
u = - 38.85 m/s^-1
negative sign indicate that the ball bounce in opposite directon
Answer:
The average speed can be calculated as the quotient between the distance travelled and the time needed to travel that distance.
To go to the school, he travels 2.4 km in 0.6 hours, then here the average speed is:
s = (2.4km)/(0.6 hours) = 4 km/h
To return to his home, he travels 2.4km again, this time in only 0.4 hours, then here the average speed is:
s' = (2.4 km)/(0.4 hours) = 6 km/h.
Now, if we want the total average speed (of going and returning) we have that the total distance traveled is two times the distance between his home and school, and the total time is 0.6 hours plus 0.4 hours, then the average speed is:
S = (2*2.4 km)/(0.6 hours + 0.4 hours)
S = (4.8km)/(1 h) = 4.8 km/h
Answer:
1.) 274.5v
2.) 206.8v
Explanation:
1.) Given that In one part of the lab activities, students connected a 2.50 µF capacitor to a 746 V power source, whilst connected a second 6.80 µF capacitor to a 562 V source.
The potential difference and charge across EACH capacitor will be
V = Voe
Where Vo = initial voltage
e = natural logarithm = 2.718
For the first capacitor 2.50 µF,
V = Vo × 2.718
746 = Vo × 2.718
Vo = 746/2.718
Vo = 274.5v
To calculate the charge, use the below formula.
Q = CV
Q = 2.5 × 10^-6 × 274.5
Q = 6.86 × 10^-4 C
For the second capacitor 6.80 µF
V = Voe
562 = Vo × 2.718
Vo = 562/2.718
Vo = 206.77v
The charge on it will be
Q = CV
Q = 6.8 × 10^-6 × 206.77
Q = 1.41 × 10^-3 C
B.) Using the formula V = Voe again
165 = Vo × 2.718
Vo = 165 /2.718
Vo = 60.71v
Q = C × 60.71
Q = C
Answer:
25.33 rpm
Explanation:
M = 100 kg
m1 = 22 kg
m2 = 28 kg
m3 = 33 kg
r = 1.60 m
f = 20 rpm
Let the new angular speed in rpm is f'.
According to the law of conservation of angular momentum, when no external torque is applied, then the angular momentum of the system remains constant.
Initial angular momentum = final angular momentum
(1/2 x M x r^2 + m1 x r^2 + m2 x r^2 + m3 x r^2) x ω =
(1/2 x M x r^2 + m1 x r^2 + m3 x r^2 ) x ω'
(1/2 M + m1 + m2 + m3) x 2 x π x f = (1/2 M + m1 + m3) x 2 x π x f'
( 1/2 x 100 + 22 + 28 + 33) x 20 = (1/2 x 100 + 22 + 33) x f'
2660 = 105 x f'
f' = 25.33 rpm
