Answer: a) E= 6.63x10^-19J
E= 3.97×10^2KJ/mol
b) E = 3.31×10^-19J
E= 18.8×10^4 KJ/mol
C) E = 1.32×10^-33J
E= 8.01×10^-10KJ/mol
Explanation:
a) E = h ×f
h= planks constant= 6.626×10^-34
E=(6.626×10^-34)×(1.0×10^15)
E=6.63×10^-19J
1mole =6.02×10^23
E=( 6.63×10^-19)×(6.02×10^23)
E=3.97×10^2KJ/mol
b) E =(6.626×10^-34)/(1.0×10^15)
E=3.13×10^-19J
E= 3.13×10^-19) ×(6.02×10^23)
E= 18.8×10^3KJ/MOL
c) E= (6.626×10^-34) /0.5
E= 1.33×10^-33J
E= (1.33×10^-33) ×(6.02×10^23)
E= 8.01×10^-10KJ/mol
Answer:
0.4113772 s
Explanation:
Given the following :
Mass of bullet (m1) = 8g = 0.008kg
Initial horizontal Velocity (u1) = 280m/s
Mass of block (m2) = 0.992kg
Maxumum distance (x) = 15cm = 0.15m
Recall;
Period (T) = 2π√(m/k)
According to the law of conservation of momentum : (inelastic Collison)
m1 * u1 = (m1 + m2) * v
Where v is the final Velocity of the colliding bodies
0.008 * 280 = (0.008 + 0.992) * v
2.24 = 1 * v
v = 2.24m/s
K. E = P. E
K. E = 0.5mv^2
P.E = 0.5kx^2
0.5(0.992 + 0.008)*2.24^2 = 0.5*k*(0.15)^2
0.5*1*5.0176 = 0.5*k*0.0225
2.5088 = 0.01125k
k = 2.5088 / 0.01125
k = 223.00444 N/m
Therefore,
Period (T) = 2π√(m/k)
T = 2π√(0.992+0.008) / 233.0444
T = 2π√0.0042910
T = 2π * 0.0655059
T = 0.4113772 s
Answer:
Explanation:
Unknown fork frequency is either
335 + 5.3 = 340.3 Hz
or
335 - 5.3 = 329.7 Hz
After we modify the known fork, the unknown fork frequency equation becomes either
(335 - x) + 8 = 340.3
(335 - x) = 332.3
x = 2.7 Hz
or
(335 - x) + 8 = 329.7
(335 - x) = 321.7
x = 13.3 Hz
IF the unknown fork frequency was 340.3 Hz,
THEN the 335 Hz fork was detuned to 335 - 2.7 = 332.3 Hz
IF the unknown fork frequency was 329.7 Hz,
THEN the 335 Hz fork was detuned to 335 - 13.3 = 321.7 Hz
Answer:
The current drawn by Horace’s reading glasses is 0.8 A.
Explanation:
Given that,
Resistance of each bulb, R = 2 ohms
Voltage of the system, V = 3.2 volts
These two bulbs are connected in series. The equivalent resistance will be 2 ohms +2 ohms = 4 ohms
Let I is the current drawn by Horace’s reading glasses. Using Ohm's law to find it such that :
So, the current drawn by Horace’s reading glasses is 0.8 A.
