Answer: 0.9264 kg
Explanation: [I'll use "cc" for cubic centimeter, instead of cm^3.
The volume is 6cm*4cm*2cm = 48 cm^3 (cc).
Density of Au is 19.3 g/cc
Mass of gold = (48 cc)*(9.3 g/cc) = 926.4 grams Au
1 kg = 1,000 g
(926.4 grams Au)*(1 kg/1,000 g) = 0.9264 kg, 0.93 kg to 2 sig figs
At gold's current price of $57,500/kg, this bar is worth $53,268. Keep it hidden from your lab partner (and instructor).
The answer is C, light cant pass through an opaque object
Answer:
Work done = 4584.9 J
Explanation:
given: q1=3.0 mC = 3.0 × 10⁻³ C, r = 20 cm = 0.20 m, q1 = 34μC = 34 × 10⁻⁶ C
Solution:
Formula for the potential difference at the center of the circle
P.E = K × q1 q2 /r (Coulomb's constant k= 8.99 × 10⁹ N·m² / C²)
P.E = 8.99 × 10⁹ N·m² / C² × 3.0 × 10⁻³ C × 34 × 10⁻⁶ C / 0.20 m
P.E = 4584.9 J = Work done
Answer:
t = 444.125 sec
Explanation:
Given data:
V = 24 volt
I = 0.1 ampere
mass of water mw = 51 gm
cr = 4.18 J/gm degree K^-1
mass of resistor = 8 gm
cr = 3.7 J/gm degree K^-1
we know that power is given as
Power P = VI
But P =E/t
so equating both side we have
solving for t
t = 444.125 sec
Answer:
6.4 Joules
Explanation:
For springs, Hooke's law states that;
F = ke
where F is a force applied, e is the extension and k is the spring constant.
Work done in a spring is the same as the potential energy stored in the spring. So that;
Work done = k
e = -
= 4 cm = 0.4 m
= 0
So that,
- = 0.4 m
Thus,
Work done = x 80 x
= 40 x 0.16
= 6.4
Work done = 6.4 J
The work done by the force on the spring is 6.4 Joules.