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2 years ago

The density of gold is 19.3 g cm³. What is the mass of a bar of gold in Kg that measures 6 cm x 4 cmx to 2 cm ?

Physics

1 answer:

klemol [59]2 years ago

3 0

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).

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For what absolute value of the phase angle does a source deliver 71 % of the maximum possible power to an RLC circuit

kherson [118]

Answer: the absolute value of the phase angle is 28°

Explanation:

taking a look at expression for the instantaneous electric power in an AC circuit;

P = VI -------let this be equation 1

p is power, v is voltage and I is current;

for maximum power

P_max = V_rms × I_rms --------let this be equ 2

where P_max is the maximum power, V_rms is the rms value of voltage and I_rms is the rms value of current.

Also for average electric power in an AC circuit

P_avg = V_rms × I_rms × cos²∅ -------let this be equ 3

where P_avg is the average power and cos∅ is the power factor

now from equation 2; P_max = V_rms × I_rms

so p_max replaces V_rms × I_rms in equation 3

we now have

P_avg = P_max × cos²∅

so we substitute

expression for the given value of the average power is

P_avg = P_max × 75%

p_avg = P_max.78/100

for the expression of the average electricity in an AC circuit

P_max.78/100 = P_max × cos²∅

78/100 = cos²∅

to get the absolute value of phase angle

∅ = cos⁻¹ ( √(78/100))

∅ = cos⁻¹ ( 0.8832)

∅ = 27.969 ≈ 28°

Therefore the absolute value of the phase angle is 28°

8 0

2 years ago

How do I do this physics problem about potential energy and kinetic energy?

larisa86 [58]

Ok i apologise for the messy working but I'll try and explain my attempt at logic

Also note i ignore any air resistance for this.

First i wrote the two equations I'd most likely need for this situation, the kinetic energy equation and the potential energy equation.

Because the energy right at the top of the swing motion is equal to the energy right in the "bottom" of the swing's motion (due to conservation of energy), i made the kinetic energy equal to the potential energy as indicated by Ek = Ep.

I also noted the "initial" and "final" height of the swing with hi and hf respectively.

So initially looking at this i thought, what the heck, there's no mass. Then i figured that using the conservation of energy law i could take the mass value from the Ek equation and use it in the Ep equation. So what i did was take the Ek equation and rearranged it for m as you can hopefully see. Then i substituted the rearranged Ek equation into the Ep equation.

So then the equation reads something like Ep = (rearranged Ek equation for m) × g (which is -9.81) × change in height (hf - hi).

Then i simplify the equation a little. When i multiply both sides by v^2 i can clearly see that there is one E on each side (at that stage i don't need to clarify which type of energy it is because Ek = Ep so they're just the same anyway). So i just canceled them out and square rooted both sides.

The answer i got was that the max velocity would be 4.85m/s 3sf, assuming no losses (eg energy lost to friction).

I do hope I'm right and i suppose it's better than a blank piece of paper good luck my dude xx

4 0

3 years ago

A sample of a gas in a rigid container has an initial pressure of 5 atm at a temperature of 254.5 k. The temperature is decrease

skelet666 [1.2K]

The gas is in a rigid container: this means that its volume remains constant. Therefore, we can use Gay-Lussac law, which states that for a gas at constant volume, the pressure is directly proportional to the temperature. The law can be written as follows:

$\frac{P_1}{T_1} = \frac{P_2}{T_2}$

Where P1=5 atm is the initial pressure, T1=254.5 K is the initial temperature, P2 is the new pressure and T2=101.8 K is the new temperature. Re-arranging the equation and using the data of the problem, we can find P2:

$P_2 = T_2 \frac{P_1}{T_1}=(101.8 K) \frac{5 atm}{254.5 K}=2 atm$