Joseph's experiment could be improved by using the same antenna at each part of the house during each trial instead of using different antenna. By doing so, he can obtain accurate results how is the signal in different part of the house under the same conditions (despite the location). So, he will see the dependence of the signal on the location. If he uses different antenna, than this antenna can also have influence of the signal.
Volume of gold in the phone = 10 cm^3
= 0.<span>00001 m^3 </span>
Density of gold = 19300 kg/m^3
1 kg mass = 2.2 pounds
Mass of 10 cm^3 of gold = 0<span>.00001 m^3 * (19300 kg/m^3)
= 0.193 kg
So
0.193 kg = 0.193 * 2.2 pounds
= 0.43 pounds
I think there is something wrong with the options given in the question.</span>
Answer:
Nicki or cardi b idek....
Answer:2.67kgm/s cube
Explanation: density = mass ÷ volume = 400 ÷ 150
Answer:
42.5W
Explanation:
To solve this problem we must go back to the calculations of a weighted average based on the time elapsed thus,
We need to calculate the average power dissipated by the 800\Omega resistor.
Our values are given by:
Aplying the values to the equation we have:
