The planet MARS is visible without a telescope on many clear nights. The planets JUPITER, MERCURY, VENUS and SATURN are also viewable without the aid of magnification.
Answer:
a) F= 0,19 [N] according to problem statement
b) F = 0,19*10⁹ [N] using the right value of K
Explanation:
The force between two electric charges is according to Coulomb´s law is:
F = K * q₁*q₂ / d² where q₁ and q₂ are the charges on body one and body 2 respectively, d is the distance between the two bodies and K is a constant K = 8,988100*10⁹ N.m²/C². The problem establishes to use K = 8,988100 N.m²/C².
NOTE: To value of is : K = 8,988100*10⁹ N.m²/C². I am going to solve the problem using K = 8,988100 N.m²/C² if that information was an error, all we need to get the right answer is multiply the result by 10⁹
Then:
F = 8,988100 * 1,2* 0,36 / (4,5)² [ N*m²/C² ] * [ C*C*/m²]
F = 3,882859/ 20,25 [N]
F= 0,19 [N]
The force is of repulsion since the two charges are positive and in the direction of the straight line which passes through the centers of the bodies
Efficiency η of a Carnot engine is defined to be:
<span>η = 1 - Tc / Th = (Th - Tc) / Th </span>
<span>where </span>
<span>Tc is the absolute temperature of the cold reservoir, and </span>
<span>Th is the absolute temperature of the hot reservoir. </span>
<span>In this case, given is η=22% and Th - Tc = 75K </span>
<span>Notice that although temperature difference is given in °C it has same numerical value in Kelvins because magnitude of the degree Celsius is exactly equal to that of the Kelvin (the difference between two scales is only in their starting points). </span>
<span>Th = (Th - Tc) / η </span>
<span>Th = 75 / 0.22 = 341 K (rounded to closest number) </span>
<span>Tc = Th - 75 = 266 K </span>
<span>Lower temperature is Tc = 266 K </span>
<span>Higher temperature is Th = 341 K</span>
Answer:
a. 165.5 V
b. 7.78 A
Explanation:
Here is the complete question
The RMS potential difference of an AC household outlet is 117 V. a) What is the maximum potential difference across a lamp connected to the outlet? b) If the RMS current through the lamp is 5.5 A, what is the maximun current through the lamp.
Solution
a. The maximum potential difference across the lamp V₀ = √2V₁ where V₁ = rms value of potential difference = 117 V
V₀ = √2V₁ = √2 × 117 V = 165.5 V
b. The maximum current through the lamp I₀ = √2I₁ where I₁ = rms value of current = 5.5 A
V₀ = √2V₁ = √2 × 5.5 A = 7.78 A
