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Anna11 [10]
3 years ago
8

At what point does the external energy enter the system?

Physics
1 answer:
Phoenix [80]3 years ago
8 0
The correct answer as the first one above !
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. The penalties for a person's second DUI conviction include completion of __________ hours of DUI school.
Ghella [55]

Answer:

21

Explanation:

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3 years ago
(a) A 1.00-μF capacitor is connected to a 15.0-V battery. How much energy is stored in the capacitor? ________ μJ (b) Had the ca
gulaghasi [49]

Answer:

(a) E_{ c} = 112.5 \mu J

(b) E'_{ c} = 18 \mu J

Solution:

According to the question:

Capacitance, C = 1.00\mu F = 1.00\times 10^{- 6} F

Voltage of the battery, V_{b} = 15.0 V

(a)The Energy stored in the Capacitor is given by:

E_{c} = \frac{1}{2}CV_{b}^{2}

E_{c} = \frac{1}{2}\times 1.00\times 10^{- 6}\times 15.0^{2}

E_{ c} = 112.5 \mu J

(b) When the voltage of the battery is 6.00 V, the the energy stored in the capacitor is given by:

E'_{c} = \frac{1}{2}CV'_{b}^{2}

E'_{c} = \frac{1}{2}\times 1.00\times 10^{- 6}\times 6.0^{2}

E'_{ c} = 18 \mu J

6 0
3 years ago
Which three characteristics make iron a good blackbody radiator?
victus00 [196]

Answer:

A, C, and D

Explanation:

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4 0
3 years ago
which planet should punch travel to if his goal is to weigh in at 118 lb? refer to the table of planetary masses and radii given
Harrizon [31]

The planet that Punch should travel to in order to weigh 118 lb is Pentune.

<h3 /><h3 /><h3>The given parameters:</h3>
  • Weight of Punch on Earth = 236 lb
  • Desired weight = 118 lb

The mass of Punch will be constant in every planet;

W = mg\\\\m = \frac{W}{g}\\\\m = \frac{236}{g}

The acceleration due to gravity of each planet with respect to Earth is calculated by using the following relationship;

F = mg = \frac{GmM}{R^2} \\\\g = \frac{GM}{R^2}

where;

  • M is the mass of Earth = 5.972 x 10²⁴ kg
  • R is the Radius of Earth = 6,371 km

For Planet Tehar;

g_T =\frac{G \times 2.1M}{(0.8R)^2} \\\\g_T = 3.28(\frac{GM}{R^2} )\\\\g_T = 3.28 g

For planet Loput:

g_L =\frac{G \times 5.6M}{(1.7R)^2} \\\\g_L = 1.94(\frac{GM}{R^2} )\\\\g_L = 1.94g

For planet Cremury:

g_C =\frac{G \times 0.36M}{(0.3R)^2} \\\\g_C = 4(\frac{GM}{R^2} )\\\\g_C = 4 g

For Planet Suven:

g_s =\frac{G \times 12M}{(2.8R)^2} \\\\g_s = 1.53(\frac{GM}{R^2} )\\\\g_s = 1.53 g

For Planet Pentune;

g_P =\frac{G \times 8.3 }{(4.1R)^2} \\\\g_P = 0.5(\frac{GM}{R^2} )\\\\g_P = 0.5 g

For Planet Rams;

g_R =\frac{G \times 9.3M}{(4R)^2} \\\\g_R = 0.58(\frac{GM}{R^2} )\\\\g_R = 0.58 g

The weight Punch on Each Planet at a constant mass is calculated as follows;

W = mg\\\\W_T = mg_T\\\\W_T = \frac{236}{g} \times 3.28g = 774.08 \ lb\\\\W_L = \frac{236}{g} \times 1.94g =457.84 \ lb\\\\ W_C = \frac{236}{g}\times 4g = 944 \ lb \\\\ W_S = \frac{236}{g} \times 1.53g = 361.08 \ lb\\\\W_P = \frac{236}{g} \times 0.5 g = 118 \ lb\\\\W_R = \frac{236}{g} \times 0.58 g = 136.88 \ lb

Thus, the planet that Punch should travel to in order to weigh 118 lb is Pentune.

<u>The </u><u>complete question</u><u> is below</u>:

Which planet should Punch travel to if his goal is to weigh in at 118 lb? Refer to the table of planetary masses and radii given to determine your answer.

Punch Taut is a down-on-his-luck heavyweight boxer. One day, he steps on the bathroom scale and "weighs in" at 236 lb. Unhappy with his recent bouts, Punch decides to go to a different planet where he would weigh in at 118 lb so that he can compete with the bantamweights who are not allowed to exceed 118 lb. His plan is to travel to Xobing, a newly discovered star with a planetary system. Here is a table listing the planets in that system (<em>find the image attached</em>).

<em>In the table, the mass and the radius of each planet are given in terms of the corresponding properties of the earth. For instance, Tehar has a mass equal to 2.1 earth masses and a radius equal to 0.80 earth radii.</em>

Learn more about effect of gravity on weight here: brainly.com/question/3908593

5 0
2 years ago
What is the speed of a wave that has a frequency of 125 Hz and a wavelength of 1.25 meters?
kirill115 [55]
Using the "v = f. λ" <span>equation...
 
Your "v" or </span>velocity = 156.25 meters/second
5 0
3 years ago
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