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elena-s [515]
3 years ago
13

A highly volatile substance has an initial mass of 1200 g and its mass is reduced by 12% each second.

Physics
1 answer:
Softa [21]3 years ago
4 0

Answer:

Explanation:

a) 1.00 - 0.12 = 0.88

m = 1200(0.88)^t

b) t = ln(m/1200) / ln(0.88)

c) m = 1200(0.88)^10 = 334.20 g

d) t = ln(10/1200) / ln(0.88) = 37.451... = 37 s

e) t = ln(1/1200) / ln(0.88) = 55.463... = 55 s

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Answer:

D) hot stars and interstellar gas, particularly hydrogen

Explanation:

Emission nebula is the nebula which is formed of the ionized gases which emit light of different wavelengths. The common source of ionization of the emission nebula is high-energy photons mostly which are emitted from hot star present nearby to it.

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3 years ago
Jordan wants to know the difference between using a 60-W and 100-W lightbulb in her lamp. She calculates the energy it would tak
Vladimir79 [104]

Answer:

B. 1200 J

Explanation:

Hello.

In this case, given the powers, we compute the energy transferred as shown below:

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What is the difference between uniform and non-uniform magnetic field lines?
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8 0
3 years ago
A solar cell generates a potential difference of 0.25 V when a 550 Ω resistor is connected across it, and a potential difference
Andre45 [30]

a) 400 \Omega

b) 0.43 V

c) 0.44 %

Explanation:

a)

For a battery with internal resistance, the relationship between emf of the battery and the terminal voltage (the voltage provided) is

V=E-Ir (1)

where

V is the terminal voltage

E is the emf of the battery

I is the current

r is the internal resistance

In this problem, we have two situations:

1) when R_1=550 \Omega, V_1=0.25 V

Using Ohm's Law, the current is:

I_1=\frac{V_1}{R_1}=\frac{0.25}{550}=4.5\cdot 10^{-4} A

2) when R_2=1000 \Omega, V_2=0.31 V

Using Ohm's Law, the current is:

I_2=\frac{V_2}{R_2}=\frac{0.31}{1000}=3.1\cdot 10^{-4} A

Now we can rewrite eq.(1) in two forms:

V_1 = E-I_1 r

V_2=E-I_2 r

And we can solve this system of equations to find r, the internal resistance. We do it by substracting eq.(2) from eq(1), we find:

V_1-V_2=r(I_2-I_1)\\r=\frac{V_1-V_2}{I_2-I_1}=\frac{0.25-0.31}{3.1\cdot 10^{-4}-4.5\cdot 10^{-4}}=400 \Omega

b)

To find the electromotive force (emf) of the solar cell, we simply use the equation used in part a)

V=E-Ir

where

V is the terminal voltage

E is the emf of the battery

I is the current

r is the internal resistance

Using the first set of data,

V=0.25 V is the voltage

I=4.5\cdot 10^{-4}A is the current

r=400\Omega is the internal resistance

Solving for E,

E=V+Ir=0.25+(4.5\cdot 10^{-4})(400)=0.43 V

c)

In this part, we are told that the area of the cell is

A=4.0 cm^2

While the intensity of incoming radiation (the energy received per unit area) is

Int.=5.5 mW/cm^2

This means that the power of the incoming radiation is:

P=Int.\cdot A=(5.5)(4.0)=22 mW = 0.022 W

This is the power in input to the resistor.

The power in output to the resistor can be found by using

P'=I^2R

where:

R=1000 \Omega is the resistance of the resistor

I=3.1\cdot 10^{-4} A is the current on the resistor (found in part A)

Susbtituting,

P'=(3.1\cdot 10^{-4})^2(1000)=9.61\cdot 10^{-5} W

Therefore, the efficiency of the cell in converting light energy to thermal energy is:

\epsilon = \frac{P'}{P}\cdot 100 = \frac{9.6\cdot 10^{-5}}{0.022}=0.0044\cdot 100 = 0.44\%

7 0
3 years ago
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