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liq [111]
2 years ago
10

A compressed spring does not have elastic potential energy.

Physics
1 answer:
Alexus [3.1K]2 years ago
6 0

Answer:

False

Explanation:

The second you let go its gonna release kinetic energy that's why it's potential

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Assessment started: undefined.
enot [183]

Answer:

Option B: Group

Explanation:

The elements fluorine, chlorine, and iodine are commonly known as halogens.

They all have 7 valence electrons in their outermost shell and thus they all belong to group 7A in the periodic table.

Correct Answer is: Group

5 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
Which of the following is a contact force?
Nutka1998 [239]
The contact force is indeed caused by "Contact". 

Air resistance is basically a type of friction, which is apparently present providing two object contact.

The rest selections are all interaction force, which is not necessarily caused by contact.
4 0
2 years ago
Read 2 more answers
Compressed gases aren't ideal. Let's consider a gas that's non-ideal only because the volume available to each of the N molecule
Colt1911 [192]

Answer:

8563732.58906 Pa

3992793.23326 Pa

5708.00923 J

Explanation:

V = Volume

N = Number of molecules = 3\times 6.023\times 10^{23}

T = Temperature = 300 K

b = 7\times 10^{-29}\ m^3

k_ = Boltzmann constant = 1.38\times 10^{-23}\ J/K

P = Pressure

We have the equation

P(V-Nb)=NkT\\\Rightarrow P=\dfrac{NkT}{V-Nb}\\\Rightarrow P=\dfrac{3\times 6.023\times 10^{23}\times 1.38\times 10^{-23}\times 300}{0.001-3\times 6.023\times 10^{23}\times 7\times 10^{-29}}\\\Rightarrow P=8563732.58906\ Pa

The pressure is 8563732.58906 Pa

For isothermal expansion

P_1(V_1-Nb)=P_2(V_2-Nb)\\\Rightarrow P_2=\dfrac{P_1(V_1-Nb)}{V_2-Nb}\\\Rightarrow P_2=\dfrac{8563732.58906(0.001-3\times 6.023\times 10^{23}\times 7\times 10^{-29})}{0.002-3\times 6.023\times 10^{23}\times 7\times 10^{-29}}\\\Rightarrow P_2=3992793.23326\ Pa

The pressure is 3992793.23326 Pa

Work done is given by

dw=Pdv\\\Rightarrow W=\int_{v_1}^{v_2}\dfrac{NkT}{V-Nb}dv\\\Rightarrow W=NkTln\dfrac{V_2-Nb}{V_1-Nb}\\\Rightarrow W=3\times 6.023\times 10^{23}\times 1.38\times 10^{-23}\times 300ln\dfrac{0.002-3\times 6.023\times 10^{23}\times 7\times 10^{-29}}{0.001-3\times 6.023\times 10^{23}\times 7\times 10^{-29}}\\\Rightarrow W=5708.00923\ J

The work done is 5708.00923 J

7 0
2 years ago
Need help with my science quiz
MaRussiya [10]

Explanation:

1.sugar

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5 0
3 years ago
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