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3 years ago

Name a characteristic property of water

Physics

2 answers:

spayn [35]3 years ago

8 0

Definite volume is a characteristic of water.

12345 [234]3 years ago

4 0

High-specific heat
High Polarity
Attraction to Polar molecules

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What is the difference in force of a car and a tractor

vodka [1.7K]

Answer:

See a tractor is more slow but has a greater force and a car is fast but has a slower force

Explanation:

so your answer is Tractors have more force then cars

6 0

3 years ago

A balloon contains 0.04m3 of air at a pressure of 1.20 x 105Pa. Calculate the pressure required to reduce its volume to 0.025m3

sdas [7]

Answer:

you can simply answer p1v1=p2v2 p2 = p1v1÷ p2 then inter what given in the formula you get it pl=1.20x 105pa vl= 0.40m3 v2=0.025m3

4 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:

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:

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

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,

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

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

Plz help me answer this is my Final Exam!!!

Ostrovityanka [42]

It's answer number four

8 0

3 years ago

If a negative charge is initially at rest in an electric ﬁeld, will it move toward a region of higher potential or lower potenti

monitta

Answer with explanation :

The negative sign means that the potential energy decreases by the movement of the electron.

negative charge at rest in an electric field moves toward the region of an electric field , so that its potential energy will diminish and change into the kinetic energy of motion. The total energy remains constant.

Positive charges will move downhill because of convention. It is to stay in accordance with other potential theories, particularly gravity, where the "charge" is mass, that moves downwards in the gravitational potential field expressed by ϕ(r)=−GM|r|ϕ(r)=−GM|r|. In an electronic system, howbeit, positive charges are fixed in their position within a component (e.g., a wire), therefore instead of the mobile,the negative charges, electrons, move uphill.

6 0

3 years ago