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2 years ago

A factor that changes in an experiment from manipulation of the independent variable is the what?

1 answer:

7 0

Manipulation of the independent variable should change the dependent variable, the value of which "depends" on the independent variable's change in value.

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1. A 250g chunk of metal is heated with 400 joules of energy and the temperature goes from 20 °C to 25°C. What is its specific h

Paha777 [63]

The specific heat capacity of this chunk of metal is equal to 0.32 J/g°C.

<u>Given the following data:</u>

Mass of metal = 250g

Quantity of energy = 400 Joules

Initial temperature = 20°C

Final temperature = 20°C

To determine the specific heat capacity of this chunk of metal:

<h3>The formula for quantity of heat.</h3>

Mathematically, quantity of heat is given by the formula;

$Q = mc\theta$

<u>Where:</u>

Q represents the quantity of heat.

m represents the mass of an object.

c represents the specific heat capacity.

∅ represents the change in temperature.

Making c the subject of formula, we have:

$c = \frac{Q}{m\theta}$

Substituting the given parameters into the formula, we have;

$c = \frac{400}{250 \times (25-20)}\\\\c = \frac{400}{250 \times 5}\\\\c = \frac{400}{1250 }$

Specific heat, c = 0.32 J/g°C.

Read more on specific heat here: brainly.com/question/2834175

5 0

1 year ago

What is the wavelength of radiation that has a frequency

White raven [17]

Answer:

$\lambda = \frac{c}{\nu}$

Explanation:

The wavelenght of a radiation is inversely proportional to its frequency. It can be estimated by the following formula:

$\lambda = \frac{c}{\nu}$

Where:

$\lambda$ is the wavelenght
$\nu$ is the frequency
$c$ is the speed of light (arroung 300000 km/s)

5 0

2 years ago

Pressure gauge at the top of a vertical oil well registers 140 bars. The oil well is 6000 m deep and filled with natural gas dow

andreyandreev [35.5K]

Explanation:

(a) The given data is as follows.

Pressure on top ( $P_{o}$ ) = 140 bar = $1.4 \times 10^{7} Pa$ (as 1 bar = $10^{5}$ )

Temperature = $15^{o}C$ = (15 + 273) K = 288 K

Density of gas = $\frac{PM}{ZRT}$

$\frac{dP}{dZ} = \rho \times g$

$\frac{dP}{dZ} = \int \frac{PM}{ZRT}$

$\int_{P_{o}}^{P_{1}} \frac{dP}{dZ} = \frac{Mg}{ZRT} \int_{0}^{4700} dZ$

$ln (\frac{P_{1}}{P_{o}}) = \frac{18.9 \times 10^{-3} \times 9.81 \times 4700 m}{0.80 \times 8.314 J/mol K \times 288 K}$

= 0.4548

$P_{1} = P_{o} \times e^{0.4548}$

= $1.4 \times 10^{7} Pa \times 1.5797$

= $2.206 \times 10^{7} Pa$

Hence, pressure at the natural gas-oil interface is $2.206 \times 10^{7} Pa$ .

(b) At the bottom of the tank,

$P_{2} = P_{1} + \rho \times g \times h$

= 2.206 \times 10^{7} Pa + 700 \times 9.81 \times (6000 - 4700)[/tex]

= $309.8 \times 10^{5} Pa$

= 309.8 bar

Hence, at the bottom of the well at $15^{o}C$ pressure is 309.8 bar.

6 0

3 years ago

What is the density of a liquid if it's volume is 125 mL and it's mass is 50g?

masha68 [24]

Answer:

0.4g/ml

Explanation:

density= mass/volume

density=50g/125ml

density=0.4g/ml

6 0

2 years ago

Hi there, I need help producing the aim and hypothesis for the question

svetlana [45]

Answer:

in supernova gold can be turned into lead.

its practical to obtain gold from lead

hope this helps you☺️☺️

4 0

2 years ago