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siniylev [52]
3 years ago
14

Imagine your teacher asks you to design an experiment where you test the effect of temperature on the growth of a plant. You hav

e 5 plants that you plant and place in different temperatures around the room. What would your variables be and what would your constants be.
Physics
1 answer:
Dmitry_Shevchenko [17]3 years ago
4 0

Answer:

The correct answer will be-

1. Dependent variable- The growth of plant in the form of height

2. Independent variable- different temperature

3. Constant variable- The amount of water, amount of sunlight, type of soil.

Explanation:

A Scientific experiment must include three types of variables which are: The independent, dependent and the constant variable.

1. Independent variable- The variable which can be modified or changed either on its own or manually. The variable directly influences the variable to be studied. In the given condition, the independent variable is the different temperature provided to the plants.

2. Dependent variable- The variable which is being studied in the experiment and directly influenced by the independent variable is the growth of the plant which is measured in the form of height.

3. Constant variable- The variable which is kept constant throughout the experiment and remains the same which could be the amount of water amount of sunlight and type of soil.

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In a domestic electric circuit (220.V),an electric kettle of 3kW power rating is operatedthat has a current rating of 4A. What r
kozerog [31]

Answer:

(i) It will take longer (75/22 times) the time to boil a given quantity of water compared to when the current is working at its rated capacity

(ii) The reasons are;

(1) For economy; most of the appliances in the home only require low power circuitry with thinner wire while a separate high power circuitry is created directly from the main supply for the high electric power rated appliances

(2) For safety; to prevent the over heating of the electric circuits when an high electric power appliance needs to be connected an high power electric power outlet has to be specified

Explanation:

(i) The power rating of the kettle = 3 kW

The voltage rating of the circuit, V = 220 V

The current rating, I = 4 A

The formula for electric power = I² × R = I × V

Therefore, we have;

Power produced = 220 V × 4 A = 880 V·A = 800 W

Hence, since the power produced is below the power rating of the electric kettle, it will take a longer time to boil a given amount of water than specified by the kettle manufacturer

The energy supplied H = V×I×t

Where:

t = Time in seconds

Therefore, we have;

3 kW = 3000 W;

3000 × t₁ = 880 × t₂

t₂/t₁ =3000/880 = 75/22

Hence the kettle will take 75/22 multiplied by the time it takes when working at rating capacity to boil a given quantity of water

(ii) This is so because the power consumption already factored in the electrical installation as well as the type of appliances utilized in the home allow for several low power rating consumption and few high power rating consumption

Therefore, for both economy and safety the electrical circuit are split to allow for the use of very thick copper or aluminium electric cables in the high power rating electric circuits to which can be plugged high electric power consuming devices such as the water heater and electric cooker

The low electric power consuming devices, such as the electric bulb and fans are connected to the low or "regular" power rating electric circuit outlets

The current required for high power and low power appliances is different and also the fuse rating required for both the appliances different, so two separate circuits are used for high power and low power appliances.

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Explanation:

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The drag force that resists the motion of a car traveling at 80 km h^- 1 is 300 N.
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The power require to keep the car traveling is 6,666 W.

The power of the engine at the given efficiency is 3,999.6 W.

<h3>What is Instantaneous power?</h3>

This the product of force and velocity of the given object.

The power require to keep the car traveling is calculated as follows;

P = Fv

P = 300\ N \ \times  \ \frac{80 \ kmh^{-1}}{3.6 \ km h^{-1}/m/s} \\\\&#10;P = 300 \ N \times 22.22 \ m/s\\\\&#10;P = 6,666 \ W

The power of the engine at the given efficiency is calculated as follows;

E = \frac{P_{out}}{P _{in}} \times 100\%\\\\&#10;60\% = \frac{P_{out}}{6,666} \times 100\%\\\\&#10;0.6 = \frac{P_{out}}{6,666} \\\\&#10;P_{out} = 3,999.6 \ W

Learn more about efficiency here: brainly.com/question/15418098

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2 years ago
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