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maks197457 [2]
3 years ago
13

WILL GIVE THE BRAINLEST

Physics
1 answer:
Leona [35]3 years ago
3 0
<h2>Question no. 1: </h2><h3>Answer: </h3>

Controlled experiment is an experiment which is designed to test a variable, which can be changed during the experiment.

<h3>Explanation: </h3>

Hence in this experiment, the importance of experiment is being tested, as it is added only in one set of plant.

A manipulated variable is an independent variable whose amount can be changed during an experiment.

Here the fertilizer is a manipulated variable.

<h2>Question 2: </h2><h3>Answer: </h3>

Because theory is the most organised statement proved by a set of independent experiments.

<h3>Explanation: </h3>

Hypothesis is the initial idea on which basis different experiments are conducted. But if 1 experiment succeeded and it supports the hypothesis, hypothesis do not become a theory.

For a theory, different experiments are necessary which supports that hypothesis.

<h2>Question no 4. </h2><h3>Answer: </h3>

Density = 1.909 kg/m3

<h2>Explanation: </h2>

Given data:

Mass of the object = 42 kg

Volume of the object = 22 m3

Density of the object = ?

Solution

Density = mass/volume

Density =42/22 = 1.909

Density = 1.909 kg/m3

<u><em>Hence the density of the object is 1.909 and its unit is kg/m3 which is system international unit of density. </em></u>

<h2>Question 5. </h2><h3>Answer: </h3>

Chlorine and water in swimming pool are mixed in such a way that there composition remains same throughout the mixture. And chlorine cannot be separated back from water. This is the homogeneous mixture.

<h3>Explanation: </h3>
  • A homogeneous mixture is a mixture of more than one substance which are mixed together in such a way that they cannot be separated and the composition of the mixture remains same throughout the mixture.
  • Example of homogeneous solution is sugar solution in water.
  • As for swimming pool cleaning chlorine is added in pool water, it mixes in the whole water by diffusion. Hence it is a homogeneous mixture.
<h2>Question no. 6. </h2><h3>Answer: </h3>

Because physical change does not include composition change, it includes change in physical parameters. And in splicing tomato only change in shape is done.

<h3>Explanation: </h3>
  • Chemical change means irreversible chemical composition change like the burning of toast.
  • Physical change means reversible or irreversible change in shape, volume density without change in chemical composition.
  • Splicing tomato comes under 2nd category because it does not include chemical change.

Remaining questions are attached in the files.

<h2 />
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Read 2 more answers
An electron enters a region with a speed of 5×10^6m/s and is slowed down at the rate of 1.25×10^-4m/s². How far does the electro
Mashutka [201]

1) The distance travelled by the electron is 1\cdot 10^{17} m

2) The time taken is 4.0\cdot 10^{10}s

Explanation:

1)

The electron in this problem is moving by uniformly accelerated motion (constant acceleration), so we can use the following suvat equation

v^2-u^2=2as

where

v is the final velocity

u is the initial velocity

a is the acceleration

s is the distance travelled

For the electron in this problem,

u=5\cdot 10^6 m/s is the initial velocity

v = 0 is the final velocity (it comes to a stop)

a=-1.25\cdot 10^{-4} m/s^2 is the acceleration

Solving for s, we find the distance travelled:

s=\frac{v^2-u^2}{2a}=\frac{0-(5\cdot 10^6)^2}{2(-1.25\cdot 10^{-4})}=1\cdot 10^{17} m

2)

The total time taken for the electron in its motion can also be found by using another suvat equation:

v=u+at

where

v is the final velocity

u is the initial velocity

a is the acceleration

t is the time taken

Here we have

u=5\cdot 10^6 m/s

v = 0

a=-1.25\cdot 10^{-4} m/s^2

And solving for t, we find the time taken:

t=\frac{v-u}{a}=\frac{0-5\cdot 10^6}{-1.25\cdot 10^{-4}}=4.0\cdot 10^{10}s

Learn more about accelerated motion:

brainly.com/question/9527152

brainly.com/question/11181826

brainly.com/question/2506873

brainly.com/question/2562700

#LearnwithBrainly

7 0
3 years ago
A 75.00 gram sample of an unknown metal initially at 99.0 degrees Celcius is added to 50.00 grams of water initially at 10.79 de
jeyben [28]

Answer:

  c_{e1} = 0.331 J / g ° C

Explanation:

We have a calorimetry exercise where all the heat yielded by one of the components is absorbed by the other.

Heat ceded          Qh = m1 ce1 (T_{h} -T_{f})

Heat absorbed     Qc = m2 ce2 (T_{f} - T₀)

Body 1 is metal and body 2 is water .  Where m are the masses of the two bodies, ce their specific heat and T the temperatures

      Qh = Qc

      m₁ c_{e1} (T_{h}- T_{f}) = m₂  c_{e2} (T_{f} - T₀)

we clear the specific heat of the metal

      c_{e1} = m₂  c_{e2} (T_{f} - T₀) / (m₁ (T_{h}-T_{f}))

     c_{e1}= 50.00 4.184 (20.15 -10.79) / (75.00 (99.0-20.15))

      c_{e1} = 209.2 (9.36) / (75 78.85)

      c_{e1} = 1958.11 / 5913.75

     c_{e1} = 0.331 J / g ° C

5 0
4 years ago
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