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valentina_108 [34]
3 years ago
14

If you add 1 proton to Carbon, it will no longer be Carbon, it will be ________________.

Physics
1 answer:
Whitepunk [10]3 years ago
4 0

Nitrogen

Explanation:

Adding one proton to a carbon atom makes Nitrogen.

A quick introspection on atoms:

  • An atom is made up of three fundamental particles.
  • They are protons, neutrons and electrons.
  • The protons are positively charged and the neutrons do not carry any charges.
  • Electrons are negatively charged.

The difference between an atom and another is the number of protons in them. This is the atomic number.

The periodic table of element is a list of elements arranged based on the number of protons they have. Every element on the table has unique number of protons which makes it differ from another.

  • Atoms do not readily lose their protons because they are held by nuclear forces in the nucleus of an atom.

When an element gains a proton, it becomes another element.

    Carbon has proton number of 6

 If a proton is added to it, it becomes 7

This is the proton or atomic number of nitrogen.

Learn more:

Atomic number brainly.com/question/5425825

#learnwithBrainly

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Answer:

This question is asking to identify the following variables:

Independent variable (IV): TYPE OF SOIL

Dependent variable (DV): HEIGHT AND NUMBER OF LEAVES

Control group: None in this experiment

Constant: SAME ROSE PLANT, SAME TIME INTERVAL (1 WEEK)

Explanation:

Independent variable in an experiment is the variable that is manipulated or changed by the experimenter in order to effect a measurable outcome. In this case, the independent variable is the TYPE OF SOIL used.

Dependent variable is the measurable variable that responds to changes made to the independent variable. In this experiment, the dependent variable is the HEIGHT AND NUMBER OF LEAVES of each rose.

Constants or control variable is the variable that is kept unchanged or constant for all groups throughout the experiment. In this experiment, the constants are SAME ROSE PLANT, SAME TIME INTERVAL (1 WEEK).

Control group are the groups that does not receive the experimental treatment. In this case, all the groups received the experimental treatment (different soil types). Hence, there is no control

4 0
3 years ago
A projectile is shot at an angle 45 degrees to the horizontalnear the surface of the earth but in the absence of air resistance.
ivann1987 [24]

Answer:

v₂ = 176.24 m/s

Explanation:

given,

angle of projectile = 45°

speed = v₁ = 150 m/s

for second trail

speed = v₂ = ?

angle of projectile = 37°

maximum height attained formula,

H_{max}= \dfrac{v^2 sin^2(\theta)}{g}

now,

H_{max}= \dfrac{v_1^2 sin^2(\theta_1)}{g}

H_{max}= \dfrac{v_2^2 sin^2(\theta_2)}{g}

now, equating both the equations

\dfrac{v_2^2}{v_1^2}=\dfrac{sin^2(\theta_1)}{sin^2(\theta_2)}

\dfrac{v_2^2}{150^2}=\dfrac{sin^2(45^0)}{sin^2(37^0)}

   v₂² = 31061.79

   v₂ = 176.24 m/s

velocity of projectile would be equal to v₂ = 176.24 m/s

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An unstable particle at rest breaks up into two fragments of unequal mass. The mass of the lighter fragment is equal to 2.90 ✕ 1
motikmotik

Answer:

The speed of the heavier fragment is 0.335c.

Explanation:

Given that,

Mass of the lighter fragment M_{l}=2.90\times10^{-28}\ kg

Mass of the heavier fragment M_{h}=1.62\times10^{-27}\ Kg

Speed of lighter fragment = 0.893c

We need to calculate the speed of the heavier fragment

Let v is the speed of the second fragment after decay

Using conservation of relativistic momentum

0=\drac{m_{1}v_{1}}{\sqrt{1-\dfrac{v_{1}^2}{c^2}}}-\drac{m_{2}v_{2}}{\sqrt{1-\dfrac{v_{1}^2}{c^2}}}

\drac{m_{1}v_{1}}{\sqrt{1-\dfrac{v_{1}^2}{c^2}}}=\drac{m_{2}v_{2}}{\sqrt{1-\dfrac{v_{1}^2}{c^2}}}

\dfrac{2.90\times10^{-28}\times0.893c}{\sqrt{1-(0.893)^2}}=\dfrac{1.62\times10^{-27}v_{2}}{\sqrt{1-\dfrac{v_{2}^2}{c^2}}}

\dfrac{v_{2}}{\sqrt{1-\dfrac{v_{2}^2}{c^2}}}=\dfrac{2.90\times10^{-28}\times0.893c}{1.62\times10^{-27}\times0.45}

\dfrac{v_{2}}{\sqrt{1-\dfrac{v_{2}^2}{c^2}}}=0.355c

\dfrac{v_{2}}{1-\dfrac{v_{2}^{2}}{c^2}}=(0.355c)^2

\dfrac{1-\dfrac{v_{2}^2}{c^2}}{v_{2}^2}=\dfrac{1}{(0.355c)}

\dfrac{1}{v_{2}^2}-\dfrac{1}{c^2}=\dfrac{1}{(0.355c)^2}

\dfrac{1}{v_{2}^2}=\dfrac{1}{c^2}+\dfrac{1}{0.126c^2}

\dfrac{1}{v_{2}^2}=\dfrac{1}{c^2}(1+\dfrac{1}{0.126})

\dfrac{1}{v_{2}^2}=\dfrac{8.93}{c^2}

v_{2}^2=\dfrac{c^2}{8.93}

v_{2}=0.335c

Hence, The speed of the heavier fragment is 0.335c.

7 0
3 years ago
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