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qaws [65]
3 years ago
14

When is a hypothesis developed in the scientific method?

Physics
2 answers:
ivanzaharov [21]3 years ago
7 0

Answer:

aren't there pics ?

Explanation:

shtirl [24]3 years ago
6 0

Answer:

D. After a question is when you do your hypothesis. I've done many science projects so I know this is right

Explanation:

The hypothesis is written as a prediction of the outcome of the experiment

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A polar bear runs at a speed of 11 m/s and has a mass of 380.2 kg. How much Kinetic energy does the bear have?
Yanka [14]

Answer:

\boxed{\sf Kinetic \ energy \ of \ the \ bear (KE) = 23002.1 \ J}

Given:

Mass of the polar bear (m) = 6.8 kg

Speed of the polar bear (v) = 5.0 m/s

To Find:

Kinetic energy of the polar bear (KE)

Explanation:

Formula:

\boxed{ \bold{\sf KE =  \frac{1}{2} m {v}^{2} }}

Substituting values of m & v in the equation:

\sf \implies KE =  \frac{1}{2}  \times 380.2 \times  {11}^{2}

\sf \implies KE = \frac{1}{ \cancel{2}}  \times  \cancel{2} \times 190.1 \times 121

\sf \implies KE = 190.1 \times 121

\sf \implies KE = 23002.1 \: J

\therefore

Kinetic energy of the polar bear (KE) = 23002.1 J

5 0
3 years ago
The decomposition of ammonia to the elements is a first-order reaction with a half-life of 200 s at a certain temperature. How l
Feliz [49]

Based on the half-life of the reaction, time taken for the partial pressure of ammonia to decrease from 0.100 atm to 0.00625 atm is 800 seconds.

<h3>What is the half-life of a substance?</h3>

The half-life of a substance is the time taken for half the amount of atoms present in that substance to decay.

The half-life of the reaction is 200 seconds.

After, one half-life, pressure reduces to 0.0500 atm

After, two half-lives, pressure reduces to 0.0250 atm

After, three half-lives, pressure reduces to 0.01250 atm

After, four half-lives, pressure reduces to 0.00625 atm

Time taken = 4 * 200 = 800 seconds

Therefore, time taken for the partial pressure of ammonia to decrease from 0.100 atm to 0.00625 atm is 800 seconds.

Learn more about half-life at: brainly.com/question/26689704

#SPJ1

7 0
1 year ago
To avoid using the variable "a"
sdas [7]

Answer:

option c .

Explanation:

..............................

6 0
2 years ago
The atomic radii of Mg2+ and F- ions are 0.072 and 0.133 nm, respectively.(a) Calculate the force of attraction between these tw
timurjin [86]

Answer:

1.09527\times 10^{-8}\ N

Explanation:

q_1 = Mg ion = +2q

q_2 = F ion = -q

q = Charge of electron = 1.6\times 10^{-19}\ C

r = Distance between ions = 0.072+0.133\ nm

k = Coulomb constant = 8.99\times 10^{9}\ Nm^2/C^2

Electrical force is given by

F=-\dfrac{kq_1q_2}{r^2}\\\Rightarrow F=-\dfrac{8.99\times 10^9\times 2\times 1.6\times 10^{-19}\times -1\times 1.6\times 10^{-19}}{[(0.072+0.133)\times 10^{-9}]^2}\\\Rightarrow F=1.09527\times 10^{-8}\ N

The attractive force is 1.09527\times 10^{-8}\ N

8 0
3 years ago
in the diagram, q2 is +34.4*10^-6 C, and q3 is -72.8*10^-6 C. The net force on q2 is 225 N to the right. What is q1? Include the
sesenic [268]

Answer:

  q₁ = -6.54 10⁻⁵ C

Explanation:

Force is a vector quantity, but since all charges are on the x-axis, we can work in one dimension, let's apply Newton's second law

                F = F₁₂ + F₂₃

the electric force is given by Coulomb's law

                F = k q₁q₂ / r₁₂²

let's write the expression for each force

                F₂₃ = k q₂ q₃ / r₂₃²

                F₂₃ = 9 10⁹ 34.4 10⁻⁶ 72.8 10⁻⁶ / 0.1²

                F₂₃ = 2.25 10³ N

               

                F₁₂ = k q₁q₂ / r₁₂²

                F₁₂ = 9 10⁹ q₁ 34.4 10⁻⁶ / 0.1²

                F₁₂ = q₁   3,096 10⁷ N

we substitute in the first equation

                225 = q₁  3,096 10⁷ +2.25 10³

                q₁ = (225 - 2.25 10³) / 3,096 10⁷

                q₁ = -6.54 10⁻⁵ C

4 0
3 years ago
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