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Alja [10]
3 years ago
10

Which of the following is an example of radiation?

Chemistry
1 answer:
Lapatulllka [165]3 years ago
5 0
I believe the best answer to that question wud be D. I cud b wrong 
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H2SO4 +HI → __ H2S+12 +H2O balance the equation
Citrus2011 [14]

Answer:

H2SO4 + 8HI → H2S + 4I2 + 4H2O

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How many hydrogen atoms are needed to form five water molecules
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2 Hydrogen Atoms Are Needed To Form Five Water Molecules
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a vehicle accelerates with 0.4 m/s. calculate tue time taken by the vehicle to increase its speed from 20m/s to 40m/s.​
Grace [21]

Answer:

\huge\boxed{\sf t = 50 \ seconds}

Explanation:

<h3><u>Given data:</u></h3>

Acceleration = a = 0.4 m/s²

Initial Speed = V_i = 20 m/s

Final Speed = V_f = 40 m/s

<h3><u>Required:</u></h3>

Time = t = ?

<h3><u>Formula:</u></h3>

\displaystyle a =\frac{V_f-V_i}{t}

<h3><u>Solution:</u></h3>

Rearranging formula for t

\displaystyle t =\frac{V_f-V_i}{a} \\\\t = \frac{40-20}{0.4} \\\\t = \frac{20}{0.4} \\\\\boxed{t = 50 \ seconds}\\\\\rule[225]{225}{2}

6 0
2 years ago
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Can you test a hypothesis that hummingbirds spend most of their waking hours feeding
Vitek1552 [10]
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6 0
3 years ago
If the distance between two objects is decreased to - of the original
Charra [1.4K]

Answer:

The new force will be \frac{1}{100} of the original force.

Explanation:

In the context of this problem, we're dealing with the law of gravitational attraction. The law states that the gravitational force between two object is directly proportional to the product of their masses and inversely proportional to the square of a distance between them.

That said, let's say that our equation for the initial force is:

F = G\frac{m_1m_2}{R^2}The problem states  that  the distance decrease to 1/10 of the original distance, this means:[tex]R_2 = \frac{1}{10}R

And the force at this distance would be written in terms of the same equation:

F_2 = G\frac{m_1m_2}{R_2^2}

Find the ratio between the final and the initial force:

\frac{F_2}{F} = \frac{G\frac{m_1m_2}{R_2^2}}{G\frac{m_1m_2}{R^2}}

Substitute the value for the final distance in terms of the initial distance:

\frac{F_2}{F} = \frac{G\frac{m_1m_2}{(\frac{R}{10})^2}}{G\frac{m_1m_2}{R^2}}

Simplify:

\frac{F_2}{F} = \frac{\frac{1}{100R^2}}{\frac{1}{R^2}}=\frac{1}{100}

This means the new force will be \frac{1}{100} of the original force.

8 0
3 years ago
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