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Solnce55 [7]
4 years ago
14

When a constant force acts upon an object, the acceleration of the object varies inversely with its mass 2kg. When a certain con

stant force acts upon an object with mass , the acceleration of the object is 26m/s^2 . If the same force acts upon another object whose mass is 13 , what is this object's acceleration
Physics
1 answer:
BARSIC [14]4 years ago
3 0

Answer:

The acceleration of the second object is 4 m/s².

Explanation:

Given that,

when the a constant force acts upon an object, the acceleration varies inversely with its mass .

mass\propto \frac{1}{acceleration}

We can rewrite the equation as

\frac{m_1}{m_2}=\frac{a_2}{a_1}

When an a certain constant force acts upon an object with mass 2 kg , the acceleration of the object is 26 m/s².

Here, m_1 = 2 kg, a_1 = 26 m/s²,m_2 = 13 kg and a_2=?

\frac{m_1}{m_2}=\frac{a_2}{a_1}

\frac{2}{13}=\frac{a_2}{26}

\Rightarrow \frac{a_2}{26}= \frac{2}{13}

\Rightarrow {a_2}= \frac{2}{13}\times 26

\Rightarrow a_2=4 m/s²

The acceleration of the second object is 4 m/s².

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How would the force between two charged particles change if one of the charges were to triple in strength (3x stronger)?
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Answer:

The correct option is (d). "3x stronger".

Explanation:

The force between two charged particle is given by :

F=k\dfrac{q_1q_2}{r^2}

If one charge is tripled, q_1'=3q

New force will be :

F'=\dfrac{kq_1'q_2}{r^2}\\\\=3\times F

It means if one of the charges were to triple in strength, then the force will become 3 times of the initial force.

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As the speed/velocity of an object increases what happens to the kinetic energy of the object
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Which best describes why scientific models change over time?
irina1246 [14]
Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.

Below are the choices and the answer is <span>B. 
</span>a.) Models can be changed by scientists who are more famous than previous scientists. 
<span>b.) As technology advances, new experiments often expose problems in accepted theories. </span>
<span>c.) Scientists tend to value new theories because they are more exciting. </span>
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6 0
4 years ago
The star Sirius has an apparent magnitude of -1.46 and appears 95-times brighter compared to the more distant star Tau Ceti, whi
dimaraw [331]

Answer:

(a) Apparent magnitude is the perceived brightness of an astronomical object

Absolute magnitude is the luminosity based on viewing an object from a 32.6 light-years distance

Bolometric magnitude is the total emitted radiation of a star

(b) The apparent magnitude of the star Tau Ceti = 3.51

(c) The distance between the Earth and Tau Ceti is 1.13 × 10¹⁴ km

Explanation:

(a) Apparent magnitude is an estimate of an astronomical objects' brightness as the object is perceived from the Earth

The absolute magnitude  is the magnitude an object appears to have when viewed from a 32.6 light-years distance while having constant transfer of its luminosity that is not affected by cosmic dust and objects present in the line of sight

The bolometric magnitude of a star is the sum total of the star's radiation released over all electromagnetic spectrum wavelengths

(b) The apparent magnitude of the star Tau Ceti is found using the following equation;

m_{2}-m_{1} = -2.512\times log\left (\dfrac{B_{2}}{B_{1}}  \right )

Where:

m₁ = Apparent magnitude of Tau Ceti

m₂ = Apparent magnitude of  Sirius = -1.46

B₁ = Brightness of Tau Ceti

B₂ = Brightness of Sirius

\left \dfrac{B_{2}}{B_{1}}  \right  = 95

Hence we have;

-1.46-m_{1} = -2.512\times log\left (95 \right )

m₁ = -1.46 + 2.512 × log(95) = 3.51

The apparent magnitude of the star Tau Ceti = 3.51

(c) The distance between the Earth and Tau Ceti is found using the following equation;

m-M = 5\times log\left (\dfrac{d}{10}  \right )

Where:

m = Apparent magnitude of Tau Ceti

M = Absolute magnitude of Tau Ceti = 5.69

d = The distance between the Earth and Tau Ceti

Which gives;

3.51-5.69 = 5\times log\left (\dfrac{d}{10}  \right )

\therefore \dfrac{d}{10} = 10^{-0.436} = 0.3664

d = 10 × 0.3664 = 3.664 parsecs = 3.664 × 3.0857 × 10¹⁶ m

d = 1.13 × 10¹⁴ km.

4 0
3 years ago
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