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Scilla [17]
3 years ago
6

A patricle is accelerated uniformly from rest, so that after 10 seconds it has a speed of 15m/s find its acceleration and the di

stance it has covered?
Physics
1 answer:
grin007 [14]3 years ago
8 0

Answer:

Its acceleration is 1.5 \frac{m}{s^{2} } and the distance it has covered is 150 m.

Explanation:

Acceleration is a quantity that indicates the variation in speed of a moving body as time passes. That is, acceleration relates changes in velocity with the time in which they occur.

The average acceleration is defined as the relationship between the variation or change in the speed of a mobile and the time used in said change:

acceleration=\frac{change in speed}{time}

In this case:

  • change in speed= final speed - initial speed= 15 m/s - 0 m/s= 15 m/s
  • time= 10 s

Replacing:

acceleration=\frac{15 \frac{m}{s} }{10 s}

and solving you get:

<u><em>acceleration= 1.5 </em></u>\frac{m}{s^{2} }<u><em></em></u>

Speed ​​is a quantity that expresses the relationship between the space traveled by an object and the time used for it. This is:

speed=\frac{distance}{time}

So the distance can be calculated as:

distance= speed* time

In this case:

  • speed= 15 m/s
  • time= 10 s

Replacing:

distance= 15 m/s* 10 s

and solving you get:

<u><em>distance= 150 m</em></u>

<u><em>Its acceleration is 1.5 </em></u>\frac{m}{s^{2} }<u><em> and the distance it has covered is 150 m.</em></u>

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Answer:

Acceleration of the ship, a=2.14\times 10^{-7}\ m/s^2

Explanation:

It is given that,

Mass of both ships, m=39000\ metric\ tons=39\times 10^6\ kg

Distance between two ships, d = 110 m

The gravitational force between two ships is given by :

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F=6.67\times 10^{-11}\ Nm^2/kg^2\times \dfrac{(39\times 10^6\ kg)^2}{(110\ m)^2}

F = 8.38 N

Let a is the acceleration. Now, using second law of motion as :

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a=\dfrac{8.38\ N}{39\times 10^6\ kg}

a=2.14\times 10^{-7}\ m/s^2

So, the acceleration of either ship due to the gravitational attraction of the other is 2.14\times 10^{-7}\ m/s^2. Hence, this is the required solution.

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work done as mass 1 moves to mass 2. the gravitational force between two point masses separated by a distance r is proportional
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gravitational force

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3 years ago
Object A has 27 J of kinetic energy. Object B has one-quarter the mass of object A.
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Answer:

the final speed of object A changed by a factor of  \frac{1}{\sqrt{3} } = 0.58

the final speed of object B changed by a factor of \sqrt{\frac{5}{3} } = 1.29

Explanation:

Given;

kinetic energy of object A, = 27 J

let the mass of object A = m_A

then, the mass of object B = m_B = \frac{m_A}{4}

work done on object A = -18 J

work done on object B = -18 J

let v_i be the initial speed

let v_f be the final speed

For object A;

K.E_A = 27\\\\\frac{1}{2} m_A v_i^2 = 27\\\\m_A v_i^2  = 54\\\\m_A = \frac{54}{v_i^2} ----Equation \ (1)\\\\Apply \ work-energy \ theorem;\\\\\delta K.E_A = -18\\\\\frac{1}{2} m_A v_f^2 - \frac{1}{2} m_A v_i^2 = -18\\\\\frac{1}{2} m_A ( v_f^2 \ -  v_i^2 )\ =- 18\\\\v_f^2 \ -  v_i^2  = -\frac{36}{m_A} ---Equation \ (2)\\\\v_f^2 \ -  v_i^2  = -\frac{36v_i^2}{54}\\\\ v_f^2 \ =v_i^2 - \frac{36v_i^2}{54}\\\\ v_f^2 = \frac{54v_i^2 -36v_i^2 }{54} \\\\v_f^2 = \frac{18v_i^2}{54} \\\\v_f^2 = \frac{v_i^2}{3} \\\\

v_f = \sqrt{\frac{v_i^2}{3} }\\\\v_f = \frac{1}{\sqrt{3} } \ v_i\\\\

Thus, the final speed of object A changed by a factor of  \frac{1}{\sqrt{3} } = 0.58

To obtain the change in the final speed of object B, apply the following equations.

K.E_B_i = \frac{1}{2} m_Bv_i^2\\\\m_B = \frac{m_A}{4} \\\\K.E_B_i = \frac{1}{2}(\frac{m_A}{4} )v_i^2\\\\K.E_B_i = \frac{m_Av_i^2}{8} \\\\But, \ m_Av_i^2 = 54 \\\\K.E_B_i = \frac{54}{8} \\\\Apply \ work-energy \ theorem ;\\\\\delta K.E = -18\\\\K.E_f -K.E_i = -18\\\\\frac{1}{2}m_Bv_f^2 - \frac{1}{2} m_Bv_i^2 = -18\\\\Recall \ m_B =  \frac{m_A}{4} \\\\\frac{1}{2}(\frac{m_A}{4} )v_f^2 - \frac{1}{2}(\frac{m_A}{4} )v_i^2 = -18\\\\\frac{1}{2}\times \frac{m_A}{4} (v_i^2 -v_f^2) = 18\\\\

\frac{1}{2}\times \frac{m_A}{4} (v_i^2 -v_f^2) = 18\\\\v_i^2 -v_f^2 = \frac{8}{m_A} \times 18\\\\v_i^2 -v_f^2 =\frac{144}{m_A} \\\\But , m_A = \frac{54}{v_i^2} \\\\v_i^2 -v_f^2 =\frac{144v_i^2}{54} \\\\v_f^2 = v_i^2 - \frac{144v_i^2}{54}\\\\v_f^2 = \frac{54v_i^2-144v_i^2}{54}\\\\ v_f^2 = \frac{-90v_i^2}{54} \\\\v_f^2 = \frac{-5v_i^2}{3} \\\\|v_f| = \sqrt{\frac{5v_i^2}{3}} \\\\|v_f| = \sqrt{\frac{5}{3}} \ v_i

Thus, the final speed of object B changed by a factor of \sqrt{\frac{5}{3} } = 1.29

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On the contrary, when the appliances are connected in parallel, they are connected in different branches, so if one of them is switched off, the other branches continue working unaffacted by it.

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