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Vlad1618 [11]
3 years ago
9

a body has weight 20N how much force is required to move it vertically upwards with an acceleration of 2m/s​

Physics
1 answer:
kramer3 years ago
3 0

<h2>Given :-</h2>

  • Weight = 20 N

  • Acceleration = 2 m/s

<h2>To Find :-</h2>

  • Force applied=?

<h2>Solution :-</h2>

As we know that,

<h3>F = ma</h3>

  • F is the Force applied
  • M is the Mass
  • A is the Acceleration

<u>According</u><u> </u><u>to</u><u> </u><u>the question</u><u>, </u>

F = 20 × 2

F = 40 N

<h3>Hence :-</h3>

Force applied is 40 N

<h2>Know More :-</h2>

  • First equation of motion
  • v = u + at

  • Second equation of motion
  • s = ut + ½at²

  • Third Equation of motion
  • v² - u² = 2as

\begin{gathered} \\ \end{gathered}

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Answer:

Final speed of the train is 7.5 m/s

Explanation:

It is given that,

Uniform acceleration of the train is, a = 1.5 m/s²

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v=0+1.5\ m/s^2\times 5\ s  

v = 7.5 m/s

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7 0
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Two glasses of water have the same thermal energy. must they have the same temperature? explain.
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The first car has twice the mass of a second car, but only half as much kinetic energy. When both cars increase their speed by 9
kirill [66]

Answer:

Speed of the car 1 =V_1=8.98m/s

Speed of the car 2 =V_2=17.96m/s

Explanation:

Given:

Mass of the car 1 , M₁ = Twice the mass of car 2(M₂)

mathematically,

M₁ = 2M₂

Kinetic Energy of the car 1 = Half the kinetic energy of the car 2

KE₁ = 0.5 KE₂

Now, the kinetic energy for a body is given as

KE =\frac{1}{2}mv^2

where,

m = mass of the body

v = velocity of the body

thus,

\frac{1}{2}M_1V_1^2=0.5\times \frac{1}{2}M_2V_2^2

or

\frac{1}{2}2M_2V_1^2=0.5\times \frac{1}{2}M_2V_2^2

or

2M_2V_1^2=0.5\times M_2V_2^2

or

2V_1^2=0.5\times V_2^2

or

4V_1^2= V_2^2

or

2V_1= V_2  .................(1)

also,

\frac{1}{2}M_1(V_1+9.0)^2=\frac{1}{2}M_2(V_2+9.0)^2

or

\frac{1}{2}2M_2(V_1+9.0)^2=\frac{1}{2}M_2(2V_1+9.0)^2

or

2(V_1+9.0)^2=(2V_1+9.0)^2

or

\sqrt{2}(V_1+9.0)=(2V_1+9.0)

or

(\sqrt{2}V_1+ \sqrt{2}\times 9.0)=(2V_1+9.0)

or

(\sqrt{2}V_1+ 12.72)=(2V_1+9.0)

or

(2V_1-\sqrt{2}V_1)=(12.72-9.0)

or

(0.404V_1)=(3.72)

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V_1=8.98m/s

and, from equation (1)

V_2=2\times 8.98m/s = 17.96m/s

Hence,

Speed of car 1 =V_1=8.98m/s

Speed of car 2 =V_2=17.96m/s

8 0
3 years ago
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Answer:

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Explanation:

d = ½ g t2

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75% sure

Hope this helps!!!

5 0
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