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2 years ago

Exercise Book: 18

Physics

1 answer:

Harlamova29_29 [7]2 years ago

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When object goes under acceleration

lidiya [134]

when object goes under acceleration

c).its velocity always increases

<h3>Additional information:-</h3>

★ Acceleration: Rate of increase in velocity.

★ Velocity: Distance travelled by a body per unit time in given direction is called velocity .

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3 years ago

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How is the kinetic energy of the particles that make up a substance different

Feliz [49]

Answer:

the kinetic energy is less after a substance changes from liquid to gas

Explanation:

the kinetic energy is greater after a substance changes from a solid to liquid

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An airplane, starting from rest, moves down the runway at constant acceleration for 18 s and then takes off at a speed of 60 m/s

ruslelena [56]

Answer:

a =3.33 m/s²

Explanation:

given,

initial speed of Plane, u = 0 m/s

final speed of plane, v = 60 m/s

time of the acceleration, t = 18 s

average acceleration of the plane, a = ?

average acceleration is equal to change in velocity per unit time.

$a = \dfrac{v - u}{t}$

$a = \dfrac{60 - 0}{18}$

$a = \dfrac{60}{18}$

a =3.33 m/s²

Hence, average acceleration of the plane is equal to a =3.33 m/s²

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2 years ago

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What is the amount of charge when 13.5a is flowing for 2 1/2 hours

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Answer:

Quick Answer:

a. 8.9Ω

b. 1.2×104 C

Explanation:

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2 years ago

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Tarzan swings on a 26.2 m long vine initially inclined at an angle of 28° from the vertical. (a) What is his speed at the bottom

Marianna [84]

Answer

given,

length of the swing = 26.2 m

inclined at an angle = 28°

let, the initial height of the Tarzan be h

h = L (1 - cos θ)

a) initial velocity v₁ = 0 m/s

final velocity of Tarzan = v_f

law of conservation of energy

PE_i + KE_i = PE_f + KE_f

$mgh_i + \dfrac{1}{2}mv_i^2= mgh_f + \dfrac{1}{2}mv_f^2$

$mgh_i + 0 = 0 + \dfrac{1}{2}mv_f^2$

$mgh_i = \dfrac{1}{2}mv_f^2$

$v_f = \sqrt{2gh_i}$

$= \sqrt{2gL(1- cos\theta)}$

$= \sqrt{2\times 9.8 \times 26.2(1- cos 28^0)}$

= 7.75 m/s

the speed tarzan at the bottom of the swing

v_f = 7.75 m/s

b)initial speed of the = 3 m/s

$mgh_i + \dfrac{1}{2}mv_i^2= mgh_f + \dfrac{1}{2}mv_f^2$

$mgh_i + 0 = 0 + \dfrac{1}{2}mv_f^2$

$mgh_i+ \dfrac{1}{2}mv_i^2 = \dfrac{1}{2}mv_f^2$

$gh_i+ \dfrac{1}{2}v_i^2 = \dfrac{1}{2}v_f^2$

$v_f = \sqrt{v_1^2+2gh_i}$

$v_f = \sqrt{3^2+2\times 9.8 \times (1- cos 28^0)}$

v_f= 11.29 m/s

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3 years ago