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3 years ago

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5. Did objects have to touch to interact? What causes this?

2 answers:

dimulka [17.4K]3 years ago

6 0

Answer:

★ For example, a useful analogy for explaining the Earth's gravity force is that the Earth can pull on objects without touching them just like a magnet can affect other objects without touching them. In Addition, the main notion to convey here is that forces can act at a distance with no perceivable substance in between.

Explanation:

Hope you have a great day :)

djverab [1.8K]3 years ago

5 0

Answer/Explanation

A useful analogy for explaining the Earth's gravity force is that the Earth can pull on objects without touching them just like a magnet can affect other objects without touching them. The main notion to convey here is that forces can act at a distance with no perceivable substance in between.

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A 350-N child is in a swing that is attached to a pair of ropes 2.10 m long. Find the gravitational potential energy of the chil

o-na [289]

Answer:

a) U = 735 J , b) U = 125.7 J , c) U = 0 J

Explanation:

The gravitational power energy is

U = mg y - mg y₀

The last value is a constant, for simplicity we can make it zero, if the lowest point is at the origin of the coordinate system, which in this case we will place in the lowest part

a) Rope is horizontal

The height in this case is the same length of the rope

y = 2.10 m

w = mg = 350 N

U = 350 2.10

U = 735 J

b) when the angle is 34º

y = L - L cos 34

y = L (1- cos34)

y = 2.10 (1- cos 34)

y = 0.359 m

U = 350 0.359

U = 125.7 J

c) in this case this point coincides with the reference system

y = 0

U = 0 J

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3 years ago

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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Answer:

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Explanation:

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3 years ago

What are the three basic parts of a circuit?

Nat2105 [25]

Answer: energy source, path, and load

Explanation:

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3 years ago

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A ball is thrown at a 60.0° angle above the horizontal across level ground. It is thrown from a height of 2.00 m above the groun

klio [65]

Answer:

Option E is correct.

Time the ball remains in the air before striking the ground is closest to 3.64 s

Explanation:

yբ = yᵢ + uᵧt + gt²/2

yբ = 0

yᵢ = 2 m

uᵧ = u sinθ = 20 sin 60 = 17.32 m/s

g = -9.8 m/s², t = ?

0 = 2 + 17.32t - 4.9t²

4.9t² - 17.32t - 2 = 0

Solving the quadratic equation,

t = 3.647 s or t = -0.1112 s

time is a positive variable, hence, t = 3.647 s. Option E.

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2 years ago