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Angelina_Jolie [31]
3 years ago
10

What are some other examples of a force acting on an object without touching it?

Physics
1 answer:
a_sh-v [17]3 years ago
8 0
An example would be gravity
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Which of the following is an example of a properly written testable hypothesis? *
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D. if it is dark, then an owl will find a mouse by the sound the mouse makes

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A mass is placed at the end of a spring. It has starting velocity of V & allowed to oscillate freely. If the mass has a star
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Equation for SHM can be written

V = w A cos w t        where w is the angular frequency and the velocity is a                                         maximum at t = 0

V1 = w1  A cos w1 t

V2 = w2 A cos w2 t

V2 / V1 = w2 / w1     since cos X t = 1 if t = zero

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1. What does a football player need to do to change the position of the football?​
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Yes, since formations aren't mentioned at all in the rules, they can be adjusted. Sometimes when making a substitution, a coach will sub in a defender for an attacker/midfielder if the team is ahead and wants to protect their lead....

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Why is the solar system commonly represented as a scale model?
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A spider begins to spin a web by first hanging from a ceiling by his fine, silk fiber. He has a mass of 0.025 kg and a charge of
Rasek [7]

Answer:

a) (5.59 × 10³) N/C

b) 0.226 N directed away from the spider.

Explanation:

a) Electric field, E, felt as a result of point charge, Q, at a distance, d away is given by

E = kQ/d²

So, magnitude of the electric field due to the charge on the second spider at the position of the first spider

Q = 4.2 µC = 4.2 × 10⁻⁶ C

k = Coulomb's constant = 8.99 × 10⁹ Nm²/C

d = 2.6 m

E = (8.99 × 10⁹ × 4.2 × 10⁻⁶)/2.6²

E = 5.59 × 10³ N/C

b) Tension in the silk fiber above the spider is the net force due to the weight of spider one and the force of repulsion due the two charges.

Force due to the two charges = Eq

where q now represents the charge of the first spider at the first point, feeling the electric field calculated in (a)

F = 5.59 × 10³ × 3.4 × 10⁻⁶ = 0.01901 N directed upwards. (That is, F = + 0.019 N)

Weight of the spider = mg = 0.025 × 9.8 = 0.245 N directed downwards. (That is, W = -0.245 N)

Net force, T = mg - F = 0.245 - 0.019 = 0.226 N (that is, 0.226 N, directed upwards, away from the spider).

8 0
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