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Andreyy89
3 years ago
10

WHAT IS THE UNBALANCED FORCE THAT SLOWS DOWN SLIDING DESKS AND MOVING CARS

Physics
2 answers:
gogolik [260]3 years ago
6 0
Friction is a forces that slows things down.
umka2103 [35]3 years ago
3 0
Friction slows it down
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When sugar is poured from the box into the sugar bowl, the rubbing of sugar grains creates a static electric charge that repels
Afina-wow [57]

Answer:

2.6×10⁻³ N

Explanation:

From coulomb's law,

F = kq'q/r²................ Equation 1

Where F = Repulsive force, q' = charge on the first sugar grain, q = charge on the second sugar grain, r = distance of separation between the sugar grain, k = proportionality constant.

From the question,

since q' = q

Then,

F = kq²/r²..................... Equation 2

Given: q = 1.79×10⁻¹¹ C, r = 3.45×10⁻⁵ m,

Constant: k = 9×10⁹ Nm²/kg².

Substitute into equation 2

F = 9×10⁹(1.79×10⁻¹¹)²/(3.45×10⁻⁵ )²

F = 9×10⁹(3.2041×10⁻²²)/(11.9025×10⁻¹⁰)

F = (28.8369×10⁻¹³)/(11.9025×10⁻¹⁰)

F = 2.6×10⁻³ N.

3 0
3 years ago
You push a cart with mass 15 kg forward , giving it an acceleration of 3 m/s ^ 2 . How much force did you apply ? O A. 0.2N O B.
Alexandra [31]

Answer: force = mass x accelegation

Explanation: 15x3 = 45 N

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The change of energy from one type to another; for example, kinetic to potential, or kinetic
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Answer:Broadly speaking, all energy in the universe can be categorized as either potential energy or kinetic energy. Potential energy is the energy associated with position, like a ball held up in the air. When you let go of that ball and let it fall, the potential energy converts into kinetic energy, or the energy associated with motion.

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3 years ago
A rugby player sits on a scrum machine that weighs 200 Newtons. Given that the coefficient of static friction is 0.67, the coeff
Trava [24]

a. 850 N is the minimum force needed to get the machine/player system moving, which means this is the maximum magnitude of static friction between the system and the surface they stand on.

By Newton's second law, at the moment right before the system starts to move,

• net horizontal force

∑ F[h] = F[push] - F[s. friction] = 0

• net vertical force

∑ F[v] = F[normal] - F[weight] = 0

and we have

F[s. friction] = µ[s] F[normal]

It follows that

F[weight] = F[normal] = (850 N) / (0.67) = 1268.66 N

where F[weight] is the combined weight of the player and machine. We're given the machine's weight is 200 N, so the player weighs 1068.66 N and hence has a mass of

(1068.66 N) / g ≈ 110 kg

b. To keep the system moving at a constant speed, the second-law equations from part (a) change only slightly to

∑ F[h] = F[push] - F[k. friction] = 0

∑ F[v] = F[normal] - F[weight] = 0

so that

F[k. friction] = µ[k] F[normal] = 0.56 (1268.66 N) = 710.45 N

and so the minimum force needed to keep the system moving is

F[push] = 710.45 N ≈ 710 N

4 0
2 years ago
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