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likoan [24]
3 years ago
9

Whenever you push an object across a frictional surface, it starts out taking a lot of force to push to get the object moving. H

owever, once the object starts moving, it is much easier to keep it moving. What does this tell you about one of the main differences between static and kinetic friction?
a.Static friction only depends on the surface material, while kinetic friction only depends on the motion

b Kinetic and Static friction are the exact same thing

c Static friction and kinetic friction are combined in one at all times, but static friction is just not seen during the motion

d Static Friction is much stronger than kinetic friction, so it takes more force to get an object moving than it does to keep it moving
Physics
1 answer:
son4ous [18]3 years ago
5 0

Answer:

The answer is A

Explanation:

Here's an example. A child is in school taking a test. They have made a mistake on a question, and want to erase it. The eraser is made out of a type of rubber, the rubber has friction, which means the eraser has something that's going to resist movement. Now the child has exerted enough force to get it moving, and it's moving, it won't stop unless the child stops exerting force to keep it moving. Both Newton's 1st and 3rd law explain the action of moving something on a surface with friction.

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In which regions can the gravitational field strength due to the two planets be zero? Explain.
yan [13]

Answer:

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3 years ago
The lawn sprinkler consists of four arms that rotate in the horizontal plane. The diameter of each nozzle is 8 mm, and the water
sashaice [31]

Answer: the constant angular velocity of the arms is 86.1883 rad/sec

Explanation:

First we calculate the linear velocity of the single sprinkler;

Area of the nozzle = π/4 × d²

given that d = 8mm = 8 × 10⁻³

Area of the nozzle = π/4 × (8 × 10⁻³)²

A = 5.024 × 10⁻⁵ m²

Now total discharge is dived into 4 jets so discharge for single jet will be;

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So using continuity equation ;

Q_single = A × V_single

V_single = Q_single/A

we substitute

V_single = (1.5 × 10⁻³) / (5.024 × 10⁻⁵)

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Now resolving the forces as shown in the second image,

Vt = Vcos30°

Vt = 29.8566 × cos30°

Vt = 25.8565 m/s

Finally we calculate the angular velocity;

Vt = rω

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from the given diagram, radius is 300mm = 0.3m

so we substitute

ω_single = 25.8565 / 0.3

ω_single = 86.1883 rad/sec

Therefore the constant angular velocity of the arms is 86.1883 rad/sec

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https://www.omnicalculator.com/physics/specific-heat
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