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Vikki [24]
3 years ago
6

The higher the value of the coefficient of friction, the _______ the resistance to sliding.

Physics
2 answers:
Tom [10]3 years ago
7 0
<span>The higher the value of the coefficient of friction, the more the resistance to sliding.  The answer is the more the resistance to sliding.  The</span> coefficient of friction<span> is a measure of how easily one object moves over another object. It is a ratio of: Force to move the object / weight of the object (or Normal Force)</span>
MrRa [10]3 years ago
5 0

Answer:

More APEX Verified

Explanation:

You might be interested in
Describe an experiment to demonstrate the principles of the parallelogram of forces
Tcecarenko [31]

Answer:

There are 4 forces because a parallelogram has 4 sides.

It is a quadilateral. Let me know if I am right or not.

5 0
3 years ago
A sled slides down a hill 42 meters high with a slope of 27 o. The total mass of the sled plus contents is 256 kg. It starts fro
kkurt [141]

Answer:

v = 28.7 m/s

Explanation:

  • According to the Work Energy Theorem, the work done by external forces on the system, is equal to the change in the kinetic energy of the system.
  • In absence of friction, the only force that does work , producing a displacement in the direction of the force, is the component of gravity parallel to the slide:

       Fg_{p } = m*g* sin \theta (1)

  • The displacement d,along the slide, can be found from the definition of the sine of an angle, as follows:

        sin \theta =\frac{h}{d}

  • Solving for d, we get:

       d = \frac{h}{sin \theta} (2)

  • Now, the work done by Fgp, is just the product of the force times the displacement, as follows:

        W = Fgp * d

  • From (1) and (2) we can find W as follows:

       W =Fg_{p }* d = m*g* sin \theta * \frac{h}{sin \theta} = m*g*h

  • This expression must be equal to ΔK, as follows:

        \Delta K = \frac{1}{2} * m *v^{2} = m*g*h

  • Simplifying common terms, we can solve for v (the velocity of the sled at the bottom of the slide), as follows:

        v =\sqrt{2*g*h} = \sqrt{2*9.8 m/s2*42 m} = 28.7 m/s

  • The velocity of the sled at the bottom of the slide is 28.7 m/s, taking as positive the direction down the slide.
7 0
3 years ago
An electric heater carries a current of 13.5 A when operating at a voltage of 120V . What is the resistance of the heater?
Aloiza [94]

An electric heater that carries a current of 13.5 A when operating at a voltage of 120V , will have  the resistance of the heater as 8.89 ohm.

<h3>What is Ohm's law?</h3>

Ohm's law  can be described as the law in physics that   focus on how voltage or potential difference between two points is directly proportional to the current or electricity that is moving in the direction of the resistance.

An this is also  directly proportional to the resistance of the circuit, hence it can be calculated using the Ohm's law is V=IR.

from V=IR

R=V/I

= 120/13.55

=8.89 ohm.

Therefore,  electric heater that carries a current of 13.5 A when operating at a voltage of 120V , will have  the resistance of the heater as 8.89 ohm.

Learn more about ohms law on:

brainly.com/question/231741

#SPJ4

4 0
1 year ago
Burl the painter weighs 600 Newtons. He stands in the exact middle of his stage. There are two scale readings attached to the tw
chubhunter [2.5K]

Answer:

T = 300 N

Explanation:

As we know that the painter is standing at the middle of the scaffold

So here the two ropes will exert same tension on the scaffold

So we will have

T_1 + T_2 = W

also we know

T_1 = T_2

so we will have

2T = W

T = \frac{W}{2}

so tension is half of the weight in both the strings

So it is

T = 300 N

6 0
3 years ago
you nose out another runner to win 100.000 m dash. if your total time for the race was 13.800 s and you aced out the other runne
sammy [17]

Your average speed was

(100 m) / (13.8 s) = 7.25 m/s .

If you finished 0.001s ahead of him, then at your average speed, that corresponds to

(7.25 m/s) x (0.001 s) = 0.00725 m

That's 7.25 millimeters ... about 0.28 of an inch !

NOTE:. I think this is only valid if your speed was a constant ~7.25 m/s all the way.

3 0
3 years ago
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