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fomenos
4 years ago
11

An avalanche takes place when ice and snow move down a slope with great speed. What effect does friction have on its rate of mov

ement? A. Friction causes the avalanche to move rapidly. B. Friction creates a curved surface that decreases the speed. C. The speed of the avalanche decreases with increased friction. D. The speed of the avalanche increases with increased friction. E. Friction does not have any effect on the avalanche.
Physics
2 answers:
vredina [299]4 years ago
5 0

The answer is; D

The friction causes the ice at the base of the avalanche to melt into water (also due to the pressure of the weight of the ice-rock above). The melted water acts as a lubricant hence reducing the drag/friction in the avalanche movement. This increases its speed down the slope hence making it more destructive.  

Mazyrski [523]4 years ago
4 0

Answer:

the answer is A

Explanation:

hope it helps :D

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A block is given a very brief push up a 20.0 degree frictionless incline to give it an initial speed of 12.0 m/s.(a) How far alo
Orlov [11]

Explanation:

(a)   Net force acting on the block is as follows.

           F_{net} = -mg Sin (\theta)

or,           ma = -mg Sin (\theta)[/tex]

                 a = -g Sin (\theta)

                    = -9.8 \times Sin (20^{o})

                    = -3.35 m/s^{2}

According to the kinematic equation of motion,

             v^{2} - v^{2}_{o} = 2as

Distance traveled by the block before stopping is as follows.

     s = \frac{v^{2} - v^{2}_{o}}{2a}

        = \frac{(0)^{2} - (12.0)^{2}_{o}}{2 \times -3.35}

        = 21.5 m

According to the kinematic equation of motion,

               v = v_{o} + at

      0 = 12.0 m/s + \frac{1}{2} \times -3.35 m/s^{2} \times t

   t_{1} = 7.16 sec

Therefore, before coming to rest the surface of the plane will slide the box till 7.16 sec.

(b)    When the block is moving down the inline then net force acting on the block is as follows.

                 F_{net} = -mg Sin (\theta)

                ma = mg Sin (\theta)

                    a = g Sin (\theta)

                       = 9.8 m/s^{2} \times Sin (20^{o})

                       = 3.35 m/s^{2}

Kinematics equation of the motion is as follows.

                   s = v_{o}t + \frac{1}{2}at^{2}

      21.5 m = 0 + \frac{1}{2} \times 3.35 m/s^{2} \times t^{2}

     t_{2} = \sqrt{\frac{2 \times 21.5 m}{3.35 m/s^{2}}}

             = 3.58 sec

Hence, total time taken by the block to return to its starting position is as follows.

               t = t_{1} + t_{2}

                 = 7.16 sec + 3.58 sec

                 = 10.7 sec

Thus, we can conclude that 10.7 sec time it take to return to its starting position.

3 0
4 years ago
A certain piece of metal (density 9.45 grams per cubic centimeter) has the shape of a hockey puck with a diameter of 13 cm and a
atroni [7]

Answer:

0.0195 m

Explanation:

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d_{p} = diameter of hockey puck = 13 cm = 0.13 m

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\rho _{m} = density of mercury = 13.6 gcm⁻³ = 13600 kgm³

d = depth of puck below surface of mercury

According to Archimedes principle, the weight of puck is balanced by the weight of mercury displaced by puck

Weight of mercury displaced = Weight of puck

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In order to use the Pythagorean theorem to find the magnitude of a resultant vector, which must be true regarding the two initia
kirill115 [55]
We know that the Pythagorean theorem applies only to right triangles, therefore the original vectors must be parallel to the legs of a right triangle.

In other words, they must be orthogonal (i.e. perpendicular to each other) in order that the Pythagorean theorem applies.

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Answer:

4. To the east

Explanation:

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Answer:

Explanation:

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