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zhuklara [117]
3 years ago
7

A block with initial velocity of 3 m/s slides 9 m across a rough horizontal surface before coming to rest. What is the coefficie

nt of kinetic
friction?
Physics
1 answer:
Yakvenalex [24]3 years ago
6 0

The coefficient of kinetic friction between the block and the horizontal surface is 0.051.

The given parameters;

  • <em>initial velocity, u = 3 m/s</em>
  • <em>distance traveled by the block, s = 9 m</em>

The acceleration of the block is calculated as follows;

v^2 = u^2 +  2(-a)s\\\\v^2 = u^2 - 2as\\\\a = \frac{u^2 - v^2}{2s} \\\\a = \frac{3^2 - 0^2}{2\times 9} \\\\a =0.5 \ m/s^2

The coefficient of kinetic friction between the block and the horizontal surface is calculated as follows;

\mu_k mg = ma\\\\\mu_k = \frac{a}{g} \\\\\mu_k = \frac{0.5}{9.8} \\\\\mu_k = 0.051

Learn more here:brainly.com/question/19454558

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Answer:

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Explanation:

F=(k)(q1q2/r^2)

F=(8.99×10^9)(3×10^-6)(2×10^-6)/(5^2)

F=0.0021576N

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2 years ago
A vector A⃗ has a magnitude of 40.0 m and points in a direction 20.0∘ below the positive x axis. A second vector, B⃗, has a magn
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The magnitude of the vector C is 96.32m

<h3>How to solve for the magnitude of vector c</h3>

Ax = AcosθA

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Bx = B cos θ B

= 75Cos50

= 48.21

By = BsinθB

= 75sin50

= 57.45

Cx = AX + Bx

= 37.59 + 48.21

= 85.8

Cy = Ay + By

= -13.65 + 57.45

= 43.77

The magnitude is solved by

|c| = \sqrt{Cx^{2}+Cy^{2}  }

= √85.8² + 43.77²

= 96.32m

The magnitude of the vector c is 96.32m

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6 0
2 years ago
At the circus, the Human Cannonball is
svlad2 [7]

The correct answer is  1.4285714.

In physics, velocity is characterised as a vector measurement of the motion's direction and speed. To be more precise, the rate of change in an object's position relative to a frame of reference and time is another way to describe velocity. The definition of velocity simply states the rate of motion of an object in a specific direction. It determines how quickly or slowly something is going.

Velocity = distance/ time  

Thus time = distance/velocity

Here velocity = 350m/s

diatnce = 500 m

time = 500/350

time = 1.42857142857

t= 200m /350m/s  = 1.4285714

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6 0
1 year ago
A 70kg man is moving with a speed of 3.2 m/s on a rough surface, his speed was decreasing uniformly until he reaches to a stop b
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Answer:

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Explanation:

6 0
3 years ago
a volleyball is hit upward with an initial velocity of 7.5 m/s. calculate the displacement of the volleyball when its final velo
Luden [163]

Answer:

The displacement of the volleyball is 2.62 m

Explanation:

Given;

initial velocity of the volleyball, u = 7.5 m/s

final velocity of the volleyball, v = 2.2 m/s

displacement of the volleyball, d = ?

Apply the following kinematic equation;

v² = u² - 2gd

2gd = u² - v²

d = \frac{u^{2}-v^{2}  }{2g}\\\\d = \frac{7.5^{2}-2.2^{2}  }{2*9.8}\\\\d = 2.62 \ m

Therefore, the displacement of the volleyball is 2.62 m

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3 years ago
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