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3 years ago

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Which applied force will allow a 7.65 kg block of ice to begin sliding on a sheet of ice? The block of ice has a kinetic coeffic

ient of friction of 0.030 and a static coefficient of friction of 0.10.

2 answers:

Debora [2.8K]3 years ago

8 0

Answer:

its D 9.0

Explanation:

i got it wrong so you could get it right ;))

Anni [7]3 years ago

7 0

Answer:

force for start moving is 7.49 N

force for moving constant velocity 2.25 N

Explanation:

given data

mass = 7.65 kg

kinetic coefficient of friction = 0.030

static coefficient of friction = 0.10

solution

we get here first weight of block of ice that is

weight of block of ice = mass × g

weight of block of ice = 7.65 × 9.8 = 74.97 N

so here Ff = Fa

so for force for start moving is

Fa = weight × static coefficient of friction

Fa = 74.97 × 0.10

Fa = 7.49 N

and

force for moving constant velocity is

Fa = weight × kinetic coefficient of friction

Fa = 74.97 × 0.030

Fa = 2.25 N

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Answer:

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Explanation:

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The distance in meters traveled by a point outside the rim is calculated as follows;

$\theta = \omega t\\\\\theta = (100 \frac{rev}{\min} \times \frac{2\pi \ rad}{1 \ rev} \times \frac{1\min}{60 s} ) \times (30 s)\\\\\theta = 100 \pi \ rad\\\\d = \theta r\\\\d = 100\pi \ \times \ 0.5m\\\\d = 50 \pi \ m = 157.1 \ m$

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A ball is thrown directly downward with an initial speed of 7.70 m/s, from a height of 30.2 m. After what time interval does it

KatRina [158]

Answer:

$t = 1.82$

Explanation:

Given

$u = 7.70m/s$ -- initial velocity

$s = 30.2m$ --- height

Required

Determine the time to hit the ground

This will be solved using the following motion equation.

$s = ut + \frac{1}{2}gt^2$

Where

$g = 9,8m/s^2$

So, we have:

$30.2 = 7.70t + \frac{1}{2} * 9.8 * t^2$

$30.2 = 7.70t + 4.9 * t^2$

Subtract 30.2 from both sides

$30.2 -30.2 = 7.70t + 4.9 * t^2 - 30.2$

$0 = 7.70t + 4.9 * t^2 - 30.2$

$0 = 7.70t + 4.9t^2 - 30.2$

$7.70t + 4.9t^2 - 30.2 = 0$

$4.9t^2 + 7.70t - 30.2 = 0$

Solve using quadratic formula:

$t = \frac{-b\±\sqrt{b^2 - 4ac}}{2a}$

Where

$a = 4.9;\ b = 7.70;\ c = -30.2$

$t = \frac{-7.70\±\sqrt{7.70^2 - 4*4.9*-30.2}}{2*4.9}$

$t = \frac{-7.70\±\sqrt{651.21}}{9.8}$

$t = \frac{-7.70\±25.52}{9.8}$

Split the expression

$t = \frac{-7.70+25.52}{9.8}$ or $t = \frac{-7.70-25.52}{9.8}$

$t = \frac{17.82}{9.8}$ or $t = -\frac{33.22}{9.8}$

Time can't be negative; So, we have:

$t = \frac{17.82}{9.8}$

$t = 1.82$

Hence, the time to hit the ground is 1.82 seconds

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