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cluponka [151]
2 years ago
7

Is the following sentence true or false? When two surfaces rub, the irreg-ularities of one surface get caught on those of the ot

her surface.
Physics
1 answer:
steposvetlana [31]2 years ago
8 0
This is in fact true!
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NeX [460]
Percentage for which columns btw if its something as easy as percentage for 4/5... you divid 4/5= .80 then.... .80*100= 80%
7 0
3 years ago
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A yoyo with a mass of m = 150 g is released from rest as shown in the figure.
avanturin [10]

(1) The linear acceleration of the yoyo is 3.21 m/s².

(2) The angular acceleration of the yoyo is 80.25 rad/s²

(3) The  weight of the yoyo is 1.47 N

(4) The tension in the rope is 1.47 N.

(5) The angular speed of the yoyo is 71.385 rad/s.

<h3> Linear acceleration of the yoyo</h3>

The linear acceleration of the yoyo is calculated by applying the principle of conservation of angular momentum.

∑τ = Iα

rT - Rf = Iα

where;

  • I is moment of inertia
  • α is angular acceleration
  • T is tension in the rope
  • r is inner radius
  • R is outer radius
  • f is frictional force

rT - Rf = Iα  ----- (1)

T - f = Ma  -------- (2)

a = Rα

where;

  • a is the linear acceleration of the yoyo

Torque equation for frictional force;

f = (\frac{r}{R} T) - (\frac{I}{R^2} )a

solve (1) and (2)

a = \frac{TR(R - r)}{I + MR^2}

since the yoyo is pulled in vertical direction, T = mg a = \frac{mgR(R - r)}{I + MR^2} \\\\a = \frac{(0.15\times 9.8 \times 0.04)(0.04 - 0.0214)}{1.01 \times 10^{-4} \ + \ (0.15 \times 0.04^2)} \\\\a = 3.21 \ m/s^2

<h3>Angular acceleration of the yoyo</h3>

α = a/R

α = 3.21/0.04

α = 80.25 rad/s²

<h3>Weight of the yoyo</h3>

W = mg

W = 0.15 x 9.8 = 1.47 N

<h3>Tension in the rope </h3>

T = mg = 1.47 N

<h3>Angular speed of the yoyo </h3>

v² = u² + 2as

v² = 0 + 2(3.21)(1.27)

v² = 8.1534

v = √8.1534

v = 2.855 m/s

ω = v/R

ω = 2.855/0.04

ω = 71.385 rad/s

Learn more about angular speed here: brainly.com/question/6860269

#SPJ1

3 0
2 years ago
The land between two normal faults moves upward to form a
Leya [2.2K]
<span>The land between two normal faults moves upward to form a

Answer:D</span><span>
fault-block mountain.</span>
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4 years ago
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Help with these questions please , 15 PTS &amp; brainliest
myrzilka [38]

For these two questions, first you need to know that the voltage across each branch of a parallel circuit is the same.

So, for Q5, we can first find out the voltage across R₂ by V=IR.

Voltage across R₂ = 2.5 × 8 = 20V

Since R₂ and R₃ are in parallel circuit, their voltage should be the same. Thus, voltage across R₃ is 20V.

So, by V=IR,

current of R₃ = \frac{20}{4} = 5A

Q6. voltage across R₁ = 2 × 4 = 8V

∴voltage across R₂ = 8V

current of R₂ = \frac{8}{8} = 1A

<h3><u>Alternative method</u></h3>

From these two examples, you can find out that the current of each branch of the parallel circuit is inversely proportional to the resistance of the branch.

ie. for Q5,

\frac{R2}{R3} = \frac{I3}{I2}

\frac{8}{4} = \frac{I3}{2.5}

I₃ = 5A

Q6. \frac{R1}{R2} = \frac{I2}{I1}

\frac{4}{8} = \frac{I2}{2}

I₂ = 1A

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3 years ago
You and your little cousin sit on a see-saw. You sit 0.5 m from the fulcrum, and your cousin sits 1.5 m from the fulcrum. You we
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Answer:

200N

Explanation:

0.5/1.5=x/600N

1/3=x/600

x=200N

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