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VashaNatasha [74]
3 years ago
10

The picture given here shows part of a skating track inside a campus. Children who are roller-skating come from both directions

as shown by the arrows. Children coming in opposite directions often bang into each other at some point(s) because they are not able to see through the thick bushes and trees on either side of the path. At which point(s) is this MOST likely to happen?

Physics
1 answer:
Savatey [412]3 years ago
3 0

Answer: 7

Explanation: 7 is the superior number

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Craters flat land or rock
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Explain one way the water cycle affects climate. Use complete sentences.
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When water vaporizes into the air, it becomes humid out.
8 0
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A cat walks along a plank with mass M= 6.00 kg. The plank is supported by two sawhorses. The center of mass of the plank is a di
dimulka [17.4K]

Answer:

d₂ = 1.466 m

Explanation:

In this case we must use the rotational equilibrium equations

        Στ = 0

        τ = F r

we must set a reference system, we use with origin at the easel B and an axis parallel to the plank , we will use that the counterclockwise ratio is positive

      + W d₁ - w_cat d₂ = 0

      d₂ = W / w d₁

      d₂ = M /m d₁

      d₂ = 5.00 /2.9    0.850

      d₂ = 1.466 m

6 0
4 years ago
A 1.35 ×103 kg car rounds a circular turn of radius 11.9 m. The acceleration of gravity is 9.81 m/s 2 . If the road is flat and
mihalych1998 [28]

Answer:

velocity = 8.84 m/s

Explanation:

given data

radius r =  11.9 m

acceleration of gravity g = 9.81 m/s²

static friction μ = 0.67

to find out

how fast can the car go without skidding

solution

we get here velocity that is car go without skidding is express as

force = \frac{mv^2}{r}

force = μ × mg

μ × mg = \frac{mv^2}{r}

v² = μ × r × g

v = \sqrt{\mu *r*g}    ..................1

v = \sqrt{0.67 *11.9*9.81}

v = 8.84 m/s

velocity = 8.84 m/s

3 0
3 years ago
Problem:
kenny6666 [7]

Answer:

a) x = 1.5 *10⁻⁴cos(524πt) m

b) v = -1.5 *10⁻⁴(524π)sin(524πt) m/s

   a =  -1.5 *10⁻⁴(524π)²cos(524πt) m/s²

c) x(1) = 1.5 *10⁻⁴ m = 1.5 *10⁻1 mm

x(0.001) = -1.13*10⁻⁵ m = -1.13*10⁻² mm

Explanation:

x = Acos(ωt)

ω = 2πf = 2π(262) = 524π rad/s

x = 1.5 *10⁻⁴cos(524πt)

v = y' = -Aωsin(ωt)

v = -1.5 *10⁻⁴(524π)sin(524πt)

a = v' = -Aω²cos(ωt)

a =  -1.5 *10⁻⁴(524π)²cos(524πt)

not sure about the last part as time is generally not given in mm

I will show at 1 second and at 0.001 s to try to cover bases

x(1) = 1.5 *10⁻⁴cos(524π(1))

x(1) = 1.5 *10⁻⁴cos(524π)

x(1) = 1.5 *10⁻⁴(1)

x(1) = 1.5 *10⁻⁴ m = 1.5 *10⁻1 mm

x(0.001) = 1.5 *10⁻⁴cos(524π(0.001))

x(0.001) = 1.5 *10⁻⁴cos(0.524π)

x(0.001) = 1.5 *10⁻⁴(-0.0753268)

x(0.001) = -1.129902...*10⁻⁵ m

x(0.001) = -1.13*10⁻⁵ m = -1.13*10⁻² mm

7 0
3 years ago
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