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Anon25 [30]
4 years ago
14

A sprinter in a 100-m race accelerates uniformly for the first 71 m and then runs with constant velocity. The sprinter’s time fo

r the first 71 m is 9.9 s. Determine his acceleration.
Physics
1 answer:
adoni [48]4 years ago
3 0

Answer:

The acceleration of the sprinter is 1.4 m/s²

Explanation:

Hi there!

The equation of position of the sprinter is the following:

x = x0 + v0 · t + 1/2 · a · t²

Where:

x = position of the sprinter at a time t.

x0 = initial position.

v0 = initial velocity.

t = time.

a = acceleration.

Since the origin of the frame of reference is located at the starting point and the sprinter starts from rest, then, x0 and v0 are equal to zero:

x = 1/2 · a · t²

At t = 9.9 s, x = 71 m

71 m = 1/2 · a · (9.9 s)²

2 · 71 m / (9.9 s)² = a

a = 1.4 m/s²

The acceleration of the sprinter is 1.4 m/s²

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A model engine accelerated forward from rest along a straight track at 10.0 m/s2 for 3.0 seconds. It then accelerates coward at
Mrac [35]

Answer:

17.2 seconds

Explanation:

Given:

v₀ = 0 m/s

a₁ = 10.0 m/s²

t₁ = 3.0 s

a₂ = 16 m/s²

t₂ = 5.0 s

a₃ = -12 m/s²

v₃ = 0 m/s

Find: t

First, find v₁:

v₁ = a₁t₁ + v₀

v₁ = (10.0 m/s²) (3.0 s) + (0 m/s)

v₁ = 30 m/s

Next, find v₂:

v₂ = a₂t₂ + v₁

v₂ = (16 m/s²) (5.0 s) + (30 m/s)

v₂ = 110 m/s

Finally, find t₃:

v₃ = a₃t₃ + v₂

(0 m/s) = (-12 m/s²) t₃ + (110 m/s)

t₃ = 9.2 s

The total time is:

t = t₁ + t₂ + t₃

t = 3.0 s + 5.0 s + 9.2 s

t = 17.2 s

Round as needed.

4 0
3 years ago
Because the silt (dirt particles) in muddy water eventually settles out, the muddy water is a ______________ .
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Because the silt (dirt particles) in muddy water eventually settles out, the muddy water is a __suspension__ .
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4 years ago
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True or False: The only type of satellites in Earth's orbit are natural satellites.
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3 years ago
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5.
Kamila [148]

Answer:  (x - 4)² + (y - 7)² = 9

<u>Explanation:</u>

The equation of a circle is: (x - h)² + (y - k)² = r²    where

  • (h, k) is the center
  • r is the radius

Given: (h, k) = (4, 7)

Find the intersection of the given equation and the perpendicular passing through (4, 7).

3x - 4y = -1

     -4y = -3x - 1

        y=\dfrac{3}{4}x-1

              m=\dfrac{3}{4}      -->      m_{\perp}=-\dfrac{4}{3}

y-y_1=m_{\perp}(x-x_1)\\\\y-7=-\dfrac{4}{3}(x-4)\\\\\\y=-\dfrac{4}{3}x+\dfrac{16}{3}+7\\\\\\y=-\dfrac{4}{3}x+\dfrac{37}{3}

Use substitution to find the point of intersection:

x=\dfrac{29}{5}=5.8,\qquad y=\dfrac{23}{5}=4.6

Use the distance formula to find the distance from (4, 7) to (5.8, 4.6) = radius

r=\sqrt{(5.8-4)^2+(4.6-7)^2}\\\\r=\sqrt{3.24+5.76}\\\\r=\sqrt9\\\\r=3

Input h = 4, k = 7, and r = 3 into the circle equation:

(x - 4)² + (y - 7)² = 3²

(x - 4)² + (y - 7)² = 9

4 0
4 years ago
Please help with any of them!!!!!!
Law Incorporation [45]
Its to blury for me to see it 
5 0
4 years ago
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