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Anon25 [30]
3 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]3 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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worty [1.4K]
T = (v-0)/a
t = (45.5)/(9.8)
= 4.64m/s.

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Concave lens cause rays to _____.
Molodets [167]
Diverge or spread apart.
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If Chris drives 27 km N and then 40 km E , what is the distance and displacement
Bas_tet [7]
The answer would be 13.
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3 years ago
An AC power source has an rms voltage of 120 V and operates at a frequency of 60.0 Hz. If a purely inductive circuit is made fro
Dmitrij [34]

Answer:

(a) 17634.24 Ω

(b) 0.0068 A

Explanation:

(a)

The formula for inductive inductance is given as

X' = 2πFL................... Equation 1

Where X' = inductive reactance, F = frequency, L = inductance

Given: F = 60 Hz, L = 46.8 H, π = 3.14

Substitute into equation 1

X' = 2(3.14)(60)(46.8)

X' = 17634.24 Ω

(b)

From Ohm's law,

Vrms = X'Irms

Where Vrms = Rms Voltage, Irms = rms Current.

make Irms the subject of the equation

Irms = Vrms/X'...................... Equation 2

Given: Vrms = 120 V, X' = 17634.24 Ω

Substitute into equation 2

Irms = 120/17634.24

Irms = 0.0068 A

5 0
3 years ago
Two converging lenses are placed 30 cm apart. The focal length of the lens on the right is 20 cm while the focal length of the l
Masja [62]

Answer:

a)   I2 = 3 (o-10) / (o- 30) , b)   h ’/h=  3 (o-10) / o (o-30)

Explanation:

The builder's equation is

          1 / f = 1 / o + 1 / i

Where f is the focal length, or e i are the distance to the object and image, respectively

As the separation between the lenses is greater than the focal distances, we must work them individually and separately. Let's start with the leftmost lens with focal length f = 15 cm

Let's calculate the position of the image of this lens

         1 / i1 = 1 / f - 1 / o

         1 / i1 = 1/15 - 1 / o

         i1 = o 15 / (o-15)

Let's calculate the distance to the image of the second lens, for this the image of the first is the distance to the object of the second

        o2 = d-i1

We write the builder equation

       1 / f2 = 1 / o2 + 1 / i2

       1 / i2 = 1 / f2 -1 / o2

       1 / i2 = 1 / f2 - 1 / (d-i1)

       1 / i2 = 1/20 - 1 / (d-i1)            (1)

Let's evaluate the last term

      d-i1 = d - 15 o / (o-15)

      d-i1 = (d (o-15) - 15 o) / (o-15)

      d- i1 = (30 or -30 15 -15 o) / (o-15)

      d-i1 = (15 or - 450) / (o- 15)

      d-i1 = = (15 or -450) / (o-15)

replace in 1

      1 / i2 = 1/20 - (or - 15) / (15 or -450)

      1 / i2 = [(15 o-450) - (o-15) 20] / (15 or -150)

      1 / i2 = (15 or - 450 - 20 or + 300) / (15 or - 150)

      1 / i2 = (-5 or -150) / (15 or -150)

      1 / i2 = (or -30) / (3 or - 30)

      I2 = 3 (o-10) / (o- 30)

Part B

The height of the image, we use the magnification equation

     m = h ’/ h = - i / o

     h ’= - h i / o

In our case

     h ’= h i2 / o

     h ’= h 3 (o-10) / o (o-30)

If they give the distance to the object it is easier

5 0
3 years ago
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