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kaheart [24]
3 years ago
6

Two runners start simultaneously from the same point on a circular 200-m track and run in the same direction. One runs at a cons

tant speed of 6.20 m/s and the other runs at a constant speed of 5.50 m/s.When will the fast one first overtake ("lap") the slower one?How far from the starting point will each have run at that instant?When will the fast one overtake the slower one for the second time?How far from the starting point will they be at that instant?
Physics
1 answer:
sladkih [1.3K]3 years ago
5 0

Answer

given,

length of the track = 200 m

speed of runner 1 = 6.20 m/s

speed of the runner 2 = 5.5 m/s

distance covered by the fastest runner, d = 6.2 t

distance covered by the slow runner,d = 5.50 t + 200

time taken to overtake for the first time

 6.2 t = 5.50 t + 200

  0.7 t = 200

       t = 285.7 s

distance covered from the starting point

 d = 6.2 t

 d = 6.2 x 285.7

 d = 1771.43 m

for next overtaking time taken will be double

 t' = 258.7 x 2 = 571.4 s

distance travel by the runner from the starting point

d' = 6.2 t'

d' = 6.2 x 571.4

d' = 3542.68 m

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Pavlova-9 [17]

Answer:

12m

Explanation:

To obtain the answer to the question given, we must observe the characteristics of image formed by a plane mirror.

The image formed by a plane mirror have the following characteristics:

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3. Same height as the object.

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With the above information, we can calculate the distance between the boy and his image as follow:

Initially:

Object distance (u) = 4m

Image distance (v) = 4m

The boy moved 2m away, therefore:

Object distance (u) = 2 + 4 = 6m

Image distanc(v) = 2 + 4 = 6m

The distance between the boy and his image will be the sum of his distance (u) and image distance (v) i.e (u + v)

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6 0
3 years ago
Two boxes, P and Q, are at rest on a frictionless horizontal surface. A light, flexible cord connects them. The mass of P is gre
elixir [45]

the answer is less than F, but greater than zero


4 0
3 years ago
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You have been hired as a technical consultant for an early-morning cartoon series for children to make sure that the science is
katen-ka-za [31]

The initial potential energy of the wagon containing gold boxes will enable

it roll down the hill when cut loose.

The Lone Ranger and Tonto have approximately <u>5.1 seconds</u>.

Reasons:

Mass wagon and gold = 166 kg

Location of the wagon = 77 meters up the hill

Slope of the hill = 8°

Location of the rangers = 41 meters from the canyon

Mass of Lone Ranger, m₁ = 65 kg

Mass of Tonto m₂ = 66 kg

Solution;

Height of the wagon above the level ground, h = 77 m × sin(8°) ≈ 10.72 m

Potential energy = m·g·h

Where;

g = Acceleration due to gravity ≈ 9.81 m/s²

Potential energy of wagon, P.E. ≈ 166 × 9.81 × 10.72 = 17457.0912

Potential energy of wagon, P.E. ≈ 17457.0912 J

By energy conservation, P.E. = K.E.

K.E. = \mathbf{\dfrac{1}{2} \cdot m \cdot v^2}

Where;

v = The velocity of the wagon a the bottom of the cliff

Therefore;

\dfrac{1}{2} \times 166 \times v^2 = 17457.0912

v = \sqrt{\dfrac{17457.0912}{\dfrac{1}{2} \times 166} } \approx 14.5

Velocity of the wagon, v ≈ 14.5 m/s

Momentum = Mass, m × Velocity, v

Initial momentum of wagon = m·v

Final momentum of wagon and ranger = (m + m₁ + m₂)·v'

By conservation of momentum, we have;

m·v = (m + m₁ + m₂)·v'

\therefore v' = \mathbf{ \dfrac{m \cdot v}{(m + m_1 + m_2)  }}

Which gives;

\therefore v' = \dfrac{166 \times 14.5}{(166 + 65 + 66)  } \approx 8.1

The velocity of the wagon after the Ranger and Tonto drop in, v' ≈ 8.1 m/s

Time = \dfrac{Distance}{Velocity}

\mathrm{The \ time \ the\ Lone \  Ranger \  and  \ Tonto \  have,  \ t} = \dfrac{41 \, m}{8.1 \, m/s} \approx 5.1 \, s

The Lone Range and Tonto have approximately <u>5.1 seconds</u> to grab the

gold and jump out of the wagon before the wagon heads over the cliff.

Learn more here:

brainly.com/question/11888124

brainly.com/question/16492221

5 0
2 years ago
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ankoles [38]

Answer:

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Explanation:

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3 0
3 years ago
A spring stretches by 0.0208 m when a 3.39-kg object is suspended from its end. How much mass should be attached to this spring
sashaice [31]

Answer:

COMPLETE QUESTION

A spring stretches by 0.018 m when a 2.8-kg object is suspended from its end. How much mass should be attached to this spring so that its frequency of vibration is f = 3.0 Hz?

Explanation:

Given that,

Extension of spring

x = 0.0208m

Mass attached m = 3.39kg

Additional mass to have a frequency f

Let the additional mass be m

Using Hooke's law

F= kx

Where F = W = mg = 3.39 ×9.81

F = 33.26N

Then,

F = kx

k = F/x

k = 33.26/0.0208

k = 1598.84 N/m

The frequency is given as

f = ½π√k/m

Make m subject of formula

f² = ¼π² •(k/m

4π²f² = k/m

Then, m4π²f² = k

So, m = k/(4π²f²)

So, this is the general formula,

Then let use the frequency above

f = 3Hz

m = 1598.84/(4×π²×3²)

m = 4.5 kg

4 0
3 years ago
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