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Genrish500 [490]
3 years ago
7

A suspicious-looking man runs as fast as he can along a moving sidewalk from one end to the other, taking 2.50 s. Then security

agents appear, and the man runs as fast as he can back along the sidewalk to his starting point, taking 13.0 s. What is the ratio of the man's running speed to the sidewalk's speed?
Physics
2 answers:
Zigmanuir [339]3 years ago
6 0
13:9 because he starts back then runs slower
snow_tiger [21]3 years ago
6 0

Answer:

The ratio of the man's running speed to the sidewalk's speed: 5: 26.

Explanation:

Let the distance between the sidewalk from one end to the other be x

Distance covered by the man = x

Time taken by the person to cover x distance = 2.50 seconds

Speed of the man while running to other end point = S

S=\frac{x}{2.50 s}..[1]

Distance covered by the man by running back to his starting point after agents appears = x

Time taken by the person to cover x distance after agents appears = 13.0 seconds

Speed of the man while running back to starting point after agents appears = S'

S'=\frac{x}{13.0 s}...[2]

The ratio of the man's running speed to the sidewalk's speed:

[2] ÷ [1]

\frac{S'}{S}=\frac{\frac{x}{13.0 s}}{\frac{x}{2.50 s}}=\frac{5}{26}

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2 years ago
Two ropes are attached to either side of a 100.0 kg wagon as shown below. The rope on the right is pulled at an angle 40.0° rela
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The acceleration of the wagon is found by applying Newton's Second Law of motion.

1. The responses for question 1 are;

  • x-component of the tension in the rope on the right is approximately <u>91.93 N</u>
  • y-component of the tension in the rope on the right is approximately <u>71.135 N</u>
  • x-component of the tension in the rope on the left is -80.0 N
  • y-component of the tension in the rope on the left is 0

2. The net force in the x-direction is approximately <u>11.93 N</u>

3. The net acceleration of the wagon in the horizontal direction is approximately <u>0.1193 m/s²</u>.

Reasons:

The given parameters are;

Mass of the wagon, m = 100.0 kg

Angle of inclination to the horizontal of the rope to the right, θ = 40.0°

Tension in the rope on the right = 120.0 N

Direction in which the rope on the left is pulled = To the west

Tension in the rope on the left = 80.0 N

1. The <em>x</em> and <em>y</em> component of the tension in the rope on the right are;

x-component = 120.0 N × cos(40.0°) ≈ <u>91.93 N</u>

y-component = 120.0 N × sin(40.0°) ≈ <u>77.135 N</u>

The <em>x</em> and <em>y</em> component of the tension in the rope on the left are;

x-component = 80.0 N × cos(180°) = <u>-80.0 N</u>

y-component = 80.0 N × sin(180°) = <u>0.0 N</u>

2. The net force in the horizontal direction, Fₓ, is found as follows;

Fₓ = The x-component of the rope on the left + The x-component of the rope on the right

Which gives;

Fₓ = 91.93 N - 80.0 N = <u>11.93 N</u>

3. The net acceleration of the block is given as follows;

According to Newton's Second Law of motion, we have;

Force in the horizontal direction, Fₓ = Mass of wagon, m × Acceleration of the wagon in the horizontal direction, aₓ

Fₓ = m × aₓ

Therefore;

\displaystyle a_x = \frac{F_x}{m}  \approx \frac{11.93 \, N}{100.0 \, kg} = \mathbf{0.1193 \ m/s^2}

  • The acceleration of the wagon in the horizontal direction, aₓ ≈ <u>0.1193 m/s²</u>.

Learn more here:

brainly.com/question/20357188

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When a certain element is excited with electricity, we see three main lines in its emission spectrum: two red lines and one oran
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The absorption spectrum would have all the wavelengths of the light source but would have black lines where the two red and one orange lines were in the emission spectrum
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Tapping the surface of a pan of water generates 17.5 waves per second. If the wavelength of each wave is 45 cm, what is the spee
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Answer:

Speed of the wave is 7.87 m/s.

Explanation:

It is given that, tapping the surface of a pan of water generates 17.5 waves per second.

We know that the number of waves per second is called the frequency of a wave.

So, f = 17.5 Hz

Wavelength of each wave, \lambda=45\ cm=0.45\ m

Speed of the wave is given by :

v=f\lambda

v=17.5\times 0.45

v = 7.87 m/s

So, the speed of the wave is 7.87 m/s. Hence, this is the required solution.

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