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sergejj
2 months ago
6

Two swimmers begin a race and Swimmer A completes each length of the pool in 30 seconds, while Swimmer B completes each length i

n 40 seconds. How far ahead of Swimmer B will Swimmer A be after five minutes?
Physics
1 answer:
Maru [3.3K]2 months ago
6 0

Answer: 1.33 times

Explanation:

A completes each pool length in 30 seconds, whereas B takes 40 seconds per length.

Let L represent the length of the pool, with V_a and V_b denoting distances swum by A and B.

After 5 minutes, A will have covered

x_a=\frac{L}{30}\times 5\times 60=10L

and

x_b=\frac{L}{40}\times 5\times 60=7.5L

showing A is ahead of B by 2.5L.

Therefore, A has swum 1.33 times the distance B has in the same 5-minute span.

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When the displacement of a mass on a spring is 1/2a the half of the amplitude, what fraction of the mechanical energy is kinetic
ValentinkaMS [3465]
Total energy associated with a spring:
E = \frac{1}{2} kx^2 + \frac{1}{2} mv^2 = \frac{1}{2} ka^2

When x = 0.5a:
\frac{1}{2} k \frac{a}{2} ^2 + \frac{1}{2} mv^2 = \frac{1}{2} ka^2 \\ \frac{1}{2} mv^2 = \frac{1}{2} ka^2 - \frac{1}{8} ka^2 = \frac{3}{8} ka^2

The ratio:
\frac{ \frac{3}{8}ka^2 }{ \frac{1}{2} ka^2} = \frac{3}{4}
5 0
19 days ago
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Keith_Richards [3271]
The ball covers a horizontal distance of 0.902 meters. The trajectory of a kicked football adheres to a quadratic equation expressed as: f(x), where f(x) indicates the vertical distance in feet, and x signifies how far the ball travels horizontally. To compute the distance the ball will advance before striking the ground, we set the condition f(x) = 0. Upon solving this quadratic equation, we find that the horizontal distance traveled by the ball is: x = -0.902 meters, leading us to conclude that it travels 0.902 meters across the field.
7 0
28 days ago
What mass needs to be attached to a spring with a force constant of 7N/m in order to make a simple harmonic oscillator oscillate
Sav [3153]

Answer:

The mass will be 4.437 kg

Explanation:

The force constant k is given as 7 N/m

The time period of oscillation T is 5 sec

Thus, angular frequency \omega =\frac{2\pi }{T}=\frac{2\times 3.14}{5}=1.256rad/sec

It is known that angular frequency is computed via

\omega =\sqrt{\frac{k}{m}}

1.256 =\sqrt{\frac{7}{m}}

Squaring both sides gives us

1.577 =\frac{7}{m}

The mass equals 4.437 kg

6 0
22 days ago
A soft drink (mostly water) flows in a pipe at a beverage plant with a mass flow rate that would fill 220 cans, 0.355 - l each,
Keith_Richards [3271]
Flow rate calculations yield 220 cans, each with a volume of 0.355 l, leading to 78.1 l/min or 1.3 l/s or 0.0013 m³/s.

At Point 2:
A2 = 8 cm² = 0.0008 m²
V2 = Flow rate/A2 = 0.0013/0.0008 = 1.625 m/s
P1 = 152 kPa = 152000 Pa

At Point 1:
A1 = 2 cm² = 0.0002 m²
V1 = Flow rate/A1 = 0.0013/0.0002 = 6.5 m/s
P1 =?
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Using Bernoulli’s principle;
P2 + 1/2 * V2² / density = P1 + 1/2 * V1² / density + density * gravitational acceleration * height
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3 0
1 month ago
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Newton’s law of cooling states that dx dt = −k(x−A) where x is the temperature,t is time, A is the ambient temperature, and k &g
Maru [3345]

Answer:

a) X= kA_o\dfrac{1}{k^2+\omega^2}\left ( kcos\omega t+\omega sin\omega t \right )+Ce^{-kt}

b) D does not influence the long-term results.

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\dfrac{dx}{dt}=-k(x-A)

A = A0 cos(ωt)

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\dfrac{dx}{dt}+kx=kA_o cos(\omega t)This is a linear equation hence the integration factor, I

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I=e^{kt}Now using the characteristics of linear equations

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b) At t= 0

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does not affect the long-term outcome.

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5 0
1 month ago
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