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Tema [17]
2 years ago
15

A girl with a mass of 27 kg is playing on a swing. There are three main forces

Physics
1 answer:
N76 [4]2 years ago
3 0

The tension in the swing's chain at the bottom of the swing is 178.35 N.

The given parameters:

  • Mass of the girl, m = 27 kg
  • Speed of the girl, v = 3 m/s
  • Radius of the circle, r = 4 m

The tension in the swing's chain at the bottom of the swing is calculated as follows;

T = mg + ma_c\\\\ T= mg + \frac{mv^2}{r} \\\\T = (12 \times 9.8) + (\frac{27 \times 3^2}{4} )\\\\T = 117.6 \ N \ + \ 60.75 \ N\\\\T = 178.35 \ N

Thus, the tension in the swing's chain at the bottom of the swing is 178.35 N.

Learn more about tension in vertical circle here: brainly.com/question/19904705

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Radar uses radio waves of a wavelength of 2.4 \({\rm m}\) . The time interval for one radiation pulse is 100 times larger than t
blondinia [14]

Answer:

120 m

Explanation:

Given:

wavelength 'λ' = 2.4m

pulse width 'τ'= 100T ('T' is the time of one oscillation)

The below inequality express the range of distances to an object that radar can detect

τc/2 < x < Tc/2 ---->eq(1)

Where, τc/2 is the shortest distance

First we'll calculate Frequency 'f' in order to determine time of one oscillation 'T'

f = c/λ (c= speed of light i.e 3 x 10^{8} m/s)

f= 3 x 10^{8} / 2.4

f=1.25 x  10^{8} hz.

As, T= 1/f

time of one oscillation T= 1/1.25 x  10^{8}

T= 8 x 10^{-9} s

It was given that pulse width 'τ'= 100T

τ= 100 x 8 x 10^{-9} => 800 x 10^{-9} s

From eq(1), we can conclude that the shortest distance to an object that this radar can detect:

x_{min}= τc/2 =>  (800 x 10^{-9} x 3 x 10^{8})/2

x_{min}=120m

8 0
4 years ago
(1 point) A mass m=4kg is attached to both a spring with spring constant k=577N/m and a dash-pot with damping constant c=4N⋅s/m
sasho [114]

Answer:

The function is x = e^(-t/2) * (0.792*sin12t + 5cos12t)

Explanation:

we have to:

m = mass = 4 kg

k = spring constant = 577 N/m

c = damping constant = 4 N*s/m

The differential equation of motion is equal to:

m(d^2x/dt^2) + c(dx/dt) + k*x = 0

Replacing values:

4(d^2x/dt^2) + 4(dx/dt) + 577*x = 0

Thus, we have:

4*x^2 + 4*x + 577 = 0

we will use the quadratic equation to solve the expression:

x = (-4 ± (4^2 - (4*4*577))^1/2)/(2*4) = (-4 ± (-9216))/8 = (1/2)  ± 12i

The solution is equal to:

x = e^(1/2) * (c1*sin12t + c2*cos12t)

x´ = (-1/2)*e^(1/2) * (c1*sin12t + c2*cos12t) + e^(-t/2) * (12*c1*cos12t - 12*c2*sin12t)

We have the follow:

x(0) = 5

e^0(0*c1 + c2) = 5

c2 = 5

x´(0) = 7

(-1/2)*e^0 * (0*c1 + c2) + e^0 * (12*c1 - 0*c2) = 7

(-1/2)*(5) + 12*c1 = 7

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c1 = 0.792

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x = e^(-t/2) * (0.792*sin12t + 5cos12t)

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3 years ago
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4vir4ik [10]

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seropon [69]

Answer:

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