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Westkost [7]
3 years ago
12

A child of mass M is swinging on a swing set. The ropes attaching the swing to the top bar have length L. Find the gravitational

potential energy of the child relative to her lowest position when the ropes are (a) hanging straight down; (b) exactly horizontal; (c) at an angle ϴ from the vertical; and (d) at an angle phi from the horizontal.
Physics
1 answer:
myrzilka [38]3 years ago
5 0

Answer:

(a) 0

(b) 10ML

(c) 10ML(1 - cos(\theta))

(d) 10ML(1 + sin(\phi))

Explanation:

(a) When hanging straight down. The child is at the lowest position. His potential energy with respect to this point would also be 0.

(b) Since the rope has length L m. When the rope is horizontal, he is at L (m) high with respect to the lowest swinging position. His potential energy with respect to this point should be

E_h = mgh = 10ML

where g = 10m/s2 is the gravitational acceleration.

(c) At angle \theta from the vertical. Vertically speaking, the child should be at a distance of Lcos(\theta) to the swinging point, and a vertical distance of L - Lcos(\theta) to the lowest position. His potential energy to this point would be:

E_{\theta} = mgh = 10M(L - Lcos(\theta)) = 10ML(1 - cos(\theta))

(d) at angle \phi from the horizontal. Suppose he is higher than the horizontal line. This would mean he's at a vertical distance of Lsin(\phi) from the swinging point and higher than it. Therefore his vertical distance to the lowest point is L + Lsin(\phi) = L(1 + sin(\phi))

His potential energy to his point would be:

E_{\phi} = mgh = 10ML(1 + sin(\phi))

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Alenkinab [10]

Answer:

Explanation:

spring constant, K = 13.1 N/m

22 oscillations in 20 seconds

time taken to complete one oscillation is called time period.

T = 20 / 22 second = 0.909 seconds

(a) let m be the mass.

The formula for the time period is

T = 2\pi \sqrt{\frac{m}{K}}

m = \frac{T^{2}K}{4\pi ^{2}}

m = \frac{0.909^{2}\times 13.1}{4\pi ^{2}}

m = 0.275 kg

(b) maximum speed, v = ω A = 2π A / T

v = ( 2 x 31.4 x 0.1) / 0.909

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3 years ago
The power rating of a 400-ΩΩ resistor is 0.800 W.
sashaice [31]

Answer:

voltage= 17.88volts

current= 0.04 amps

Explanation:

Step one:

given data

resistance R=400 ohms

Power P= 0.8W

a.  What is the maximum voltage that can be applied across this resistor without damaging it?

the expression relating power and voltage is

P=V^2/R

substituting we have

0.8=V^2/400

V^2=0.8*400

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V=√320

V=17.88 volts

the maximum voltage is 17.88volts

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we know that from Ohm' law

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I=17.88/400

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El peso de un cuerpo es de 392.2 N ¿Cuál es su masa?
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Answer:

⇒ To find the mass, we apply the following formula:

                         \boxed{ \boxed{\mathbf{m=\dfrac{w}{g} }}}

<u>Data</u>:

➢  \textrm{Weight(w) = 392.2 Newtons}

➢  \mathrm{Gravity(g) = \ 9,81\ m/s^{2} }

➢  \textrm{Mass(m) =\ ?}

∴ We replace and develop:

\mathrm{m=\dfrac{w}{g} }

\mathrm{Mass=\dfrac{392.2}{9.81} }

\mathrm{Mass =39.9\ kg }

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