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3 years ago

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A 40.0 kg child is in a swing that is attached to ropes 2.00 m long. Find the gravitational potential energy associated with the

child relative to the child’s lowest position under the following conditions:
a. when the ropes are horizontal
b. when the ropes make a 30.0° angle with the vertical
c. at the bottom of the circular arc

1 answer:

o-na [289]3 years ago

8 0

Answer:

A. As the ropes are horizontal the child has travelled 2m of vertical displacement from his lowest position.

Gpe @ A=mgh=40*9.81*2=784.8J

B. At 30degree vertical angle the vertical displacement from lowest position is given by

2-2cos(30)=2-1.73=0.27m

Gpe @B= 40*9.81*0.27=106 J

C: at the bottom of circular arc it's Gpe is zero relative to lowest position as bottom of arc itself is lowest position.

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Klio2033 [76]

Frequency=v=4.75×10^12Hz
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We know

$\boxed{\sf \lambda=\dfrac{C}{V}}$

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3 years ago

Argon in the amount of 1.5 kg fills a 0.04-m3 piston cylinder device at 550 kPa. The piston is now moved by changing the weights

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Answer:

275 kPa

Explanation:

mass of the gas=m=1.5 kg

initial volume if the gas=V₁=0.04 m³

initial pressure of the gas= P₁=550 kPa

as the condition is given final volume is double the initial volume

V₂=final volume

V₂=2 V₁

As the temperature is constant

T₁=T₂=T

$\frac{P1V1}{T1}$ = $\frac{P2 V2}{T2}$

putting the values in the equation.

$\frac{P1V1}{T1}$ = $\frac{P2 *2V1}{T2}$

P₂= $\frac{P1}{2}$

P₂= $\frac{550}{2}$

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4 years ago

What's an adjective to describe Isaac Newton?

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3 years ago

Two Earth satellites, A and B, each of mass m, are to be launched into circular orbits about Earth's center. Satellite A is to o

Juliette [100K]

(a) 0.473

The potential energy of a satellite orbiting around Earth is given by

$U=-\frac{GMm}{R+h}$

where

G is the gravitational constant

M is the Earth's mass

m is the satellite's mass

R is the Earth's radius

h is the altitude of the satellite above the Earth's surface

So the potential energy of satellite A is

$U_A=-\frac{GMm}{R+h_A}$

while potential energy of satellite B is

$U_B=-\frac{GMm}{R+h_B}$

Therefore the ratio of the potential energy of satellite B to that of satellite A is

$\frac{U_B}{U_A}=\frac{R+h_A}{R+h_B}$

and using

hA = 5920 km

hB = 19600 km

R = 6370 km

we find

$\frac{U_B}{U_A}=\frac{6370+5920}{6370+19600}=0.473$

(b) 0.473

The kinetic energy of a satellite orbiting around Earth instead is given by

$K=\frac{GMm}{2(R+h)}$

So the kinetic energy of satellite A is

$K_A=\frac{GMm}{2(R+h_A)}$

while kinetic energy of satellite B is

$K_B=\frac{GMm}{2(R+h_B)}$

Therefore the ratio of the kinetic energy of satellite B to that of satellite A is

$\frac{K_B}{K_A}=\frac{R+h_A}{R+h_B}$

which is identical to before, so it gives

$\frac{K_B}{K_A}=\frac{6370+5920}{6370+19600}=0.473$

(c) Satellite B

The total energy of a satellite in orbit is given by

$E=U+K = -\frac{GMm}{R+h}+\frac{GMm}{2(R+h)}=-\frac{GMm}{2(R+h)}$

We see that the total energy is:

1) negative (because the satellite is on a bound orbit)

2) inversely proportional to the distance of the satellite from the Earth's center, R+h

So the magnitude of the fraction in the equation is larger for the satellite which is closer to the Earth's surface (satellite A), but since the energy is negative, this means that the total energy of this satellite is smaller than that of satellite B. So, satellite B has a greater total energy.

(d) $1.03\cdot 10^7 J$

We have to calculate the total energy of each satellite.

Given:

$G=6.67\cdot 10^{-11}$

$M=5.98\cdot 10^{24} kg$

m = 12.0 kg

$R+h_A = 6370 km+5920 km=12290 km = 12.3 \cdot 10^6 m$

$R+h_B = 6370 km+19600 km=25970 km = 26.0 \cdot 10^6 m$

We find:

$E_A = - \frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24})(12.0)}{2(12.3\cdot 10^6)}=-1.95\cdot 10^{7} J$

$E_B = - \frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24})(12.0)}{2(26.0\cdot 10^6)}=-9.2\cdot 10^{6} J$

So the difference in total energy is

$E_B-E_A = -9.2\cdot 10^6 - (-1.95\cdot 10^7) =1.03\cdot 10^7 J$

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A 1200 w microwave oven transforms 1.8x10(to the power of 5) J of energy while reheating some food. Calculate how long the food

Fofino [41]

1200 watts = 1200 joules per second

(1.8 x 10⁵ joules) / (1,200 joules/sec) = 150 sec = <em>2.5 minutes

</em>

5 0

3 years ago