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likoan [24]
3 years ago
7

The weight of a body above sea level varies inversely with the square of the distance from the center of Earth. If a woman weigh

s 131 pounds when she is at sea​ level, 3960 miles from the center of​ Earth, how much will she weigh when she is at the top of a​ mountain, 4.8 miles above sea​ level?
Physics
1 answer:
Serjik [45]3 years ago
4 0

Answer:

89.16pounds

Explanation:

The equation that defines this problem is as follows

W=k/X^2

where

W=Weight

K= proportionality constant

X=distance from the center of Earth

first we must find the constant of proportionality, with the first part of the problem

k=WX^2=131x3960^2=2054289600pounds x miles^2

then we use the equation to calculate the woman's weight with the new distance

W=2054289600/(4800)^2=89.16pounds

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When aluminum foil is formed into a loose ball, it can float on water. But when the ball of foil is pounded flat with a hammer,
docker41 [41]
Air caught in the ball of foil makes the ball less dense than water
8 0
2 years ago
A Person whose weight is 5.20 x 10^2 N is being pulledup
Gnoma [55]

Answer:

Explanation:

Given

Weight of Person W=5.20\times 10^{2} N

Cave is h=35.1 m deep

Breaking stress T=569 N

Net Force on Person

F_{net}=569-520=49 N

a_{net}=\frac{F_{net}}{\frac{W}{g}}

a_{net}=\frac{49}{\frac{520}{9.8}}

a_{net}=0.923 m/s^2

The shortest time such that the person can be taken out of cave

h=ut+\frac{1}{2}at^2

where

h=distance moved

t=time

a=acceleration

35.1=0+\frac{1}{2}(0.923)(t)^2

t^2=76.05

t=\sqrt{76.05}

t=8.72\ s                    

6 0
2 years ago
4. Una cuerda de acero de piano mide 1.60 m de longitud y 0.20 cm de diámetro. ¿Cuál es la tensión en la cuerda si se estira 0.2
bazaltina [42]

Answer:

1030.83\ \text{N}

Explanation:

\Delta L = Cambio en la longitud de la cuerda = 0.25 cm

T = tensión en cuerda

A = Área de la cadena = \dfrac{\pi}{4}d^2

d = Diámetro de la cuerda = 0.2 cm

L = Longitud original de la cuerda = 1.6 m

El cambio de longitud de una cuerda viene dado por

\Delta L=\dfrac{TL}{AE}\\\Rightarrow T=\dfrac{\Delta LAE}{L}\\\Rightarrow T=\dfrac{0.25\times 10^{-2}\times \dfrac{\pi}{4}(0.2\times 10^{-2})^2\times 210\times 10^9}{1.6}\\\Rightarrow T=1030.84\ \text{N}

La tensión en la cuerda es 1030.84\ \text{N}.

8 0
2 years ago
A frictionless piston-cylinder device contains 10 kg of superheated vapor at 550 kPa and 340oC. Steam is then cooled at constant
Alla [95]

Answer:

a) the work (W) done during the process is -2043.25 kJ

b) the work (W) done during the process is -2418.96 kJ

Explanation:

Given the data in the question;

mass of water vapor m = 10 kg

initial pressure P₁ = 550 kPa

Initial temperature T₁ = 340 °C

steam cooled at constant pressure until 60 percent of it, by mass, condenses; x = 100% - 60% = 40% = 0.4

from superheated steam table

specific volume v₁ = 0.5092 m³/kg

so the properties of steam at p₂ = 550 kPa, and dryness fraction

x = 0.4

specific volume v₂ = v_f + xv_{fg

v₂ = 0.001097 + 0.4( 0.34261 - 0.001097 )

v₂ = 0.1377 m³/kg

Now, work done during the process;

W = mP₁( v₂ - v₁ )

W = 10 × 550( 0.1377 - 0.5092 )

W = 5500 × -0.3715

W = -2043.25 kJ

Therefore, the work (W) done during the process is -2043.25 kJ

( The negative, indicates work is done on the system )

b)

What would the work done be if steam were cooled at constant pressure until 80 percent of it, by mass, condenses

x₂ = 100% - 80% = 20% = 0.2

specific volume v₂ = v_f + x₂v_{fg

v₂ = 0.001097 + 0.2( 0.34261 - 0.001097 )

v₂ = 0.06939 m³/kg

Now, work done during the process will be;

W = mP₁( v₂ - v₁ )

W = 10 × 550( 0.06939 - 0.5092 )

W = 5500 × -0.43981

W = -2418.96 kJ

Therefore, the work (W) done during the process is -2418.96 kJ

3 0
2 years ago
You are standing on a train station platform as a train goes by close to you. As the train approaches, you hear the whistle soun
mojhsa [17]

Answer:

Kindly check explanation

Explanation:

Given the following :

As train approaches ; frequency, f1 = 94Hz

As train recedes; frequency, f2 = 71Hz

Speed of sound in air ; v = 340m/s

A) speed of sound source (speed of train) = vs

From doppler effect :

As the train recedes ;

f2 = fs [v / (v + vs)] - - - - (1)

As train approaches :

f1 = fs [v / (v - vs)] ----- (2)

To find vs equate (1) and (2)

fs [v / (v - vs)] = fs [v / (v + vs)]

f1/f2 = v / (v - vs) ÷ v / (v + vs)

f1 / f2 = v / (v - vs) × (v + vs) / v

f1 / f2 = (v + vs) / (v - vs)

Let f1 / f2 = f

f = (v + vs) / (v - vs)

f (v - vs) = v + vs

fv - fvs = v + vs

fv - v = vs + fvs

v(f - 1) = vs(1 + f)

v(f - 1) / (1 + f) = vs

B)

v(f - 1) / (1 + f) = vs

f = f1 / f2 = 94/71 = 1.32 Hz

340(1.324 - 1) / (1 + 1.324) = vs

vs = 340(0.324) / 2.324

vs = 110.16 / 2.324

vs = 47.40 m/s

C.) To calculate fs, frequency of train, substitute vs into our equation.

f2 = fs [v / (v + vs)]

Following our substitikn we obtain:

fs = (2f / (f + 1))f2

D)

fs = (2f / (f + 1))f2

fs = 2(1.324) / (1.324 +1)) × 71

fs = (2.648 / 2.324) × 71

fs = 1.1394148 × 71

fs = 80.898450

fs = 80.90 Hz

3 0
3 years ago
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