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schepotkina [342]
3 years ago
11

Calcula la distancia que debe haber entre un pisa papel y una biblia tipo masas son de 250g y 650g con una fuerza cuya magnitud

es de 1.85 x 10-6 dinas.
Physics
1 answer:
ruslelena [56]3 years ago
8 0

Answer:

  r = 0.765 m

Explanation:

For this exercise let's use the law of universal gravitation

        F = G \ \frac{m_1m_2}{r^2}

        r = \sqrt{G \frac{m_1m_2}{F} }

let's reduce the magnitudes to the SI system

         m₁ = 250 g (1 kg / 1000g) = 0.250 kg

         m₂ = 650 g (1 kg / 1000g) = 0.650

         F = 1.85 10-6 dynes (1 N / 10⁵ dynes) = 1.85 10-11 N

let's calculate

        r = \sqrt{ \frac{ 6.67 \ 10^{-11} \ 0.250 \ 0.650}{1.85 \ 10^{-11}} }

        r = √0.58587

        r = 0.765 m

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4 0
3 years ago
Read 2 more answers
The acceleration due to gravity near Earth ... Select one: a. varies inversely with the distance from the center of Earth. by. v
vladimir1956 [14]

Answer:

b. varies inversely with the square of the distance from the center of Earth.

Explanation:

Comparing the Newton's law of universal gravitation and second law of motion;

from Newton's second law of motion,

F = ma ............. 1

from New ton's law of universal gravitation,

F = \frac{GMm}{r^{2} } ........... 2

Equating 1 and 2, we have;

mg =  \frac{GMm}{r^{2} }

g = \frac{GM}{r^{2} }

Therefore, the acceleration due to gravity near Earth, g, is inversely proportional to the square of the distance from the center of Earth.

7 0
3 years ago
12.51 A parallel RLC circuit, which is driven by a variable frequency 2-A current source, has the following values: R = 1 kΩ, L
Anastaziya [24]

Answer:

BW = 100 rad/s

wlow = 452.49 rad/s

whigh = 552.49 rad/s

V(jwlow) =1414.21 < 45°V

V(jwhigh) =1414.21 <-45°V

Explanation:

To calculate bandwidth we have formula

BW = 1/RC

BW = 1/ 1000x10x10^¯6

BW = 100 rad/s

We will first calculate resonant frequency and quality factor for half power frequencies.

For resonant frequency

wo = 1/(SQRT LC)

wo = 1/SQRT 400×10¯³ × 10×10^¯6

wo = 500 rad/s

For Quality

Q = wo / BW

Q = 500/100

Q = 5

wlow = wo [-1/2Q+ SQRT (1/2Q)² + 1]

wlow = 500 [-1/2×5 + SQRT (1/2×5)² + 1]

wlow = 452.49 rad/s

whigh = wo [1/2Q+ SQRT (1/2Q)² + 1]

whigh = 500 [1/2×5 + SQRT (1/2×5)² + 1]

whigh = 552.49 rad/s

We will start with admittance at lower half power frequency

Y(jwlow) = (1/R) + (1/jwlow L) + (jwlow C)

Y(jwlow) = (1/1000) + (1/j×452.49×400×10¯³) + (j×452.49×10×10^¯6)

Y(jwlow) = 0.001 - j5.525×10¯³ + j4.525×10¯³

Y(jwlow) = (1-j).10¯³ S

Voltage across the network is calculated by ohm's law

V(jwlow) = I/Y(jwlow)

V(jwlow) = 2/(1-j).10¯³

V(jwlow) = 1414.2 < 45°V

Now we will calculate the admittance at higher half power frequency

Y(jwhigh) = (1/R) + (1/jwhigh L) + (jwhigh C)

Y(jwhigh) = (1/1000) + (1/j×552.49×400×10¯³) + (j×552.49×10×10^¯6)

Y(jwhigh) = 0.001 - j4.525×10¯³ + j5.525×10¯³

Y(jwhigh) = (1+j).10¯³ S

Voltage across network will be calculated by ohm's law

V(jwhigh) = I/Y(jwhigh)

V(jwhigh) = 2/(1+j).10¯³

V(jwhigh) = 1414.2 < - 45°V

6 0
3 years ago
ANOTHER PHYSICS QUESTION PLEASE HELP!
Sergio039 [100]
Answer


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5 0
3 years ago
A scuba diver at 70 m below the surface of a lake, where the temperature is 4 degrees C, releases an air bubble with a volume of
posledela

Answer:

121.3 cm^3

Explanation:

P1 = Po + 70 m water pressure (at a depth)

P2 = Po (at the surface)

T1 = 4°C = 273 + 4 = 277 K

V1 = 14 cm^3

T2 = 23 °C = 273 + 23 = 300 K

Let the volume of bubble at the surface of the lake is V2.

Density of water, d = 1000 kg/m^3

Po = atmospheric pressure = 10^5 N/m^2

P1 = 10^5 + 70 x 1000 x 10 = 8 x 10^5 N/m^2

Use the ideal gas equation

\frac{P_{1}V_{1}}{T_{1}}=\frac{P_{2}V_{2}}{T_{2}}

By substituting the values, we get

\frac{8\times 10^5\times 14}{277}=\frac{10^{5} \timesV_{2}}{300}

V2 = 121.3 cm^3

Thus, the volume of bubble at the surface of lake is 121.3 cm^3.

6 0
3 years ago
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