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Stolb23 [73]
3 years ago
13

Si pudieras viajar a la Luna:

Physics
1 answer:
Nostrana [21]3 years ago
4 0

Answer:

i) Distancia, ii) La cinta métrica es impracticable.

Explanation:

i) El concepto físico que se construye únicamente del punto de salida y el punto de llegada a la Luna es el concepto de desplazamiento, definido como la distancia en línea recta de un punto en el espacio con respecto a un punto de referencia (la Tierra en este caso).

La distancia puede involucrar trayectorias curvilíneas entre los puntos mencionados.

ii) Por último, el uso de una cinta métrica es impracticable debido a la cantidad de material a utilizar y los efectos gravitacionales, electromagnéticos y mecánicos que inducen a una deflexión o una ruptura de esa cinta debido a la magnitud de la distancia entre las superficies del planeta y el satélite, respectivamente.

En este caso, es mejor utilizar la medición con tecnología láser, basadas en el fenómeno del electromagnetismo.

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A 4 cm diameter "bobber" with a mass of 3 grams floats on a pond. A thin, light fishing line is tied to the bottom of the bobber
Tasya [4]

Answer:

Explanation:

Calculate the volume of the lead

V=\frac{m}{d}\\\\=\frac{10g}{11.3g'cm^3}

Now calculate the bouyant force acting on the lead

F_L = Vpg

F_L=(\frac{10g}{11.3g/cm^3} )(1g/cm^3)(9.8m/s^2)\\\\=8.673\times 10^{-3}N

This force will act in upward direction

Gravitational force on the lead due to its mass  will act in downward direction

Hence the difference of this two force

T=mg-F_L\\\\=(10\times10^{-3}kg(9.8m/s^2)-8.673\times 10^{-3}\\\\=8.933\times10^{-3}N

If V is the volume submerged in the water then bouyant force on the bobber is

F_B=V'pg

Equate bouyant force with the tension and gravitational force

F_B=T_mg\\\\V'pg=\frac{(8.933\times10^{-2}N)+mg}{pg} \\\\V'=\frac{(8.933\times10^{-2}N)+mg}{pg}

Now Total volume of bobble is

\frac{V'}{V^B} =\frac{\frac{(8.933\times10^{-2})+Mg}{pg} }{\frac{4}{3} \pi R^3 }\times100\\\\=\frac{\frac{(8.933\times10^{-2})+(3)(9.8)}{(1000)(9.8)} }{\frac{4}{3} \pi (4.0\times10^{-2})^3 }\times100\\\\

=\large\boxed{4.52 \%}

7 0
3 years ago
What is the Orbital Notation for Radon
Marina CMI [18]

Answer:

[Rn]7s2

Explanation:

3 0
3 years ago
Please anyone post any class 9 physics question​
aalyn [17]

Answer:

I will but can you just wait for some minutes cus I am in a hurry now.

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6 0
2 years ago
A baseball hits a car, breaking its window and triggering its alarm which sounds at a frequency of 1210 Hz. What frequency (in H
IRINA_888 [86]

Answer:

The frequency of sound heard by the boy is 1181 Hz.

Explanation:

Given that,

Frequency of sound from alarm  f_{0} = 1210\ Hz

Speed = -8.25 m/s

Negative sign show the boy riding away from the car

Speed of sound = 343

We need to calculate the heard frequency

Using formula of frequency

f = f_{0}(\dfrac{v+v_{0}}{v-v_{s}})

Where, f_{0} = frequency of source

v_{0} = speed of observer

v_{s} = speed of source

v = speed of sound

Put the value into the formula

f=1210\times\dfrac{343+(-8.25)}{343-0}

here, source is at rest

f=1180.8\ Hz

f=1181\ Hz

Hence, The frequency of sound heard by the boy is 1181 Hz.

8 0
3 years ago
Please help me out someone nothing helping me out please someone
Zolol [24]
Well i think the best answer would be A

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