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nasty-shy [4]
2 years ago
11

How do you reconcile the law of falling bodies (that all objects fall to earth at the same acceleration despite their weight) wi

th the Second Law of Motion which states that the acceleration of a body subjected to an applied force is inversely proportional to its mass?
Physics
1 answer:
ASHA 777 [7]2 years ago
5 0

From the gravity acceleration theorem due to a celestial body or planet, we have that the Force is given as

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

Where,

F = Strength

G = Universal acceleration constant

M = Mass of the planet

m = body mass

r = Distance between centers of gravity

The acceleration by gravity would be given under the relationship

g = \frac {F} {m}

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

Here the acceleration is independent of the mass of the body m. This is because the force itself depended on the mass of the object.

On the other hand, the acceleration of Newton's second law states that

a = \frac {F} {m}

Where the acceleration is inversely proportional to the mass but the Force does not depend explicitly on the mass of the object (Like the other case) and therefore the term of the mass must not necessarily be canceled but instead, considered.

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Calculate the wavelength of each frequency of electromagnetic radiation: a. 100.2 MHz (typical frequency for FM radio broadcasti
Natalka [10]

Answer:

a). 100.2 MHz (typical frequency for FM radio broadcasting)

The wavelength of a frequency of 100.2 Mhz is 2.99m.

b. 1070 kHz (typical frequency for AM radio broadcasting) (assume four significant figures)

The wavelength of a frequency of 1070 khz is 280.3 m.

c. 835.6 MHz (common frequency used for cell phone communication)

The wavelength of a frequency of 835.6 Mhz is 0.35m.

Explanation:

The wavelength can be determined by the following equation:

c = \lambda \cdot \nu  (1)

Where c is the speed of light, \lambda is the wavelength and \nu is the frequency.  

Notice that since it is electromagnetic radiation, equation 1 can be used. Remember that light propagates in the form of an electromagnetic wave.

<em>a). 100.2 MHz (typical frequency for FM radio broadcasting)</em>

Then, \lambda can be isolated from equation 1:

\lambda = \frac{c}{\nu} (2)

since the value of c is 3x10^{8}m/s. It is necessary to express the frequency in units of hertz.

\nu = 100.2 MHz . \frac{1x10^{6}Hz}{1MHz} ⇒ 100200000Hz

But 1Hz = s^{-1}

\nu = 100200000s^{-1}

Finally, equation 2 can be used:

\lambda = \frac{3x10^{8}m/s}{100200000s^{-1}}

\lambda = 2.99 m

Hence, the wavelength of a frequency of 100.2 Mhz is 2.99m.

<em>b. 1070 kHz (typical frequency for AM radio broadcasting) (assume four significant figures)</em>

<em> </em>

\nu = 1070kHz . \frac{1000Hz}{1kHz} ⇒ 1070000Hz

But  1Hz = s^{-1}

\nu = 1070000s^{-1}

Finally, equation 2 can be used:

\lambda = \frac{3x10^{8}m/s}{1070000s^{-1}}

\lambda = 280.3 m

Hence, the wavelength of a frequency of 1070 khz is 280.3 m.

<em>c. 835.6 MHz (common frequency used for cell phone communication) </em>

\nu = 835.6MHz . \frac{1x10^{6}Hz}{1MHz} ⇒ 835600000Hz

But  1Hz = s^{-1}

\nu = 835600000s^{-1}

Finally, equation 2 can be used:

\lambda = \frac{3x10^{8}m/s}{835600000s^{-1}}

\lambda = 0.35 m

Hence, the wavelength of a frequency of 835.6 Mhz is 0.35m.

6 0
3 years ago
What are the periodic variations in Earth's rotation and orbit around the sun that alter the way solar radiation is distributed
Dahasolnce [82]

Answer:

1. The precession of the equinoxes.

2. Changes in the tilt angle of Earth’s rotational axis relative to the plane of Earth’s orbit around the Sun.

3. Variations in the eccentricity

Explanation:

These variations listed above;  the precession of the equinoxes (refers, changes in the timing of the seasons of summer and winter), this occurs on  a roughly about 26,000-year interval; changes in the tilt angle of Earth’s rotational axis relative to the plane of Earth’s orbit around the Sun, this occurs roughly in a 41,000-year interval; and changes in the eccentricity (that is a departure from a perfect circle) of Earth’s orbit around the Sun, occurring on a roughly 100,000-year timescale. which influences the mean annual solar radiation at the top of Earth’s atmosphere.

5 0
3 years ago
Please help i'm going to throw up from stress
Eddi Din [679]

Answer:

Explanation:

First of all, I used the specific heat of water as 4182 J/(kgC) and the specific heat of ethyl alcohol (EtOH) as 2440 J/(kgC); that means that we need the masses in kg, not g.

120.g = .1200 kg of ethyl alcohol. Now for the formula:

t_f=\frac{(m_{H2O}*spheat_{H2O}*temp_{H2O})+(m_{EtOH}*spheat_{EtOH}*temp_{EtOH})}{(m_{H2O}*spheat_{H2O})+(m_{EtOH}*spheat_{EtOH})} where spheat is specific heat.

Filling that horrifying-looking formula in with some values:

16.0=\frac{(x*4182*20.0)+(.1200*2440*10.0)}{(x*4182)+(.1200*2440)} and

16.0=\frac{83640x+2928}{4182x+292.8} and

16(4182x + 292.8) = 83640x + 2928 and

66912x + 4684.8 = 83640x + 2928 and

1756.8 = 16728x so

x = .105 kg and the amount of water added is 105 g

4 0
2 years ago
Please help it’s urgent
Sidana [21]

Answer:

I believe it's sound energy.

Explanation:

Sound can move through air, whereas electric and radiant energy don't have to.

6 0
3 years ago
A coil of wire consists of 20 turns, each of which has an area of 0.0015 m2. A magnetic field is perpendicular to the surface wi
stepan [7]

Answer:

magnitude of the induced emf in the coil is 0.0153 V

Explanation:

Given data

no of turns = 20

area = 0.0015 m²

magnitude  B1 = 4.91 T/s

magnitude  B2 = 5.42 T/s

to find out

the magnitude of the induced emf in the coil

solution

we know here

emf = -n A d∅ /dt

so here n = 20 and

A = 0.0015

and d∅ = B2 - B1 = 5.42 - 4.91

d∅ = 0.51 T and dt  at 1 sec

so  put all value

emf = -n A d∅ /dt

emf = -20 (0.0015) 0.51 / 1

emf = - 0.0153

so magnitude of the induced emf in the coil is 0.0153 V

7 0
2 years ago
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