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nasty-shy [4]
3 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]3 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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3 years ago
A wire is stretched right to its breaking point by a 5000 N force. A longer wire made of the same material has the same diameter
leva [86]

Answer:

Equal to 5000N

Explanation:

The stress on the material is defined by force per unit of cross-sectional area. So it depends on the force and the diameter of the wire, which is the same for both wires. The material that defines the breaking point, is also the same. Therefore, both wires have their breaking point the same at 5000N. The wire length plays no role in here.

4 0
3 years ago
A train whistle is heard at 300 Hz as the train approaches town. The train cuts its speed in half as it nears the station, and t
givi [52]

To solve this problem we will apply the concepts related to the Doppler effect. The Doppler effect is the change in the perceived frequency of any wave movement when the emitter, or focus of waves, and the receiver, or observer, move relative to each other. Mathematically it can be described as,

f = f_0 (\frac{v_0}{v_0-v})

Here,

f_0 = Frequency of Source

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So if the speed of the train after slowing down will be v/2, we can do a system equation of 2x2 at the two moments, then,

The first equation is,

f = f_0 (\frac{v_0}{v_0-v})

300 = f_0 (\frac{343}{343-v})

(300*343) - 300v = 343f_0

Now the second expression will be,

f' = f_0 (\frac{v_0}{v_0-v/2})

290 = (343)(\frac{v_0}{343-v/2})

290*343-145v = 343f_0

Dividing the two expression we have,

\frac{(300*343) - 300v}{290*343-145v} = 1

Solving for v, we have,

v = 22.12m/s

Therefore the speed of the train before and after slowing down is 22.12m/s

6 0
3 years ago
What is the frequency of this wave?
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A 100 kg cart on a roller coaster has 1800J of kinetic energy. How fast is it going?<br> KE= 1/2mv2
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Answer:

the speed is equal to 6 m/s

Explanation:

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