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max2010maxim [7]
3 years ago
7

What did Newton's book Mathematical Principles of Natural Philosophy argue about the role of mathematics in understanding the na

tural world?. A. Newton's book argued that mathematics and physical science were separate and could not explain each other. B. Newton's book argued that gravity could not be explained by mathematics alone. C. Newton's book made the bold argument that physical science required the use of faith to understand the natural world. D. Newton's book argued that mathematical principles could be applied to our understanding of the natural world.
Physics
2 answers:
Andrew [12]3 years ago
7 0

Answer:

D. Newton's book argued that mathematical principles could be applied to our understanding of the natural world.

Explanation:

IN his book Mathematical Principles of Natural Philosophy Newton established and proposed that physics and mathematics were able to explain the physical rules that govern the whole universe, and unified the physics laws on earth to those on the outer space, while also stating that the natural would could be understood through mathematical principles.

quester [9]3 years ago
5 0
"Newton's book argued that mathematical principles could be applied to our understanding of the natural world" is what <span>Newton's book Mathematical Principles of Natural Philosophy argue about the role of mathematics in understanding the natural world. The correct option among all the options that are given in the question is the last option or option "D". I hope the answer has come to your help.</span>
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Answer:

 R = 4.24 x 10⁻⁴ m

Explanation:

given,

mass of the person = 60.3-kg

mass of the bullet = 10 gram = 0.01 Kg

velocity of bullet = 389 m/s

angle made with the horizontal = 45°

using conservation of momentum.

M v  + m u  = ( M + m ) V

60.3 x 0 + 0.01 x 389 = (60.3 + 0.01) V

V = \dfrac{3.89}{60.31}

V = \dfrac{3.89}{60.31}

V = 0.0645 m/s

for calculation of range

R = \dfrac{V^2sin 2 \theta}{g}

R = \dfrac{0.0645^2sin 2 (45^0)}{9.8}

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6 0
4 years ago
What is the acceleration of a 113 kg object exposed to a 110 N force. (Round to one decimal place.)
Norma-Jean [14]

Answer:

\boxed {\boxed {\sf a \approx  1.0 \ m/s^2}}

Explanation:

We are asked to find the acceleration of an object.

According to Newton's Second Law of Motion, force is the product of mass and acceleration.

F= m \times a

We know the object has a mass of 113 kilograms and it is exposed to a 110 Newton force.

Let's convert the units of force to make the problem easier later. 1 Newton is equal to 1 kilogram meter sper second squared. So, the force of 110 Newtons is equal to 110 kilogram meters per second squared.

  • F= 110 kg*m/s²
  • m= 113 kg

Substitute the values into the formula.

110 \ kg*m/s^2=113 \ kg * a

We are solving for a, the acceleration, so we isolate this variable. It is being multiplied by 113 kilograms. The inverse operation of multiplication is division, so we divide both sides by 113 kg.

\frac {110 \ kg*m/s^2 }{113 \ kg}= \frac{113 \ kg * a}{113 \ kg}

\frac {110 \ kg*m/s^2 }{113 \ kg}=a

The units of kilograms (kg) cancel.

\frac {110 m/s^2 }{113 }=a

0.973451327 \ m/s^2 = a

We are asked to round to one decimal place or the tenths place. The 7 in the hundredths place tells us to round the 9 up to a 0, but then we must also round the 0 in the ones place to a 1.

1.0 \ m/s^2 \approx a

The acceleration of the object is approximately <u>1.0 meter per second squared.</u>

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Northrop aircraft developed and built a deceleration sled to test the effects of the extreme forces on humans and equipment. In
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(a) before the submarines pass each other, sub b quickly turns around and speeds up to 12. 4 m/s. What is the change in frequenc
Wewaii [24]

The change in frequency noticed by observers on sub b is : 2.66 Hz before the submarines pass each other.

<u>Given data:</u>

New speed of Sub b ( Vb ) = 12.4 m/s

F₁ = 1416.44 Hz

Frequency of sonar wave ( Fₐ ) = 1400 Hz

Speed of sound in water ( V ) = 1533 m/s

Speed of submarine A ( Va ) = 8.4 m/s

<h3>Determine the change in frequency observed on sub B </h3>

The change in frequency can be calculated using the formula below

F₁' = Fₐ ( V + Vb / V - Vₐ )

    = 1400 ( 1533 + 12.4 / 1533 - 8.4 )

    = 1400 ( 1545.4 / 1524.6 )

    = 1419.1 Hz

Final step : <u>determine the change in frequency </u>

ΔF = F₁' - F₁

    = 1419.1 - 1416.44

    = 2.66 Hz.

Hence we can conclude that The change in frequency noticed by observers on sub b is : 2.66 Hz.

learn more about change in frequency : brainly.com/question/254161

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