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satela [25.4K]
2 years ago
11

Based on details in The Riddle of the Rosetta Stone, which statement best describes Jean-Baptiste Fourier's influence on Jean-Fr

ançois Champollion?
Physics
2 answers:
max2010maxim [7]2 years ago
6 0

Answer:

C

Explanation:

On edge

Ira Lisetskai [31]2 years ago
5 0

Answer:

hello your question is incomplete below is the missing part of the  complete question

which statement best describes Jean-Baptiste Fourier’s influence on Jean-François Champollion? Fourier shared his research about mathematics with Champollion, which aided him in his research. Fourier was the first to introduce Champollion to the hieroglyphs on the Rosetta Stone, which fascinated him. Fourier shared his findings about the Rosetta Stone with Champollion, which helped him in his research. Fourier was the first to introduce Champollion to a variety of languages, which prompted his love of languages.

answer :  Fourier was the first to introduce Champollion to a variety of languages, which prompted his love of languages.

Explanation:

Jean-Baptiste Fourier's influence on Jean-Francois Champollion is that Fourier was the first to introduce Champollion to a variety of languages, which prompted his love of languages. also Fourier was a mathematician  who pleaded on the behalf of Champollion when he was about to be adopted into the military.

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A student standing on a stationary skateboard tosses a textbook with a mass of mb = 1.5 kg to a friend standing in front of him.
RSB [31]

Answer:

a) v_s=\frac{m_bv_bcos\theta}{m_c}

b)  v_s=0.0385 m/s

c) P_e= 1.952 kg.m/s

Explanation:

Given:

Mass of the skateboard, m_b = 1.5 kg

Combined mass, m_c = 105 kg

velocity of the book, v_b = 2.97 m/s

angle, θ = 26°

thus,

horizontal component of the velocity = vcosθ

vertical component of velocity = vsinθ

a) Applying the concept of conservation of momentum

m_bv_bcos\theta=m_cv_s

where,

v_s is the velocity of the the student

thus,

v_s=\frac{m_bv_bcos\theta}{m_c}      ............(1)

b) on substituting the values, in the equation we get the magnitude

v_s=\frac{1.5\times2.97cos26^o}{105}

or

v_s=0.0385 m/s

c) Now, the momentum (P) transferred is given as:

P = mass × velocity

on substituting the values, we get

P_e= mass × velocity

or

P_e= m_b\times v_b\times \sin\theta

on substituting the values, we get

P_e= 1.5\times 2.97\times \sin26^o

or

P_e= 1.952 kg.m/s

5 0
3 years ago
A hot iron ball is dropped into 200.0 g of cool water. The water temperature increases by 2.0 C and the temperature of the ball
Bess [88]
0.2 KG, just search the question and it's on yahoo answers 
7 0
2 years ago
When air expands adiabatically (without gaining or losing heat), its pressure P and volume V are related by the equation PV1.4=C
DiKsa [7]

Answer:

\frac{dV}{dt}=21.21cm^3/min

Explanation:

We are given that

PV^{1.4}=C

Where C=Constant

\frac{dP}{dt}=-7KPa/minute

V=420 cubic cm and P=99KPa

We have to find the rate at which the  volume increasing at this instant.

Differentiate w.r.t t

V^{1.4}\frac{dP}{dt}+1.4V^{0.4}P\frac{dV}{dt}=0

Substitute the values

(420)^{1.4}\times (-7)+1.4(420)^{0.4}(99)\frac{dV}{dt}=0

1.4(420)^{0.4}(99)\frac{dV}{dt}=(420)^{1.4}\times (7)

\frac{dV}{dt}=\frac{(420)^{1.4}\times (7)}{1.4(420)^{0.4}(99)}

\frac{dV}{dt}=21.21cm^3/min

5 0
3 years ago
Read 2 more answers
What did Thomson’s model of the atom include that Dalton’s model did not have?
astraxan [27]

Answer: Option (d) is the correct answer.

Explanation:

Dalton's model of atom states that every matter is made up of atoms and these atoms are indivisible in nature.

On the other hand, Thomson's model of atom states that there are small particles present in an atom that has positive or negative charges.

Thomson's model of atom is also known as plum pudding model where negatively charged particles are represented by plum and positively charged particles are represented by pudding.

Thus, we can conclude that Thomson’s model of the atom include smaller particles that Dalton’s model did not have.

4 0
3 years ago
Read 2 more answers
A missile is moving 1350 m/s at a 25.0° angle. It needs to hit a target 23,500 m away in a 55.0° direction in 10.20 s. What is d
QveST [7]

Answer:

  The target's velocity is about 1320 m/s in the direction 265.7°.

Explanation:

In order for there to be a collision between missile and target, we must have ...

  (target starting position) + (target movement) = (missile movement)

assuming the missile starts from the origin of all measurements. The missile moves 10.2 seconds before impact, so moves a distance of ...

  (10.2 s)(1350 m/s) = 13,770 m

__

We are interested in the target movement, so we can solve for that:

  (target movement) = (missile movement) - (target starting position)

In terms of meters, this is ...

  (target movement) = 13770∠25° - 23500∠55° ≈ 13467.74∠-94.3°

The target covers this distance in the same 10.2 seconds before collision, so its speed is (13467.74 m)/(10.2 s) ≈ 1320.4 m/s.

As a positive angle, the target's direction is ...

  -94.3° +360° = 265.7°

The direction of the target's velocity is 265.7°.

_____

If you're calculating this by hand, there are a couple of ways you can do it. You can convert to rectangular coordinates and back (perhaps least confusing), or you can use the law of cosines to solve the triangle, then translate angles back to the x-y coordinate plane.

Using rectangular coordinates, we have ...

  13770∠25° = 13770(cos(25°), sin(25°)) ≈ (12479.9, 5819.45)

  23500∠55° = 23500(cos(55°), sin(55°)) ≈ (13479.0, 19250.1)

Then the difference is ...

  (12479.9, 5819.45) -(13479.0, 19250.1) ≈ (-999.188, -13430.6)

and the (3rd-quadrant) angle is ...

  target direction = arctan(-13430.6/-999.188) ≈ -94.3° = 265.7°

__

The target's speed is found by dividing the distance it covers by the time it takes.

  √(13430.6² +999.188²)/10.2 ≈ 1320.36 . . . m/s

3 0
3 years ago
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