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Murljashka [212]
2 years ago
11

About 50 000 years ago, in an area located outside Flagstaff, Arizona, a giant

Physics
1 answer:
Leto [7]2 years ago
7 0

Answer:

F_A =5.625*10^1^6N

Explanation:

From the question we are told that

50,000 years ago,

A giant 4.5 107-kg meteor

180-m-deep hole

20,000 m/s

Generally for this problem the energy change is given as

\triangle E=\frac{1}{2} mv^2 +mgd△E=

2

1

mv

2

+mgd

Having the potential and kinetic energy in place

Mathematically solving for Average forceF_AF

A

\triangle E=F_a*d△E=F

a

∗d

F_A =\frac{1/2* 4.5*10^7(20,000)^2-kg+4.5*10^7*9.81*160}{160}F

A

=

160

1/2∗4.5∗10

7

(20,000)

2

−kg+4.5∗10

7

∗9.81∗160

Therefore Average force F_AF

A

is given by

F_A =5.625*10^1^6N

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fgiga [73]

Answer: 2. Solution A attains a higher temperature.

Explanation: Specific heat simply means, that amount of heat which is when supplied to a unit mass of a substance will raise its temperature by 1°C.

In the given situation we have equal masses of two solutions A & B, out of which A has lower specific heat which means that a unit mass of solution A requires lesser energy to raise its temperature by 1°C than the solution B.

Since, the masses of both the solutions are same and equal heat is supplied to both, the proportional condition will follow.

<em>We have a formula for such condition,</em>

Q=m.c.\Delta T.....................................(1)

where:

  • \Delta T= temperature difference
  • Q= heat energy
  • m= mass of the body
  • c= specific heat of the body

<u>Proving mathematically:</u>

<em>According to the given conditions</em>

  • we have equal masses of two solutions A & B, i.e. m_A=m_B
  • equal heat is supplied to both the solutions, i.e. Q_A=Q_B
  • specific heat of solution A, c_{A}=2.0 J.g^{-1} .\degree C^{-1}
  • specific heat of solution B, c_{B}=3.8 J.g^{-1} .\degree C^{-1}
  • \Delta T_A & \Delta T_B are the change in temperatures of the respective solutions.

Now, putting the above values

Q_A=Q_B

m_A.c_A. \Delta T_A=m_B.c_B . \Delta T_B\\\\2.0\times \Delta T_A=3.8 \times \Delta T_B\\\\ \Delta T_A=\frac{3.8}{2.0}\times \Delta T_B\\\\\\\frac{\Delta T_{A}}{\Delta T_{B}} = \frac{3.8}{2.0}>1

Which proves that solution A attains a higher temperature than solution B.

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¿El salario es un costo fijo o variable?
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3 years ago
You have a 100 ohm resistor. How
sp2606 [1]

Answer:

R2 = 300 Ohms

Explanation:

Let the two resistors be R1 and R2 respectively.

RT is the total equivalent resistance.

Given the following data;

R1 = 100 Ohms

RT = 75 Ohms

To find R2;

Mathematically, the total equivalent resistance of resistors connected in parallel is given by the formula;

RT = \frac {R1*R2}{R1 + R2}

Substituting into the formula, we have;

75 = \frac {100*R2}{100 + R2}

Cross-multiplying, we have;

75 * (100 + R2) = 100R2

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allsm [11]
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Convert these to units of moles by dividing the molar masses.

Tin (Sn)= 5.28 g / 118.71 g/mol = 0.044 mol
Fluorine (F) = 3.37 g / 19.00 g/mol = 0.177 mol

Divide both by the least number of moles of the two.


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Therefore, the empirical formula would be:
SnF4
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