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Rom4ik [11]
3 years ago
7

4 NaF + Br2 + Al(H3Li)2 -->

Physics
1 answer:
Agata [3.3K]3 years ago
5 0

Answer:

Molecules are substances that contain two or more elements, if these elements are different, then the substance is also a compound.

However, considering only the definition of a molecule, it is evident that both CH4 and NaF are molecules as they contain more than one element.

Explanation:

hope this helps!!

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When a resistor with resistance R is connected to a 1.50-V flashlight battery, the resistor consumes 0.0625 W of electrical powe
Grace [21]

Explanation:

the answer is in the above image

8 0
3 years ago
Suppose a 49-N sled is resting on packed snow. The coefficient of kinetic friction is 0.11. If a person weighing 585 N sits on t
Andre45 [30]

Answer:

69.74 N

Explanation:

We are given that

Weight of sled=49 N

Coefficient of kinetic friction,\mu_k=0.11

Weight of person=585 N

Total weight==mg=49+585=634 N

We know that

Force needed to pull the sled across the snow at constant speed,F=Kinetic friction

F=\mu_k N

Where N= Normal=mg

F=0.11\times 634=69.74 N

Hence, the force is needed  to pull the sled across the snow at constant speed=69.74 N

7 0
3 years ago
Law of Conservation of Energy.
Tems11 [23]
Energy cannot be created nor be destroyed
3 0
3 years ago
Write the equation of a function h(t) that represents the amount of heat in joules required to heat the bar to a temperature of
bearhunter [10]
The initial temperature of the bar is 25. To get to the t temperature you need to add (t-25) degrees Celsius.

for 1 degree................... 7 Joules
      y given degree........  p Joules

p=7y

In our case y=(t-25) .

h(t) = 7(t-25) which is the final answer.

8 0
3 years ago
Three charges lie along the x axis. The positive charge q1 = 15 μC is at x= 2.0 m and the positive charge q2 = 6.0μC is at the o
Delicious77 [7]

Answer:

x = 0.775m

Explanation:

Conceptual analysis

In the attached figure we see the locations of the charges. We place the charge q₃ at a distance x from the origin. The forces F₂₃ and F₁₃ are attractive forces because the charges have an opposite sign, and these forces must be equal so that the net force on the charge q₃ is zero.

We apply Coulomb's law to calculate the electrical forces on q₃:

F_{23} = \frac{k*q_{2}*q_{3}}{x^2} (Electric force of q₂ over q₃)

F_{13} = \frac{k*q_{1}*q_{3}}{(2-x)^2} (Electric force of q₁ over q₃)

Known data

q₁ = 15 μC = 15*10⁻⁶ C

q₂ = 6 μC = 6*10⁻⁶ C

Problem development

F₂₃ = F₁₃

\frac{k*q_{2}*q_{3}}{x^2} = \frac{k*q_{1}*q_{3}}{(2-x)^2} (We cancel k and q₃)

\frac{q_2}{x^2}=\frac{q_1}{(2-x)^2}

q₂(2-x)² = q₁x²

6×10⁻⁶(2-x)² = 15×10⁻⁶(x)² (We cancel 10⁻⁶)

6(2-x)² = 15(x)²

6(4-4x+x²) = 15x²

24 - 24x + 6x² = 15x²

9x² + 24x - 24 = 0

The solution of the quadratic equation is:

x₁ = 0.775m

x₂ = -3.44m

x₁ meets the conditions for the forces to cancel in q₃

x₂ does not meet the conditions because the forces would remain in the same direction and would not cancel

The negative charge q₃ must be placed on x = 0.775 so that the net force is equal to zero.

6 0
3 years ago
Read 2 more answers
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