All you need to do is count the number of Oxygen the Products:
In 6H2O, you have 6 oxygen, and in 6CO2, you have 6 Dioxygen so 12 Oxygen. So a total of 12 + 6 = 18 atoms of Oxygen, or 9 atoms of dioxygen.
So the coefficient that should be placed in front of O2 to balance the reaction is 9:
C6H12 + 9O2 ---> 6H2O + 6CO2
Hope this Helps! :)
Answer:
The resistance for the second resistor is R2 = 240 Ohms and the equivalent resistance is Requivalent = 280 Ohms.
Explanation:
The resistance of a ohmic resistor is influeced by the type of it's material and by the it's construction. The longer the wire the greater the resistance and the greater the cross-sectional the lower the resistance. This can be expressed by the following equation:
R = (p*L)/A
Where p is a constant for the material of the resistor, L is the length of the wire and A is the area of the cross-sectional. In our case we have a resistor R1 that has a resistance of 40 Ohms, while a second resistor R2 made with the same material but with double length and half cross sectional. If we say that R1 is:
R1 = (p*L)/A
Then R2 must be:
R2 = (p*3*L)/(A/2)
Because the only things that changed were the length and area of the cross-sectional. We can now relate both resistors to find the second resistance, using the equation for R2. So we have:
R2 = [3*(p*L)/A]*2 = 6*(p*L)/A = 6*R1
We know that R1 is 40 Ohms so R2 = 6*40 = 240 Ohms.
The equivalent resistance of a series connection is the sum of the individual resistances, so we have:
Requivalent = R1 + R2 = 40 + 240 = 280 Ohms.
Explanation:
Formula for calculating the area of a rectangle A = Length *width
For statement A;
Given area of a rectangle with measured length = 2.536 mm and width = 1.4 mm.
Area of the rectangle = 2.536mm * 1.4mm
Area of the rectangle = 3.5504mm²
The rule of significant figures states that we should always convert the answer to the least number of significant figure amount the given value in question. Since 1.4mm has 2 significant figure, hence we will convert our answer to 2 significant figure.
Area of the rectangle = 3.6mm² (to 2sf)
For statement B;
Given area of a rectangle with measured length = 2.536 mm and width = 1.41 mm.
Area of the rectangle = 2.536mm * 1.41mm
Area of the rectangle = 3.57576mm²
Similarly, Since 1.41mm has 3 significant figure compare to 2.536 that has 4sf, hence we will convert our answer to 3 significant figure.
Area of the rectangle = 3.58mm² (to 3sf)
Based on the conversion, it can be seen that 3.6mm² is greater than 3.58mm², hence the area of rectangle in statement A is greater than the area of the rectangle in statement B.
Answer:
B. τ = 16 Nm
Explanation:
In order to find the torque exerted by the weight attached to the heel of man's foot, when his leg is stretched out. We use following formula:
τ = Fd
here,
τ = Torque = ?
F = Force exerted by the weight = Weight = mg
F = mg = (4 kg)(10 m/s²) = 40 N
d = distance from knee to weight = 40 cm = 0.4 m
Therefore,
τ = (40 N)(0.4 m)
<u>B. τ = 16 Nm</u>
Answer:
-209.42J
Explanation:
Here is the complete question.
A balky cow is leaving the barn as you try harder and harder to push her back in. In coordinates with the origin at the barn door, the cow walks from x = 0 to x = 6.9 m as you apply a force with x-component Fx=−[20.0N+(3.0N/m)x]. How much work does the force you apply do on the cow during this displacement?
Solution
The work done by a force W = ∫Fdx since our force is variable.
Since the cow moves from x₁ = 0 m to x₂ = 6.9 m and F = Fx =−[20.0N+(3.0N/m)x] the force applied on the cow.
So, the workdone by the force on the cow is
W = ∫₀⁶°⁹Fx dx = ∫₀⁶°⁹−[20.0N+(3.0N/m)x] dx
= ∫₀⁶°⁹−[20.0Ndx - ∫₀⁶°⁹(3.0N/m)x] dx
= −[20.0x]₀⁶°⁹ - [3.0x²/2]₀⁶°⁹
= -[20 × 6.9 - 20 × 0] - [3.0 × 6.9²/2 - 3.0 × 0²/2]
= -[138 - 0] - [71.415 - 0] J = (-138 - 71.415) J
= -209.415 J ≅ -209.42J