All categories

1 month ago

PART D

Physics

1 answer:

ValentinkaMS [3.4K]1 month ago

5 0

Answer:

Solution:

D = -4/7
C = 17/7

Explanation:

To solve the system via substitution, proceed as follows:

1. Choose the variable with the smallest coefficient and isolate it on one side.
2. Divide both sides by that variable's coefficient.
3. Substitute this expression into the other equation.
4. Solve for the remaining variable.

Applying these steps to the system:

3C + 4D = 5

2C + 5D = 2

Step 1: Isolate the variable with the smaller coefficient:

2C = 2 - 5D

Step 2: Divide both sides by 2

C = (2 - 5D)/2

Step 3: Substitute into the first equation

3 * (2 - 5D)/2 + 4D = 5

Step 4: Solve the equation

Apply distributive property: multiply 3 by each term inside the parentheses

(6 - 15D)/2 + 4D = 5

Divide each term within the parentheses by 2:

3 - (15D)/2 + 4D = 5

Move 3 to the right side by subtracting:

-(15D)/2 + 4D = 5 - 3

Combine like terms:

(-15D + 8D) / 2 = 2

-7D = 4

D = -4/7

Plug D back into the equation for C:

C = (2 - 5D)/2 = (2 - 5(-4/7))/2

C = (2 + 20/7)/2

C = (14/7 + 20/7)/2

C = 34/7 / 2

C = 34/14

C = 17/7

Final solution:

D = -4/7
C = 17/7

You might be interested in

Calculate the average charge on arginine when ph=9.20. (hint : find the average charge for each ionizable group and sum these to

Sav [3153]

Arginine is classified as a basic amino acid since it has two amino groups alongside a single acid group. At a low pH level, all ionizable groups are protonated. As the pH rises slightly, the acid group loses its proton. When the pH increases further, one of the amino groups also loses a proton. At considerably high pH levels, none of the ionizable groups remain protonated.

Pkas

pka1 = 1.82 pka2 = 8.99 pka3 = 12.48 Thus, 9.20 is above the second pKa and below the third pKa. This indicates that the acid has already lost its proton, as has one of the amino groups, while the second amino group remains protonated. When an acid is not protonated, it carries a negative charge. An unprotonated amino group is neutral, whereas when protonated, the amino group bears a positive charge. Therefore, this amino acid exhibits one positive charge (from one of the amino groups) and one negative charge (from the acid), resulting in an overall neutral charge.

4 0

10 days ago

An airplane flies with a velocity of 55.0 m/s [35o N of W] with respect to the air (this is known as air speed). If the velocity

ValentinkaMS [3465]

V - wind speed;
53° - 35° = 18°
v² = 55² + 40² - 2 · 55 · 40 · cos 18°
v² = 3025 + 1600 - 2 · 55 · 40 · 0.951
v² = 440.6
v = √440.6
v = 20.99 ≈ 21 m/s
Conclusion: The wind speed calculates to 21 m/s.

5 0

15 days ago

Each shot of the laser gun most favored by Rosa the Closer, the intrepid vigilante of the lawless 22nd century, is powered by th

Yuliya22 [3333]

U = 1794.005 × 10⁶ J. Explanation: Information provided indicates that the capacitance of the original capacitor is C = 1.27 F, and the potential difference applied to it is V = 59.9 kV, or 59.9 × 10³ V. The potential energy (U) for the capacitor is determined by the formula: U = (1/2) × C × V². Substituting the respective values, we find U = (1/2) × 1.27 × (59.9 × 10³)², resulting in U = 1794.005 × 10⁶ J.

7 0

13 days ago

Three pendulums with strings of the same length and bobs of the same mass are pulled out to angles θ1, θ2, and θ3, respectively,

Sav [3153]

Answer:

All three pendulums will have the same angular frequencies.

Explanation:

For a simple pendulum, the time period using the approximation $sin(\theta )\approx \theta$ is expressed as:

$T=2\pi\sqrt{\frac{l}{g}}$

The angular frequency $\omega$ is defined as

$\omega =\frac{2\pi }{T}\\\\\omega =\frac{2\pi}{2\pi \sqrt{\frac{l}{g}}}\\\\\therefore \omega =\sqrt{\frac{g}{l}}$

Since the angular frequency remains unaffected by the initial angle (valid strictly for small angle approximations), we deduce that the angular frequencies of the three pendulums are identical.

6 0

24 days ago

A spin bike has a flywheel in two parts—a 12.5 kg disk with radius 0.23 m, and a 7.0 kg ring with mass concentrated at the outer

Keith_Richards [3271]

Response:

Clarification:

Provided

weight of disk $m=12.5 kg$