Question 8 of 11

To convert degrees Fahrenheit (F) into degrees Celsius (C) use the formula 2003-05-04-00-00_files/i0150000.jpg. Rewrite the equa

mihalych1998 [28]

To covert degrees Fahrenheit (F) to degrees Celsius (C) we use the equation C = (F - 32) × 5/9 . To convert degrees Celsius (C) to degrees Fahrenheit (F) we have to rewrite the equation.
First step: Divide both sides by 5/9. The equation would look like this:
C(9/5) = (F - 32)

Second step: Add 32 on both sides.
32 + (9/5)C = F or F = 32 + (9/5)C

6 0

3 years ago

Is it possible to have a triangle with angles of 62, 104 and 110

NISA [10]

Answer:

No, all angles in any triangle must add up to 180.

5 0

3 years ago

Read 2 more answers

which equation, written in standard form correctly represents this scenario and indicates what the variables represent?

pishuonlain [190]

The answer is the third option.

The explanation is shown below:

1. You must keep on mind the information given in the exercise:

- She bough 3 pounds of coffee.

- She bought 2 pounds of chocolate.

- She spent a total of $24.

2. Therefore, you can call the price per pound of coffee $x$ and the price per pound of chocolate $y$ .

3. When you multiply 3 pounds by the price per pound of coffee, you obtain the amount of money she spent for 3 pounds. When you multiply 2 pounds by the price per pound of chocolate, you obtain the amount of money she spent for 2 pounds. If you add both amounts, you obtain the total spent, which was $24.

4. Therefore, you can express this as following:

$3x+2y=24$

5 0

4 years ago

Factor each completely? A2-7a-18

alisha [4.7K]

$a^{2} -7a-18$ (-9 + 2= -7, -9*2= -18)

(a - 9)(a + 2)

6 0

3 years ago

In a circus performance, a monkey is strapped to a sled and both are given an initial speed of 3.0 m/s up a 22.0° inclined track

Aloiza [94]

Answer:

Approximately $0.31\; \rm m$ , assuming that $g = 9.81\; \rm N \cdot kg^{-1}$ .

Step-by-step explanation:

Initial kinetic energy of the sled and its passenger:

$\begin{aligned}\text{KE} &= \frac{1}{2}\, m \cdot v^{2} \\ &= \frac{1}{2} \times 14\; \rm kg \times (3.0\; \rm m\cdot s^{-1})^{2} \\ &= 63\; \rm J\end{aligned}$ .

Weight of the slide:

$\begin{aligned}W &= m \cdot g \\ &= 14\; \rm kg \times 9.81\; \rm N \cdot kg^{-1} \\ &\approx 137\; \rm N\end{aligned}$ .

Normal force between the sled and the slope:

$\begin{aligned}F_{\rm N} &= W\cdot \cos(22^{\circ}) \\ &\approx 137\; \rm N \times \cos(22^{\circ}) \\ &\approx 127\; \rm N\end{aligned}$ .

Calculate the kinetic friction between the sled and the slope:

$\begin{aligned} f &= \mu_{k} \cdot F_{\rm N} \\ &\approx 0.20\times 127\; \rm N \\ &\approx 25.5\; \rm N\end{aligned}$ .

Assume that the sled and its passenger has reached a height of $h$ meters relative to the base of the slope.

Gain in gravitational potential energy:

$\begin{aligned}\text{GPE} &= m \cdot g \cdot (h\; {\rm m}) \\ &\approx 14\; {\rm kg} \times 9.81\; {\rm N \cdot kg^{-1}} \times h\; {\rm m} \\ & \approx (137\, h)\; {\rm J} \end{aligned}$ .

Distance travelled along the slope:

$\begin{aligned}x &= \frac{h}{\sin(22^{\circ})} \\ &\approx \frac{h\; \rm m}{0.375}\end{aligned}$ .

The energy lost to friction (same as the opposite of the amount of work that friction did on this sled) would be:

$\begin{aligned} & - (-x)\, f \\ = \; & x \cdot f \\ \approx \; & \frac{h\; {\rm m}}{0.375}\times 25.5\; {\rm N} \\ \approx\; & (68.1\, h)\; {\rm J}\end{aligned}$ .

In other words, the sled and its passenger would have lost (approximately) $((137 + 68.1)\, h)\; {\rm J}$ of energy when it is at a height of $h\; {\rm m}$ .

The initial amount of energy that the sled and its passenger possessed was $\text{KE} = 63\; {\rm J}$ . All that much energy would have been converted when the sled is at its maximum height. Therefore, when $h\; {\rm m}$ is the maximum height of the sled, the following equation would hold.

$((137 + 68.1)\, h)\; {\rm J} = 63\; {\rm J}$ .

Solve for $h$ :

$(137 + 68.1)\, h = 63$ .

$\begin{aligned} h &= \frac{63}{137 + 68.1} \approx 0.31\; \rm m\end{aligned}$ .

Therefore, the maximum height that this sled would reach would be approximately $0.31\; \rm m$ .

7 0

3 years ago