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2 years ago

The sum of the speed of two trains is 722.7 miles per hour. If the speed of the first train is 3.3 mph faster than the second tr

ain find the speeds of each

Mathematics

1 answer:

stellarik [79]2 years ago

7 0

$x-speed\ of\ first\ train\\x-3.3-speed\ of\ second\ train\\\\x+x-3.3=722.7\\x+x=722.7+3.3\\2x=726\\x=363\\\\x-3.3=363-3.3=359.7\\\\Speed\ of\ first\ train\ is\ 363 mph,\ speed\ of\ second\ train\ is\ 359.7.$

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Mr. Gill is draining his pool. The pump he is using changes the water level by −2 1/4 inches per hour. A stronger pump would dra

professor190 [17]

Answer:

-5.625 in/hour

Step-by-step explanation:

−2 1/4 * 2 1/2 = -5.625 in/hour

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3 years ago

7 people are having dinner at a restaurant and decide to split the cheque evenly between them. If the check came to a total of $

DanielleElmas [232]

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The answer will be $54.5

Step-by-step explanation:

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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 $0.31\; \rm m$ .

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2 years ago

What is 21/56 simplified

lozanna [386]

$21=3\cdot7\\56=8\cdot7\\\\\frac{21}{56}=\frac{3\cdot\not7^1}{8\cdot\not7_1}=\boxed{\frac{3}{8}}$

3 0

3 years ago

Read 2 more answers

A motorcycle cost $12,000 when it was purchased. The value of a motorcycle decreases by 6% each year. Find the rate of decay eac

dalvyx [7]

the rate of decay each month can calculated using the following formula:

6%/year × 1 year/(12 months) = 0.5%/month
0.06 x 1/12 = 0.005/month
So the 0.5% is the right answer.
I hope it helped.

4 0

3 years ago