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Question
Math
Posted 7 months ago

*5. 一长方体水箱的长为 14 cm14 \mathrm{~cm}, 宽为 10 cm10 \mathrm{~cm} 且里面装有 34\frac{3}{4} 满的水。小华丢入一颗体积为 390 cm3390 \mathrm{~cm}^{3} 的石头进入水缸, 水缸内的水位增加直到 40 ml40 \mathrm{~m} l 的水溢出。问在丢入石头前, 水的深度为多少 cm\mathrm{cm} ?
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Answer from Sia
Posted 7 months ago
Solution by Steps
step 1
计算水箱的体积。水箱的长为 14cm14 \, \text{cm},宽为 10cm10 \, \text{cm},深度为 hcmh \, \text{cm}。水箱的体积公式为 V=length×width×heightV = \text{length} \times \text{width} \times \text{height}
step 2
计算水箱中水的体积。水箱中装有 34\frac{3}{4} 满的水,所以水的体积为 Vwater=34×14×10×hV_{\text{water}} = \frac{3}{4} \times 14 \times 10 \times h
step 3
石头的体积为 390cm3390 \, \text{cm}^3
step 4
水溢出的体积为 40ml40 \, \text{ml},即 40cm340 \, \text{cm}^3
step 5
由于石头的体积导致水溢出,所以水箱中水的体积增加了 390cm340cm3=350cm3390 \, \text{cm}^3 - 40 \, \text{cm}^3 = 350 \, \text{cm}^3
step 6
计算水的深度增加。增加的体积 350cm3350 \, \text{cm}^3 等于 14×10×Δh14 \times 10 \times \Delta h,所以 Δh=35014×10=2.5cm\Delta h = \frac{350}{14 \times 10} = 2.5 \, \text{cm}
step 7
设丢入石头前水的深度为 hh,丢入石头后水的深度为 h+2.5cmh + 2.5 \, \text{cm}
step 8
由于水箱中水的体积为 34\frac{3}{4} 满,所以 14×10×h=34×14×10×(h+2.5)14 \times 10 \times h = \frac{3}{4} \times 14 \times 10 \times (h + 2.5)
step 9
解方程 14×10×h=34×14×10×(h+2.5)14 \times 10 \times h = \frac{3}{4} \times 14 \times 10 \times (h + 2.5),得到 h=34×(h+2.5)h = \frac{3}{4} \times (h + 2.5)
step 10
解方程 h=34h+34×2.5h = \frac{3}{4}h + \frac{3}{4} \times 2.5,得到 h34h=34×2.5h - \frac{3}{4}h = \frac{3}{4} \times 2.5,即 14h=34×2.5\frac{1}{4}h = \frac{3}{4} \times 2.5
step 11
计算 h=3×2.5=7.5cmh = 3 \times 2.5 = 7.5 \, \text{cm}
Answer
7.5 cm
Key Concept
体积计算
Explanation
通过计算水箱的体积和水的体积,并考虑石头的体积和溢出的水量,可以求出丢入石头前水的深度。

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