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b). All the water is poured into a cylindrical container of circular radius height of 35cm, calculate the height of the cylinder above the water with water.

Mathematics

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2 parts
a

Vw=325607 cm3≈4651.43 cm3V_w = \frac{32560}{7} \text{ cm}^3 \approx 4651.43 \text{ cm}^3

b

Habove=11.875 cmH_{above} = 11.875 \text{ cm}

Step-by-step solution

5 steps

Sawa blessed, hebu tutatue swali hili la jiometri.

Part a· 3 steps

**Kokotoa ujazo wa maji kwenye ndoo (frustum).

Ndoo ni frustum yenye:
Kipenyo cha juu (RR) = 8 cm
Kipenyo cha chini (rr) = 6 cm
Urefu wa frustum (hfh_f) = 30 cm

Ili kukokotoa ujazo wa frustum, tunaweza kutumia fomula ya ujazo wa koni kubwa kutoa ujazo wa koni ndogo iliyokatwa.
Kwanza, tunahitaji kupata urefu wa koni kubwa (HLH_L) na koni ndogo (HSH_S).
Kwa kutumia pembetatu zinazofanana:

RHL=rHS\frac{R}{H_L} = \frac{r}{H_S} 8HL=6HS\frac{8}{H_L} = \frac{6}{H_S} 8HS=6HL8H_S = 6H_L HS=68HL=34HLH_S = \frac{6}{8}H_L = \frac{3}{4}H_L

Tunajua kuwa urefu wa frustum ni HL−HS=30H_L - H_S = 30 cm.
Badilisha HSH_S:

HL−34HL=30H_L - \frac{3}{4}H_L = 30 14HL=30\frac{1}{4}H_L = 30 HL=120 cmH_L = 120 \text{ cm}

Sasa tafuta HSH_S:

HS=34×120=90 cmH_S = \frac{3}{4} \times 120 = 90 \text{ cm}

  1. Kokotoa ujazo wa koni kubwa.

    VL=13πR2HLV_L = \frac{1}{3}\pi R^2 H_L VL=13×227×(8 cm)2×(120 cm)V_L = \frac{1}{3} \times \frac{22}{7} \times (8 \text{ cm})^2 \times (120 \text{ cm}) VL=13×227×64×120V_L = \frac{1}{3} \times \frac{22}{7} \times 64 \times 120 VL=22×64×407=563207 cm3V_L = \frac{22 \times 64 \times 40}{7} = \frac{56320}{7} \text{ cm}^3

  2. Kokotoa ujazo wa koni ndogo.

    VS=13πr2HSV_S = \frac{1}{3}\pi r^2 H_S VS=13×227×(6 cm)2×(90 cm)V_S = \frac{1}{3} \times \frac{22}{7} \times (6 \text{ cm})^2 \times (90 \text{ cm}) VS=13×227×36×90V_S = \frac{1}{3} \times \frac{22}{7} \times 36 \times 90 VS=22×12×907=237607 cm3V_S = \frac{22 \times 12 \times 90}{7} = \frac{23760}{7} \text{ cm}^3

  3. Kokotoa ujazo wa maji (ujazo wa frustum).

    Vw=VL−VSV_w = V_L - V_S Vw=563207−237607V_w = \frac{56320}{7} - \frac{23760}{7} Vw=56320−237607=325607 cm3V_w = \frac{56320 - 23760}{7} = \frac{32560}{7} \text{ cm}^3 Vw=325607 cm3≈4651.43 cm3\boxed{V_w = \frac{32560}{7} \text{ cm}^3 \approx 4651.43 \text{ cm}^3}

Part b

Habove=11.875 cmH_{above} = 11.875 \text{ cm}

**Maji yote yamemiminwa kwenye chombo cha cylindrical chenye kipenyo cha duara 8 cm na urefu wa 35 cm. Kokotoa urefu wa silinda juu ya maji.

Chombo cha cylindrical kina:
Kipenyo (rcr_c) = 8 cm
Urefu (HcH_c) = 35 cm

  1. Kokotoa urefu wa maji (hwh_w) kwenye silinda.

    Ujazo wa maji (VwV_w) ni sawa na ujazo wa frustum uliokokotolewa hapo juu.

    Vw=πrc2hwV_w = \pi r_c^2 h_w 325607=227×(8 cm)2×hw\frac{32560}{7} = \frac{22}{7} \times (8 \text{ cm})^2 \times h_w 325607=227×64×hw\frac{32560}{7} = \frac{22}{7} \times 64 \times h_w

    Kuzidisha pande zote mbili kwa 7:

    32560=22×64×hw32560 = 22 \times 64 \times h_w 32560=1408×hw32560 = 1408 \times h_w hw=325601408h_w = \frac{32560}{1408} hw=23.125 cmh_w = 23.125 \text{ cm}

  2. Kokotoa urefu wa silinda juu ya maji.

    Urefu juu ya maji = Urefu wa silinda - Urefu wa maji

    Habove=Hc−hwH_{above} = H_c - h_w Habove=35 cm−23.125 cmH_{above} = 35 \text{ cm} - 23.125 \text{ cm} Habove=11.875 cmH_{above} = 11.875 \text{ cm} Habove=11.875 cm\boxed{H_{above} = 11.875 \text{ cm}}
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