Tafuta kani ya jumla ya chemchemi (effective spring constant) kwa chemchemi zilizounganishwa sambamba.

Kani ya chemchemi moja $k = 1000 \, \text{Nm}^{-1}$.
Kwa chemchemi mbili sambamba, kani ya jumla $k_{jumla} = k_1 + k_2$.

$$k_{jumla} = 1000 \, \text{Nm}^{-1} + 1000 \, \text{Nm}^{-1} = 2000 \, \text{Nm}^{-1}$$

Nguvu $F = mg$, ambapo $m = 0.2 \, \text{kg}$ na $g = 9.8 \, \text{ms}^{-2}$.

$$F = 0.2 \, \text{kg} \times 9.8 \, \text{ms}^{-2} = 1.96 \, \text{N}$$

Sheria ya Hooke inasema $F = k_{jumla}x$.

$$x = \frac{F}{k_{jumla}} = \frac{1.96 \, \text{N}}{2000 \, \text{Nm}^{-1}} = 0.00098 \, \text{m}$$

Urefu wa kunyooshwa ni:

$$\boxed{\text{}0.00098 \, \text{m}}$$

11. Mchoro unaonyesha chupa ya koni yenye maji hadi kiwango kilichoonyeshwa.

a) Eleza mabadiliko katika utulivu wa chupa wakati maji zaidi yameongezwa ndani yake.
Utulivu wa chupa utaongezeka.

b) Toa sababu ya jibu lako katika a) hapo juu.
Kuongeza maji kunapunguza kituo cha mvuto (center of gravity) cha chupa na maji, na kituo cha mvuto cha chini huongeza utulivu.

12. a) Gari lilianza kusonga kwa kasi ya $10 \, \text{ms}^{-1}$ na kuongeza kasi kwa usawa kwa $1 \, \text{ms}^{-2}$ hadi kufikia kasi ya $15 \, \text{ms}^{-1}$. Kokotoa:

i) Muda uliotumika.
Tunatumia fomula ya mwendo: $v = u + at$.
Hapa, $u = 10 \, \text{ms}^{-1}$, $v = 15 \, \text{ms}^{-1}$, na $a = 1 \, \text{ms}^{-2}$.

$$15 \, \text{ms}^{-1} = 10 \, \text{ms}^{-1} + (1 \, \text{ms}^{-2})t$$ $$15 - 10 = t$$ $$t = 5 \, \text{s}$$

Muda uliotumika ni:

$$\boxed{\text{}5 \, \text{s}}$$

ii) Umbali uliosafirishwa wakati wa kuongeza kasi.
Tunatumia fomula ya mwendo: $s = ut + \frac{1}{2}at^2$.
Hapa, $u = 10 \, \text{ms}^{-1}$, $t = 5 \, \text{s}$, na $a = 1 \, \text{ms}^{-2}$.

$$s = (10 \, \text{ms}^{-1})(5 \, \text{s}) + \frac{1}{2}(1 \, \text{ms}^{-2})(5 \, \text{s})^2$$ $$s = 50 \, \text{m} + \frac{1}{2}(1 \, \text{ms}^{-2})(25 \, \text{s}^2)$$ $$s = 50 \, \text{m} + 12.5 \, \text{m}$$ $$s = 62.5 \, \text{m}$$

Umbali uliosafirishwa ni:

$$\boxed{\text{}62.5 \, \text{m}}$$

iii) Kasi iliyofikiwa mita 100 kutoka mahali ambapo kuongeza kasi kulianza.
Tunatumia fomula ya mwendo: $v^2 = u^2 + 2as$.
Hapa, $u = 10 \, \text{ms}^{-1}$, $a = 1 \, \text{ms}^{-2}$, na $s = 100 \, \text{m}$.

$$v^2 = (10 \, \text{ms}^{-1})^2 + 2(1 \, \text{ms}^{-2})(100 \, \text{m})$$ $$v^2 = 100 \, \text{m}^2\text{s}^{-2} + 200 \, \text{m}^2\text{s}^{-2}$$ $$v^2 = 300 \, \text{m}^2\text{s}^{-2}$$ $$v = \sqrt{300} \, \text{ms}^{-1}$$ $$v \approx 17.32 \, \text{ms}^{-1}$$

Kasi iliyofikiwa ni:

$$\boxed{\text{}17.32 \, \text{ms}^{-1}}$$