– The diameter of the hose $d_1 = 9.6 text{ cm}$

– The speed of water through the hose $v_1 = 1.3 ; frac{text{m}}{text{s}}$

– The final speed of the water $v_2 = 21 ; frac{text{m}}{text{s}}$

Using the Equation of Continuity we can find the required diameter $d_2$:

$$
begin{equation*}
v_1 , A_1 = v_2 , A_2
end{equation*}
$$

Where the areas are found as follows:

$$
begin{align*}
A_1 &= left(frac{d_1}{2} right)^2 , pi \
A_2 &= left(frac{d_2}{2} right)^2 , pi
end{align*}
$$

So we have:

$$
begin{align*}
v_1 , d_1^2 = v_2 , d_2^2
end{align*}
$$

rearranging for $d_2$ we have:

$$
begin{align*}
d_2 = d_1 cdot sqrt{frac{v_1}{v_2}}
end{align*}
$$

Plugging in the values we have:

$$
begin{align*}
d_2 &= 9.6text{ cm} cdot sqrt{frac{1.3 ; frac{text{m}}{text{s}}}{21 ; frac{text{m}}{text{s}} }} \
d_2 &approx 2.39text{ cm} = 2.4text{ cm}
end{align*}
$$

– The diameter of the hose $d_1 = 3.4text{ cm}$

– The speed of water through the hose $v_1 = 1.1 ; frac{text{m}}{text{s}}$

– The diameter of the nozzle $d_2 = 0.57text{ cm}$

Using the Equation of Continuity we can find the required speed $v_2$:

$$
begin{equation*}
v_1 , A_1 = v_2 , A_2
end{equation*}
$$

Where the areas are found as follows:

$$
begin{align*}
A_1 &= left(frac{d_1}{2} right)^2 , pi \
A_2 &= left(frac{d_2}{2} right)^2 , pi
end{align*}
$$

So we have:

$$
begin{align*}
v_1 , d_1^2 = v_2 , d_2^2
end{align*}
$$

rearranging for $v_2$ we have:

$$
begin{align*}
v_2 = v_1 , left(frac{d_1}{d_2} right)^2
end{align*}
$$

Plugging in the values we have:

$$
begin{align*}
v_2 &= 1.1 ; frac{text{m}}{text{s}} cdot left(frac{3.4text{ cm}}{0.57text{ cm}} right)^2 \
v_2 &= 39.14 ; frac{text{m}}{text{s}}
end{align*}
$$