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1 year ago

What factors might affect how the watermill works?

2 answers:

7 0

<span>Since a watermill is powered by a water wheel the strength of the current and volume of the water passing the mill may alter the amount of power provided. Seasonal and climatic changes could contribute to changes in the current.</span>

erastova [34]1 year ago

4 0

Answer:

Watermill is an energy generating turbine that runs by water. The energy generation of a windmill is dependent upon several factors like efficiency of blades of wind turbine, wind speed and consistency of wind. The energy output of a taller wind turbine is more because wind speeds are greater at higher altitudes.

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For the first 10 seconds a squirrel runs 3 m/s to look for an acorn. The next 5 seconds he eats an acorn that he finds. Afterwar

Gala2k [10]

Distance covered by the squirrel to look for an acorn :

d = ( 3 m/s ) × 10 s = 30 m.

Time taken to eat an Acron is 5 seconds.

Time taken to cover distance of 30 m with 2 m/s speed is :

$T=\dfrac{30}{2}\ s= 15 \ s$

Therefore, total time take to get back to where he started is ( 10+5+15 ) = 30 s.

Hence, this is the required solution.

7 0

1 year ago

Water exits a garden hose at a speed of 1.2 m/s. If the end of the garden hose is 1.5 cm in diameter and you want to make the wa

Katarina [22]

Answer:

A 93%

Explanation:

$P_1=P_2$ = Pressure will be equal at inlet and outlet

$\rho$ = Density of water = 1000 kg/m³

g = Acceleration due to gravity = 9.81 m/s²

$v_1$ = Velocity at inlet = 1.2 m/s

$v_2$ = Velocity at outlet

$r_1$ = Radius of inlet = $\dfrac{1.5}{2}=0.75\ cm$

$r_2$ = Radius of outlet

From Bernoulli's relation

$P_1+\dfrac{1}{2}\rho v_1^2+\rho gh_1=P_2+\dfrac{1}{2}\rho v_2^2+\rho gh_2\\\Rightarrow \dfrac{1}{2}\rho v_1^2+\rho gh_1=\dfrac{1}{2}\rho v_2^2+\rho gh_2\\\Rightarrow v_2=\sqrt{2(\dfrac{1}{2}v_1^2+gh_1-gh_2)}\\\Rightarrow v_2=\sqrt{2(\dfrac{1}{2}1.2^2+9.81\times 15)}\\\Rightarrow v_2=17.19709\ m/s$

From continuity equation

$A_1v_1=A_2v_2\\\Rightarrow \pi r_1^2v_1=\pi r_2^2v_2\\\Rightarrow r_2=\sqrt{\dfrac{r_1^2v_1}{v_2}}\\\Rightarrow r_2=\sqrt{\dfrac{0.0075^2\times 1.2}{17.19709}}\\\Rightarrow r_2=0.00198\ m$

The fraction would be

$\dfrac{A_1-A_2}{A_1}\times 100=\dfrac{r_1^2-r_2^2}{r_1^2}\times 100\\ =\dfrac{0.0075^2-0.00198^2}{0.0075^2}\times 100\\ =93.0304\ \%$

The fraction is 93.0304%

8 0

1 year ago

The researcher requires the force on the wire to point upward with a magnitude of 4.2x10-4 N. The length of the wire that is in

lorasvet [3.4K]

Answer:

The magnetic force can be found by using the following equation:

$F_B = IL\times B$

Let’s first deal with the magnitude, then we will figure out the direction by using the right hand rule.

$4.2\times 10^{-4} = (30\times 10^{-3})\times (50\times10^{-3}) \times B\\4.2 \times 10^{-4} = 1500 \times 10^{-6}\times B\\(28\times 10^{-4})\times 10^2 = B\\B = 28\times 10^{-2}$

The direction of the L is equal to the direction of the current. So, the direction of the L is to the right. The required force is to be pointed upward.

By the right hand rule the magnetic field should be directed towards ‘into the plane’.

Explanation:

In the questions regarding the magnetic field, the cross product and the directions of the vectors are crucial. So, you should use right-hand rule efficiently.

Right-hand rule: $A \times B = C$