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denpristay
1 month ago
14

What factors might affect how the watermill works?

Physics
2 answers:
kicyunya [3.2K]1 month ago
7 0
As a watermill operates through a water wheel, variations in the current's strength and the volume of water flowing past the mill can influence the amount of power generated. Changes in seasons and climate may also affect the current.
Keith_Richards [3.2K]1 month ago
4 0
A watermill is a turbine that generates energy harnessed from water flow. The energy production of a windmill relies on various factors, such as the efficiency of its blades, wind velocity, and the steadiness of the wind. Taller wind turbines generate more energy due to stronger wind currents at elevated heights.
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Simone is walking her dog on a leash. The dog is pulling with a force of 32 N to the right and Simone is pulling backward with a
ValentinkaMS [3465]

Conclusion:

The total net force acting on the objects is 16 N, directed towards the right.

Clarification:

It is stated that,

The force exerted by the dog, F_1 = 32\ N (to the right)

The force exerted by Simone, F_2 = -16\ N (backward)

Here, assume the backward direction is negative and the right direction is positive.

The net force will move in the direction where the larger force is present. The net force can be calculated as:

F=F_1+F_2

F=32+(-16)

F = 16 N

Thus, the net force amounts to 16 N, acting towards the right.

6 0
2 months ago
Read 2 more answers
A helicopter starting on the ground is rising directly into the air at a rate of 25 ft/s. You are running on the ground starting
serg [3582]

Response:

The speed at which the distance from the helicopter to you is changing (in ft/s) after 5 seconds is \sqrt{725} ft/ sec

Clarification:

Provided:

h(t) = 25 ft/sec

x(t) = 10 ft/ sec

h(5) = 25 ft/sec. 5 = 125 ft

x(5) = 10 ft/sec. 5 = 50 ft

At this point, we can determine the distance between the individual and the helicopter utilizing the Pythagorean theorem

D(t) = \sqrt{h^2 + x^2}

Now, let's calculate the derivative of distance in relation to time

\frac{dD}{dt} (t) = \frac{2h \cdot \frac{dh}{dt} +2x \cdot\frac{dx}{dt}} {2\sqrt{h^2 + x^2}}

By plugging in the values for h(t) and x(t) and simplifying, we arrive at,

\frac{dD}{dt}(t) = \frac{50t \cdot \frac{dh}{dt} + 20 \cdot \frac{dx}dt}{2\sqrt{625\cdot t^2 + 100 \cdot t^2}}

\frac{dh}{dt} = 25ft/sec

\frac{dx}{dt} = 10 ft/sec

\frac{Dd}{dt} (t) = \frac{1250t +200t}{2\sqrt{725}t} = \frac{725}{\sqrt{725}} = \sqrt{725} ft / sec

5 0
1 month ago
1200 N-m of torque is used to drive a gear (A) of diameter 25 cm, which in turn drives another gear (B) of diameter 52 cm. What
Ostrovityanka [3204]

Response:

2.5kN.m

Details:

Torque relates directly to the pitch diameter

= Ta/Tb= Da/Db

For 120/Tb= 0.25/0.5

This gives Tb= 2.469kN.m, roughly 2.5kN.m


8 0
2 months ago
Read 2 more answers
Two thin lenses with a focal length of magnitude 12.0cm, the first diverging and the second converging, are located 9.00cm apart
kicyunya [3294]
b ) The first lens is a concave lens with a focal length of f₁ = - 12 cm and an object distance of u = - 20 cm. Using the lens formula, 1 / v - 1 / u = 1 / f, we get 1 / v + 1 / 20 = -1 / 12. This leads to 1 / v = - 1 / 20 - 1 / 12, which simplifies to 1 / v = -0.05 - 0.08333, yielding v = -7.5 cm. Consequently, the first image is formed before the first lens, near the object side, which becomes the object for the second lens with a distance of 16.5 cm from the second lens. c ) For the second lens, object distance is u = -16.5 cm, and focal length f₂ = + 12 cm (convex lens). Using the lens formula leads to 1 / v + 1 / 16.5 = 1 / 12, and this results in 1 / v = 1 / 12 - 1 / 16.5, which simplifies to 1 / v = 0.08333 - 0.0606. Finally, we find v = 44 cm (approximately). This image will be formed on the other side of the convex lens, which is 53 cm from the first lens. Magnification by the first lens is v / u = -7.5 / -20 = 0.375. For the second lens, it is v / u = 44 / - 16.5 = -2.67. d ) The total magnification becomes 0.375 x - 2.67 = - 1.00125. The height of the final image is then calculated as 2.50 mm x 1.00125 = 2.503 mm. e ) The final image will be inverted compared to the object since the total magnification is negative.
6 0
1 month ago
A car accelerates from rest to a velocity of 5 meters/second in 4 seconds. What is its average acceleration over this period of
Ostrovityanka [3204]

The average acceleration is

\bar a=\dfrac{5\,\frac{\mathrm m}{\mathrm s}-0\,\frac{\mathrm m}{\mathrm s}}{4\,\mathrm s}=1.25\,\dfrac{\mathrm m}{\mathrm s^2}

5 0
2 months ago
Read 2 more answers
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