The properties of metals depend mainly on the number and arrangement of neutrons<br> True<br> False

How does a wind turbine transform mechanical energy into electrical energy?

Alex777 [14]

Answer:

"A turbine takes the kinetic energy of a moving fluid, air in this case, and converts it to a rotary motion. As wind moves past the blades of a wind turbine, it moves or rotates the blades. These blades turn a generator."

4 0

2 years ago

A stone is dropped into a river from a bridge at a height h above the water. Another stone is thrown vertically down at a time t

Mumz [18]

Answer:

$v_{y_0} = \frac{\frac{g}{2}t(t - 2\sqrt{\frac{2h}{g}})}{\sqrt{\frac{2h}{g}} - t}$

Explanation:

We will apply the equations of kinematics to both stones separately.

First stone:

Let us denote the time spent after the second stone is thrown as 'T'.

$y - y_0 = v_{y_0}(t+T) + \frac{1}{2}a(t+T)^2\\0 - h = 0 + \frac{1}{2}(-g)(t+T)^2\\(t+T)^2 = \frac{2h}{g}\\T = \sqrt{\frac{2h}{g}}-t$

Second stone:

$y - y_0 = v_{y_0}T + \frac{1}{2}aT^2\\0 - h = v_{y_0}T -\frac{1}{2}gT^2\\-h = v_{y_0}(\sqrt{\frac{2h}{g}} - t) - \frac{g}{2}(\sqrt{\frac{2h}{g}} - t)^2\\-h = v_{y_0}(\sqrt{\frac{2h}{g}} - t) - \frac{g}{2}(\frac{2h}{g} + t^2 - 2t\sqrt{\frac{2h}{g}})\\-h = v_{y_0}\sqrt{\frac{2h}{g}} - v_{y_0}t - h -\frac{g}{2}t^2 + gt\sqrt{\frac{2h}{g}}\\v_{y_0}(\sqrt{\frac{2h}{g}} - t) = \frac{g}{2}t^2 - gt\sqrt{\frac{2h}{g}}\\v_{y_0} = \frac{\frac{g}{2}t(t - 2\sqrt{\frac{2h}{g}})}{\sqrt{\frac{2h}{g}} - t}$

6 0

3 years ago

Read 2 more answers

When scientists observed the light from stars and galaxies, they noticed that their color shifted toward the end of the visible

lakkis [162]

Color shifted towards the end is C red

7 0

3 years ago

Read 2 more answers

An object at 27°C has its temperature increased to 37°C.

Firlakuza [10]

Answer:

b. 14

Explanation:

$T_{i}$ = Initial temperature = 27 °C = 27 + 273 = 300 K

$T_{f}$ = Final temperature = 37 °C = 37 + 273 = 310 K

$P_{i}$ = Initial Power radiated by the object

$P_{f}$ = Final Power radiated by the object

We know that the power radiated is directly proportional to fourth power of the temperature. hence

$\frac{P_{f}}{P_{i}} = \frac{T_{f}^{4} }{T_{i}^{4} }\\\frac{P_{f}}{P_{i}} = \frac{(310)^{4} }{(300)^{4} }\\\frac{P_{f}}{P_{i}} = 1.14\\P_{f} = (1.14) P_{i}$