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3 years ago

People living near wind farms sometimes object to them because of visual pollution and what other

Physics

1 answer:

ddd [48]3 years ago

4 0

Answer:

Wind turbines are a source of clean renewable energy, but some people who live nearby describe the shadow flicker, the audible sounds and the subaudible sound pressure levels as "annoying." They claim this nuisance negatively impacts their quality of life

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On what principle does a bottle opener work

Vsevolod [243]

Answer:

Bottle opener works on a fulcrum.

Explanation:

A bottle opener is a second-class lever because the pivot point is at one end of the opener and the load is in the middle.

8 0

3 years ago

an op amp in unity gain configuration (buffer) with slew rate of 5v/us is used to amplify a sinusoidal signal with a frequency o

ZanzabumX [31]

Answer:

The maximum amplitude ( $V_{max}$ ) will be 7.96 V.

Explanation:

We know, for distortion free operation, the slew rate (S) of an OPAMP is written as

$S = 2 \pi f V_{max}$

where ' $f$ ' is the highest frequency signal.

Therefore, from the above equation we can write,

$&&5 \frac{V}{\mu s} = 2 \pi 100 kHz \times V_{max}\\&or,& V_{max} = \frac{5V}{10^{-6} s \times 2 \pi 100 \times 10^{3} Hz}\\&or,& V_{max} = \frac{5}{2 \pi \times 10^{-1}} V = 7.96 V$

3 0

3 years ago

Newton's second law problem. The spring constant can be determined from the balance of forces in equilibrium. Suppose the spring

mario62 [17]

Answer:

K =6.697 Kg/s²

Explanation:

Given:

delta m =41 g = 0.041 kg

delta x = 6cm = 0.06m

g = 9.8 m/s²

according to the given formula

K = delta m g /delta x

K = (0.041 kg × 9.8 m/s²) / 0.06m

K =6.697 Kg/s²

6 0

3 years ago

Question about generators and electrical currents. Please, no absurd answers! Thank you

Kamila [148]

The first answer should be correct if not then the second one

3 0

3 years ago

Read 2 more answers

A 62.0-kg athlete leaps straight up into the air from a trampoline with an initial speed of 9.6 m/s. The goal of this problem is

pochemuha

Answer:

2856.96 J

$\frac{1}{2}mv_i^2+mgh_i=\frac{1}{2}mv_f^2+mgh_f$

6.78822 m/s

Explanation:

$v_i$ = Initial velocity = 9.6 m/s

g = Acceleration due to gravity = 9.81 m/s²

h = Height

The athlete only interacts with the gravitational potential energy. Air resistance is neglected.

At height y = 0

Kinetic energy

$K=\frac{1}{2}mv^2\\\Rightarrow K=\frac{1}{2}\times 62\times 9.6^2\\\Rightarrow K=2856.96\ J$

At height y = 0 the potential energy is 0 as

$P=mgy\\\Rightarrow P=mg0=0$

At maximum height her velocity becomes 0 so the kinetic energy becomes zero.

As the the potential and kinetic energy are conserved

The general equation

$K_i+P_i=K_f+P_f\\\Rightarrow \frac{1}{2}mv_i^2+mgh_i=\frac{1}{2}mv_f^2+mgh_f$

Half of maximum height

$\\\Rightarrow mgh_i+\frac{1}{2}mv_f^2=mg\frac{h_i}{2}+\frac{1}{2}mv^2\\\Rightarrow gh_i=g\frac{h_i}{2}+\frac{1}{2}v^2\\\Rightarrow g\frac{h_i}{2}=\frac{1}{2}v^2\\\Rightarrow v=\sqrt{gh}$

$h_i=\frac{v_i^2}{2g}$

$v=\sqrt{gh}\\\Rightarrow v=\sqrt{g\times \frac{v_i^2}{2g}}\\\Rightarrow v=\sqrt{\frac{v_i^2}{2}}\\\Rightarrow v=\sqrt{\frac{9.6^2}{2}}\\\Rightarrow v=6.78822\ m/s$

The velocity of the athlete at half the maximum height is 6.78822 m/s

8 0

3 years ago