Hello i need help with this!!

A wind turbine takes in energy from wind with the goal of converting it into electrical energy. Much of the wind energy is also

kondor19780726 [428]

The efficiency of the turbine is 50%.

The given parameters:

Kinetic energy of the wind, E = 3000 J
Output electrical energy, E(out) = 750 J
Energy lost to heat, E(lost) = 750 J
Kinetic energy of the turbine, E(in) = ?

The kinetic energy of the turbine which is the input energy is calculated as follows;

E(in) = 3000 - (750 + 750)

E(in) = 1500 J

The efficiency of the turbine is calculated as follows;

$E_f_f = \frac{E_{(out)}}{E_{(in)}} \times 100\%\\\\E_f_f = \frac{750}{1500} \times 100\%\\\\E_f_f = 50 \%$

Thus, the efficiency of the turbine is 50%.

Learn more about efficiency of turbine here: brainly.com/question/2009210

5 0

3 years ago

What is the equation describing the motion of a mass on the end of a spring which is stretched 8.8 cm from equilibrium and then

Sedbober [7]

Answer:

$x=(0.088m)\cos(\sqrt{\frac{k}{m} } t)$

Explanation:

We first identify the elements of this simple harmonic motion:

The amplitude A is 8.8cm, because it's the maximum distance the mass can go away from the equilibrium point. In meters, it is equivalent to 0.088m.

The angular frequency ω can be calculated with the formula:

$\omega =\sqrt{\frac{k}{m}}$

Where k is the spring constant and m is the mass of the particle.

Now, since the spring starts stretched at its maximum, the appropriate function to use is the positive cosine in the equation of simple harmonic motion:

$x=A\cos(\omega t)$

Finally, the equation of the motion of the system is:

$x=(0.088m)\cos(\omega t)$

or

$x=(0.088m)\cos(\sqrt{\frac{k}{m} } t)$

7 0

4 years ago

A cosmic ray muon with mass mμ = 1.88 ✕ 10−28 kg impacting the Earth's atmosphere slows down in proportion to the amount of matt

anyanavicka [17]

Answer:

a. the magnitude of the force experienced by the muon is 2.55 × 10⁻¹⁹N

b. this force compare to the weight of the muon; the force is 1.38 × 10⁸ greater than muon

Explanation:

F= ma

v²=u² -2aS

(1.56 ✕ 10⁶)²=(2.40 ✕ 10⁶)²-2a(1220)

a=1.36×10⁹m/s²

recall

F=ma

F = 1.88 ✕ 10⁻²⁸ kg × 1.36×10⁹m/s²

F= 2.55 × 10⁻¹⁹N

the magnitude of the force experienced by the muon is 2.55 × 10⁻¹⁹N

b. this force compare to the weight of the muon

F/mg= 2.55 × 10⁻¹⁹/ (1.88 ✕ 10⁻²⁸ × 9.8)

= 1.38 × 10⁸

6 0

3 years ago

A wire of length 6cm makes an angle of 20° with a 3 mT

Crazy boy [7]

Answer:

Approximately $7.3 \times 10^{-3}\; \rm A$ (approximately $7.3\; \rm mA$ ) assuming that the magnetic field and the wire are both horizontal.

Explanation:

Let $\theta$ denote the angle between the wire and the magnetic field.

Let $B$ denote the magnitude of the magnetic field.

Let $l$ denote the length of the wire.

Let $I$ denote the current in this wire.

The magnetic force on the wire would be:

$F = B \cdot l \cdot I \cdot \sin(\theta)$ .

Because of the $\sin(\theta)$ term, the magnetic force on the wire is maximized when the wire is perpendicular to the magnetic field (such that the angle between them is $90^\circ$ .)

In this question:

$\theta = 20^\circ$ (or, equivalently, $(\pi / 9)$ radians, if the calculator is in radian mode.)
$B = 3\; \rm mT = 3 \times 10^{-3}\; \rm T$ .
$l = 6\; \rm cm = 6 \times 10^{-2}\;\rm m$ .
$F = 1.5\times 10^{-4}\; \rm N$ .

Rearrange the equation $F = l \cdot I \cdot \sin(\theta)$ to find an expression for $I$ , the current in this wire.

$\begin{aligned} I &= \frac{F}{l \cdot \sin(\theta)} \\ &= \frac{3\times 10^{-3}\; \rm T}{6 \times 10^{-2}\; \rm m \times \sin \left(20^{\circ}\right)} \\ &\approx 7.3 \times 10^{-3}\; \rm A = 7.3 \; \rm mA\end{aligned}$ .

5 0

3 years ago

During the day, susan notices that the wind is blowing onshore at the beach. What is this called? land breeze land breeze sea br

In-s [12.5K]

The correct option is (b) Sea breeze

During the day, susan notices that the wind is blowing onshore at the beach called sea breeze.

What is sea breeze?

Any wind that flows from a big body of water onto or onto a landmass is called a sea breeze or an onshore breeze.
Sea breezes form as a result of changes in air pressure brought on by the different heat capacities of water and dry land. Sea breezes are therefore more confined than prevailing winds.
A sea wind is frequently seen along coasts after sunrise because land warms up far more quickly than water does when exposed to solar radiation.
The sea wind front is significant because it can serve as a catalyst for afternoon thunderstorms and provide welcome cooling along the coast.

Learn more about the sea breeze with the help of the given link:

brainly.com/question/13015619

#SPJ4

8 0

2 years ago

Hello i need help with this!!​

Hello i need help with this!!