All categories

3 years ago

Which of these is a possible disadvantage of using wind as a source of

Physics

2 answers:

vitfil [10]3 years ago

4 0

Answer:

D D D D D D D D D D D D D D D D D D D D D D D D D D D D D D D D D DD

Explanation:

gizmo_the_mogwai [7]3 years ago

3 0

Answer:

D. Wind turbines take up a lot of space.

Explanation:

In wind turbines the kinetic energy received by the air molecules is converted into electrical energy by the use of turbines

So here in order to get more kinetic energy from air we need more crossectional area of the wind mill to interact with the air

So here we need the large size of turbines

so this is the main disadvantage of the wind turbines because it needs large area to install the whole setup also the efficiency of this turbine is small so it needs large number of wind mills to setup good output power

so correct answer will be

D. Wind turbines take up a lot of space.

You might be interested in

What happens to the particles in water as the water is heated and turns to vapor

lions [1.4K]

Hey There!

Your answer is that the particles will move more faster!

In a IceCube the particles are badly moving in a liquid they move more but not fast and in this case as the particles are heating changing into a gas the particles are moving faster!

If you need anymore help just ask me!

Hope this helps!

6 0

3 years ago

Read 2 more answers

Express 79 m in units of (a) centimeters

dem82 [27]

Answer: a) 7,00 centimeters

(b) 259. 19 feet

(d) 0.049 miles

Explanation:

(a) We know that 1 meter = 100 centimeters

Therefore,

$79\ m= 7,900\text{ centimeters}$

(b)Since 1 meter = 3.28084 feet

Then, $79\ m= 79\times3.28084=259.18636\approx 259.19\text{ feet}$

(c) Since, 1 feet = 12 inches.

$79\ m=259.19\text{ feet}=259.19\times12=3110.28\text{ inches}$

(d) $\text{Since 1 feet= }\dfrac{1}{5280}\text{ mile}$

$79\ m=259.19\text{ feet}=\dfrac{259.19}{5280}= 0.0490890151515\approx0.049\text{ miles}$

8 0

3 years ago

A mass of 5 kg stretches a spring 20 cm. The mass is acted on by an external force of 10 sin t 6 N (newtons) and moves in a medi

nordsb [41]

Answer:

u" + 40u' + 49u = 2 sin(t/6)

upp + 40up + 49u = 2 sin(t/6)

Explanation:

Step 1: Data given

mass = 5 kg

L = 20 cm = 0.2 m

F = 10 sin(t/6)N

Fd(t) = - 6 N

u(0) = 0.03 m/s

u(0) = 0

u'(0) = 3 cm/s

Step 2:

ω =kL

k = ω/L = m*g /L = (5*9.8)/0.2 = 245 kg/s²

Since Fd(t) = -γu'(t) we know:

γ =- Fd(t) / u'(t) = 6N/ 0.03 m/s = 200 Ns/m

The initial value problem which describes the motion of the mass is given by

5u" + 200u' + 245u = 10 sin(t/6) u(0) = 0 ; u'(0) = 0.03

This is equivalent to:

u" + 40u' + 49u = 2 sin(t/6) u(0) = 0 ; u'(0) = 0.03

upp + 40up + 49u = 2 sin(t/6)

With u in m and t in s

5 0

3 years ago

Does a resistor resist current or voltage

KIM [24]

The answer is current

4 0

3 years ago

An object is placed 18 cm in front of spherical mirror.if the image is formed at 4cm to the right of the mirror, calculate it's

ivolga24 [154]

1) Focal length

We can find the focal length of the mirror by using the mirror equation:
$\frac{1}{f}= \frac{1}{d_o}+ \frac{1}{d_i}$ (1)
where
f is the focal length
$d_o$ is the distance of the object from the mirror
$d_i$ is the distance of the image from the mirror

In this case, $d_o = 18 cm$ , while $d_i=-4 cm$ (the distance of the image should be taken as negative, because the image is to the right (behind) of the mirror, so it is virtual). If we use these data inside (1), we find the focal length of the mirror:
$\frac{1}{f}= \frac{1}{18 cm}- \frac{1}{4 cm}=- \frac{7}{36 cm}$
from which we find
$f=- \frac{36}{7} cm=-5.1 cm$

2) The mirror is convex: in fact, for the sign convention, a concave mirror has positive focal length while a convex mirror has negative focal length. In this case, the focal length is negative, so the mirror is convex.

3) The image is virtual, because it is behind the mirror and in fact we have taken its distance from the mirror as negative.

4) The radius of curvature of a mirror is twice its focal length, so for the mirror in our problem the radius of curvature is:
$r=2f=2 \cdot 5.1 cm=10.2 cm$

3 0

4 years ago