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2 years ago

Problems with solar energy include _____.

2 answers:

8 0

Third choice: inconsistencies in the availability of the resource
I think this is it. If we come to depend on solar energy, then we're expectedly out of luck at night, and we may unexpectedly be out of luck during long periods of overcast skies.

patriot [66]2 years ago

7 0

The answers are:

the inability of current technology to capture large amounts of the Sun's energy

the inability of current technology to store captured solar energy

inconsistencies in the availability of the resource

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If a wave is traveling at 200 m/s and its wavelength is 0.5 meters, what is the frequency of the

anyanavicka [17]

The frequency of the wave = 400/s

<h3>Further explanation</h3>

Given

v=200 m/s

λ=0.5 m

Required

The frequency of the wave

Solution

Frequency (f): number of waves in one second

The frequency is inversely proportional to the wavelength

$\tt f=\dfrac{v}{\lambda}\\\\f=\dfrac{200~m/s}{0.5~m}\\\\f=400/s$

7 0

2 years ago

A small ball is attached to one end of a spring that has an unstrained length of 0.169 m. The spring is held by the other end, a

Nadya [2.5K]

The ball moves around in circular motion. The centripetal force keeping it in circular motion is given by:

F = mv²/r

F = centripetal force, m = mass of ball, v = velocity of ball, r = radius of motion

The force the spring exerts on the ball is given by:

F = kΔx

F = spring force, k = spring constant, Δx = change of spring length

The spring provides the centripetal force that keeps the ball in circular motion, so set the spring force equal to the centripetal force:

kΔx = mv²/r

Let's do some algebra:

m/k = rΔx/v²

Now if we attach the same spring to the ceiling with the ball still fixed to one end and let the ball hang in static equilibrium, the spring force would balance the ball's weight:

kΔx' = mg

k = spring constant, Δx' = new change in spring length, m = mass, g = gravitational acceleration

Let's do some more algebra:

Δx' = gm/k

Substitute m/k with our previous result, rΔx/v²:

Δx' = grΔx/v²

Given values:

g = 9.81m/s²

r = 0.169m + 0.0131m = 0.1821m

Δx = 0.0131m

v = 3.15m/s

Plug in the values and solve for Δx':

Δx' = 9.81(0.1821)(0.0131)/3.15²

Δx' = 0.0024m

5 0

2 years ago

The pilot of an airplane reads the altitude 6400 m and the absolute pressure 46 kPa when flying over a city. Calculate the local

olga nikolaevna [1]

Answer:

1. 6.672 kPa

2. 49.05 mm of mercury

Explanation:

h = 6400 m

Absolute pressure, p = 46 kPa = 46000 Pa

density of air, d = 0.823 kg/m^3

density of mercury, D = 13600 kg/m^3

(a) Absolute pressure = Atmospheric pressure + pressure due to height

46000 = Atmospheric pressure + h x d x g

Atmospheric pressure = 46000 - 6400 x 0.823 x 10 = 6672 Pa = 6.672 kPa

(b) To convert the pressure into mercury pressure

Atmospheric pressure = H x D x g

Where, H is the height of mercury, D be the density of mercury, g be the acceleration due to gravity

6672 = H x 13600 x 10

H = 0.04905 m

H = 49.05 mm of mercury

4 0

2 years ago

Math hard for me now <br> Help me please please help I bean do it a lot

wlad13 [49]

5 is a prime number because it only has two factors which are 1 and 5.

4 0

1 year ago

Remember to identify all your data, write the equation, and show your work.

Paraphin [41]

Answer:

I believe it's 8.09 seconds, but I'm rusty on my physics.

Explanation:

The equation for solving the time it takes for an object to fall is $\sqrt{2d/g}$

So multiply the distance times 2, and you get 642 meters. Then you divide by gravities acceleration constant, 9.8, and you get 65.51. Finally, $\sqrt{65.51}$ , and you get 8.09 seconds.

I pulled the equation off of wikipedia and I'm unsure if it's the correct one, so hopefully this is correct. :/

6 0

2 years ago