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3 years ago

Problems with solar energy include ____.

Physics

2 answers:

lana66690 [7]3 years ago

8 0

The inability of current technology to store captured solar energy

MArishka [77]3 years ago

5 0

You have to save up some of what you get during the daytime, because you never get much at night.

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Pls help will mark brain

BartSMP [9]

Answer: 100 units

Explanation:

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3 years ago

5. Describe the relationship between frequency and wavelength.

Flura [38]

Frequency is the number of waves that passes at a given point every second. The lower the frequency, the less waves there are. The higher the frequency, the more waves. A wavelength is the distance between two frequency waves. The higher the frequency, the less distance from two waves. The lower the frequency, the longer distance from two waves.

8 0

3 years ago

Read 2 more answers

A Ferris wheel with radius 14.0 m is turning about a horizontal axis through its center. The linear speed of a passenger on the

Marina86 [1]

Answer:

a) The acceleration experimented by the passenger when she passes through the lowest point of her circular motion is: $a_{R} = 2.571\,\frac{m}{s^{2}}$ , $\angle = 90^{\circ}$ .

b) Hence, the acceleration experimented by the passenger when she passes through the highest point of her circular motion is: $a_{R} = 2.571\,\frac{m}{s^{2}}$ , $\angle = 270^{\circ}$ .

c) The Ferris wheel takes 14.646 seconds to make a revolution.

Explanation:

a) An object that rotates at constant angular velocity reports a centripetal acceleration and no tangential acceleration. When passenger passes through the lowest point in her circular motion, centripetal acceleration goes up to the center.

In addition, centripetal acceleration is determined by the following expression:

$a_{R} = \frac{v^{2}}{R}$ (Eq. 1)

Where:

$a_{R}$ - Centripetal acceleration, measured in meters per square second.

$v$ - Linear speed, measured in meters per second.

$R$ - Radius of the Ferris wheel, measured in meters.

If we know that $v = 6\,\frac{m}{s}$ and $R = 14\,m$ , the magnitude of radial acceleration is:

$a_{R} = \frac{\left(6\,\frac{m}{s} \right)^{2}}{14\,m}$

$a_{R} = 2.571\,\frac{m}{s^{2}}$

The acceleration experimented by the passenger when she passes through the lowest point of her circular motion is: $a_{R} = 2.571\,\frac{m}{s^{2}}$ , $\angle = 90^{\circ}$ .

b) In the highest point the magnitude of radial acceleration is the same but direction is the opposed to that at lowest point. That is, centripetal acceleration goes down to the center.

Hence, the acceleration experimented by the passenger when she passes through the highest point of her circular motion is: $a_{R} = 2.571\,\frac{m}{s^{2}}$ , $\angle = 270^{\circ}$ .

c) At first we need to calculate the angular velocity of the Ferris wheel ( $\omega$ ), measured in radians per second, by using the following expression:

$\omega = \frac{v}{R}$ (Eq. 2)

If we know that $v = 6\,\frac{m}{s}$ and $R = 14\,m$ , then the angular velocity of the Ferris wheel is:

$\omega = \frac{6\,\frac{m}{s} }{14\,m}$

$\omega = 0.429\,\frac{rad}{s}$

Now we proceed to obtain the period of the Ferris wheel ( $T$ ), measured in seconds, which is the time needed by that wheel to make on revolution:

$T = \frac{2\pi}{\omega}$ (Eq. 3)

Where $\omega$ is the angular velocity of the Ferris wheel, measured in radians per second.

If we get that $\omega = 0.429\,\frac{rad}{s}$ , then:

$T = \frac{2\pi}{0.429\,\frac{rad}{s} }$

$T = 14.646\,s$

The Ferris wheel takes 14.646 seconds to make a revolution.

8 0

3 years ago

A) A 5.75 mL sample of mercury has a measured mass of 77.05 g. The density is ___________.

Kitty [74]

Answer:

Density of the sample will be 13.4 kg/L

Explanation:

We have given volume of the sample $V=5.75mL=5.75\times 10^{-3}L$

Mass of the sample $M=77.05gram =77.05\times 10^{-3}kg$

We have to find the density of the sample

Density of the sample is given by

$Density=\frac{mass}{volume}=\frac{77.05\times 10^{-3}}{5.75\times 10^{-3}}=13.4kg/L$

So density of the sample will be 13.4 kg/L

4 0

4 years ago

What portion of the electromagnetic spectrum is most absorbed by the chlorophyll pigments in the thylakoid membranes?

nadya68 [22]

Answer: The visible part of the electromagnetic spectrum, specifically blue and red wavelength

Explanation:

Let's begin by explaining that pigments generally absorb more light than they reflect (they absorb certain wavelengths and reflect others). Therefore, the color that a given object seems to have depends on which parts of the visible electromagnetic spectrum are reflected and which parts are absorbed.

In this sense, in the thylakoid membranes there are two types of chlorophyll pigments:

-clorophyll b that absorbs the blue light of the electromagnetic spectrum

-clorophyll a that bsorbs the red light of the electromagnetic spectrum

That is why we generally see plants in green color and not in other colors.

4 0

3 years ago