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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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What is the speed of a car traveling 100 m in 20 s?

Ludmilka [50]

Answer: To answer this question, we will need the following equation: SPEED = DISTANCE/TIME (A multiplication and division triangle will be shown)i) The speed of the car is calculated by doing 100 metres/ 20 seconds which gives us 5 metres per second. ii) Rearranging the equation earlier, we can make the distance the subject of the equation so that we get SPEED x TIME = DISTANCE. We worked out the speed and the time was given as 1 minute 40 seconds but we cannot plug in the numbers yet as the time has to be converted to units of seconds (because our speed is in meters per second). 1 minute 40 seconds = 60 seconds + 40 seconds = 100 secondsWe then plug in the numbers to get the distance travelled = 5 metres per second x 100 seconds = 500 metres.

Explanation:

3 0

3 years ago

The photeselestric effect is observed when light of a sufficiently high frequency is focused onto a polished metal surface, emit

Helga [31]

Answer:

$3.4\cdot 10^{-19} J$

Explanation:

In order to convert the work function of cesium from electronvolts to Joules, we must use the following conversion factor:

$1 eV = 1.6 \cdot 10^{-19} J$

In our problem, the work function of cesium is

$E=2.1 eV$

so, we can convert it into Joules by using the following proportion:

$1 eV : 1.6\cdot 10^{-19} J = 2.1 eV : x\\x=\frac{(1.6\cdot 10^{-19} J)(2.1 eV)}{1 eV}=3.4\cdot 10^{-19} J$

8 0

3 years ago

A positively charged particle Q1 = +35nC is held fixed at the origin. A second charge of mass m = 3.5ug is floating a distance d

cupoosta [38]

Answer:

Explanation:

Q1 = 35 nC = 35 x 10^-9 C

m = 3.5 micro gram = 3.5 x 10^-9 Kg

d = 35 cm = 0.35 m

(a) The electrostatic force between the two charges is balanced by the weight of another charge.

F = m g

$\frac{1}{4\pi \epsilon _{0}}\frac{Q_{1}Q_{2}}{d^{2}}=mg$

$Q_{2}=\frac{4\pi \epsilon _{0}mgd^{2}}{Q_{1}}$

(b) By substituting the values

$Q_{2}=\frac{3.5\times 10^{-9}\times 9.8\times 0.35\timees 0.35}{9\times 10^{9}\times 35\times 10^{9}}$

Q2 = 13.34 x 10^-12 C

Q2 = 0.0134 nC

4 0

3 years ago

43 kg bear slides, from rest, 15 m down a lodgepole pine tree, moving with a speed of 5.5 m/s just before hitting the ground. (a

olga2289 [7]

Answer:

-6327.45 Joules

650.375 Joules

378.47166 N

Explanation:

h = Height the bear slides from = 15 m

m = Mass of bear = 43 kg

g = Acceleration due to gravity = 9.81 m/s²

v = Velocity of bear = 5.5 m/s

f = Frictional force

Potential energy is given by

$P=mgh\\\Rightarrow P=43\times -9.81\times 15\\\Rightarrow P=-6327.45\ J$

Change that occurs in the gravitational potential energy of the bear-Earth system during the slide is -6327.45 Joules

Kinetic energy is given by

$K=\frac{1}{2}mv^2\\\Rightarrow K=\frac{1}{2}\times 43\times 5.5^2\\\Rightarrow K=650.375\ J$

Kinetic energy of the bear just before hitting the ground is 650.375 Joules

Change in total energy is given by

$\Delta E=fh=-(\Delta K+\Delta P)\\\Rightarrow fh=-(650.375-6327.45)\\\Rightarrow fh=5677.075\\\Rightarrow f=\frac{5677.075}{h}\\\Rightarrow f=\frac{5677.075}{15}\\\Rightarrow f=378.47166\ N$

The frictional force that acts on the sliding bear is 378.47166 N

5 0

3 years ago

Read 2 more answers

A swimming duck paddles the water with its feet once per time interval of 1.6 s, producing surface waves with this period. The d

soldier1979 [14.2K]

Answer:

a. 0.275m/s

b. 1.08m

Explanation:

Since the duck the paddle the water at an interval of 1.6sec, we can determine the frequency of the wave formed using the equation

f=1/T

Where T is the period.

f=1/1.6

f=0.625Hz.

Also from the equation used in determining the speed of a wave

V=fλ,

v=0.625*0.2

v=0.125m/s

in the question it was stated that that the duck produce a wave moving at a speed of 0.40m/s.

Hence the speed of the duck is

v(duck)=0.40-0.125

v(duck)=0.275m/s

b. The distance between the crest behind the duck is the wavelength of the waves.

To determine this, the wavelength is expressed as

λ=v/f

but the speed in this case is the speed of the duck and the surface wave,as this account for the wave speed behind the duck,

Hence we have

λ=(0.40+0.275)/0.625

λ=1.08m.

The wavelength behind the duck is 1.08m

4 0

3 years ago