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vlada-n [284]
3 years ago
9

How solar panels are effective for sprayers?​

Physics
1 answer:
vichka [17]3 years ago
6 0

Answer:

Solar pesticide sprayer can give less tariff or price in effective spraying. Solar energy is absorbed by the solar panel which contains photovoltaic cells. ... This converted energy utilizes to store the voltage in the DC battery and that battery further used for driving the spray pump.

Explanation:

here is your answer Hope you will enjoy and mark me as brainlist

thank you

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A car starts from rest and travels for 5.8 s with a uniform acceleration of 1.6 m/s² in the negative direction. What is the fina
elena-s [515]

Answer:

Final velocity of the car will be -9.28 m/sec        

Explanation:

We have given that the car starts from the rest so initial velocity of the car u = 0 m /sec

Acceleration of the car a=1.6m/sec^2 in negative direction so acceleration will be a=-1.6m/sec^2

From first equation of motion we know that

v = u+at

So v=0+(-1.6)\times 5.8=-9.28m/sec

So final velocity will be -9.28 m/sec

8 0
3 years ago
an empty density bottle weighed 23.5 and 48.4g when filled with water .The empty bottle was partially filled with sand until and
gavmur [86]

Answer:

Density of Sand is 2.653g/cm^{3}.

Explanation:

Given Empty Density bottle weighs 23.5gm(W=23.5gm)

Weight of bottle when completely filled water=48.4gm

So amount of water required to fill the bottle=Weight of bottle filled with water-W

Amount of water required to fill the bottle(w_{max})=48.4gm-23.5gm

w_{max}=24.9g

Since we know density of water d_{w} =1g/cm^{3} and w_{max}

We can calculate volume of empty space in the bottle(V).

w_{max}=d_{w}\timesV

V=\frac{w_{max} }{d_{w} }

V=\frac{24.9}{1}

V=24.9 g/cm^{3}

Now bottle is partially filled with sand,and weight of bottle is (w_{s})36.5gm

So,

Amount of sand added (m_{s})=36.5-Weight of the bottle

m_{s}=13g

After filling the bottle with water again,the weight of the bottle becomes (W_{2}=56.5g)

Therefore,

amount of water added to the bottle of sand in grams = W_{2}-36.5gm

amount of water added =56.5g-36.5g

amount of water added =20g

As the density of water = 1g/cm^{3}

Amount of water (in grams )=Volume of water occupied

20=volume of water added

Therfore volume of water added to the sand filled bottle(V_{w})=20cm^{3}

As we know the total volume of the water bottle(V),

Volume of the sand occupied in the water bottle=V-V_{w}

V_{s}=24.9g-20g

V_{s}=4.9g

We know,

Density=Mass/Volume

Therefore,

density of sand = \frac{m_{s} }{V_{s} }

density of sand =\frac{13}{4.9}

density of sand = 2.653g/cm^{3}

8 0
2 years ago
Through which material does the light travel the fastest?
eimsori [14]
The answer is A because it gives you a straight line which would make it easy for you go just go to end fast as a car .
6 0
3 years ago
Equations E = 1 2πε0 qd z3 and E = 1 2πε0 P z3 are approximations of the magnitude of the electric field of an electric dipole,
tamaranim1 [39]

Answer:

The ratio of E_{app} and E_{act} is 0.9754

Explanation:

Given that,

Distance z = 4.50 d

First equation is

E_{act}=\dfrac{qd}{2\pi\epsilon_{0}\times z^3}

E_{act}=\dfrac{Pz}{2\pi\epsilon_{0}\times (z^2-\dfrac{d^2}{4})^2}

Second equation is

E_{app}=\dfrac{P}{2\pi\epsilon_{0}\times z^3}

We need to calculate the ratio of E_{act} and E_{app}

Using formula

\dfrac{E_{app}}{E_{act}}=\dfrac{\dfrac{P}{2\pi\epsilon_{0}\times z^3}}{\dfrac{Pz}{2\pi\epsilon_{0}\times (z^2-\dfrac{d^2}{4})^2}}

\dfrac{E_{app}}{E_{act}}=\dfrac{(z^2-\dfrac{d^2}{4})^2}{z^3(z)}

Put the value into the formula

\dfrac{E_{app}}{E_{act}}=\dfrac{((4.50d)^2-\dfrac{d^2}{4})^2}{(4.50d)^3\times4.50d}

\dfrac{E_{app}}{E_{act}}=0.9754

Hence, The ratio of E_{app} and E_{act} is 0.9754

8 0
3 years ago
High-level clouds usually occur above 20,000 ft. Which is the best description of the clouds’ appearance?
PSYCHO15rus [73]
D. Thin and composed of Ice crystals
8 0
3 years ago
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