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nirvana33 [79]
3 years ago
7

s person is deciding whether to install solar panels on her house. lt and explain three pieces of info she needs to consider to

make her decision
Physics
1 answer:
Marrrta [24]3 years ago
8 0
Efficiency- How much energy is wasted and how much useful energy is produced
Money- How expensive to install them
Location- She has to consider the location where she places them so that there is enough sunlight and light energy for the solar panels, obviously she should have solar panels if the location receives a sufficient amount of sunlight
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A car accelerates from rest for 8.0 s, and reaches a speed of 66 m/s. How far will the
PSYCHO15rus [73]

Answer: C

Explanation:

Find the acceleration using this kinematic equation:

v_f=v_i+at\\66=0+a*8\\a=8.25m/s^2

Now use this kinematic equation to find the displacement:

v_f^2=v_i^2+2a*d\\66^2=0^2+2(8.25)d\\4356=16.5d\\d=264m

3 0
3 years ago
Brainliest please help<br><br>tell me if am right <br>if not correct me <br><br><br>​please
REY [17]

Answer:

See the answers below

Explanation:

To solve this problem we must use the following equation of kinematics.

v_{f}=v_{o}+a*t

where:

Vf = final velocity [m/s]

Vo = initial velocity [m/s]

a = acceleration [m/s²]

t = time [s]

<u>First case</u>

Vf = 6 [m/s]

Vo = 2 [m/s]

t = 2 [s]

6=2+a*2\\4=2*a\\a=2[m/s^{2} ]

<u>Second case</u>

Vf = 25 [m/s]

Vo = 5 [m/s]

a = 2 [m/s²]

25=5+2*t\\t = 10 [s]

<u>Third case</u>

Vo =4 [m/s]

a = 10 [m/s²]

t = 2 [s]

v_{f}=4+10*2\\v_{f}=24 [m/s]

<u>Fourth Case</u>

Vf = final velocity [m/s]

Vo = initial velocity [m/s]

a = acceleration [m/s²]

t = time [s]

v_{f}=5+8*10\\v_{f}=85 [m/s]

<u>Fifth case</u>

Vf = final velocity [m/s]

Vo = initial velocity [m/s]

a = acceleration [m/s²]

t = time [s]

8=v_{o}+4*2\\v_{0}=8-8\\v_{o}=0

8 0
3 years ago
Why does the large number of hydrogen atoms in the universe suggest that other elements?
lidiya [134]

Answer:

Explanation:

The abundance of the chemical elements is a measure of the occurrence of the chemical elements relative to all other elements in a given environment. Abundance is measured in one of three ways: by the mass-fraction (the same as weight fraction); by the mole-fraction (fraction of atoms by numerical count, or sometimes fraction of molecules in gases); or by the volume-fraction. Volume-fraction is a common abundance measure in mixed gases such as planetary atmospheres, and is similar in value to molecular mole-fraction for gas mixtures at relatively low densities and pressures, and ideal gas mixtures. Most abundance values in this article are given as mass-fractions.

For example, the abundance of oxygen in pure water can be measured in two ways: the mass fraction is about 89%, because that is the fraction of water's mass which is oxygen. However, the mole-fraction is about 33% because only 1 atom of 3 in water, H2O, is oxygen. As another example, looking at the mass-fraction abundance of hydrogen and helium in both the Universe as a whole and in the atmospheres of gas-giant planets such as Jupiter, it is 74% for hydrogen and 23–25% for helium; while the (atomic) mole-fraction for hydrogen is 92%, and for helium is 8%, in these environments. Changing the given environment to Jupiter's outer atmosphere, where hydrogen is diatomic while helium is not, changes the molecular mole-fraction (fraction of total gas molecules), as well as the fraction of atmosphere by volume, of hydrogen to about 86%, and of helium to 13%.[Note 1]

The abundance of chemical elements in the universe is dominated by the large amounts of hydrogen and helium which were produced in the Big Bang. Remaining elements, making up only about 2% of the universe, were largely produced by supernovae and certain red giant stars. Lithium, beryllium and boron are rare because although they are produced by nuclear fusion, they are then destroyed by other reactions in the stars.[1][2] The elements from carbon to iron are relatively more abundant in the universe because of the ease of making them in supernova nucleosynthesis. Elements of higher atomic number than iron (element 26) become progressively rarer in the universe, because they increasingly absorb stellar energy in their production. Also, elements with even atomic numbers are generally more common than their neighbors in the periodic table, due to favorable energetics of formation.

The abundance of elements in the Sun and outer planets is similar to that in the universe. Due to solar heating, the elements of Earth and the inner rocky planets of the Solar System have undergone an additional depletion of volatile hydrogen, helium, neon, nitrogen, and carbon (which volatilizes as methane). The crust, mantle, and core of the Earth show evidence of chemical segregation plus some sequestration by density. Lighter silicates of aluminum are found in the crust, with more magnesium silicate in the mantle, while metallic iron and nickel compose the core. The abundance of elements in specialized environments, such as atmospheres, or oceans, or the human body, are primarily a product of chemical interactions with the medium in which they reside.

4 0
3 years ago
A blacksmith heats a 1,540 g iron horseshoe to a temperature of 1445°c before dropping it into 4,280 g of water at 23.1°c. if th
Marta_Voda [28]
Given:
m₁ = 1540 g, mass of iron horseshoe
T₁ = 1445 °C, initial temperature of horseshoe
c₁ = 0.4494 J/(g-°C), specific heat

m₂ = 4280 g, mass of water
T₂ = 23.1 C, initial temperature of water
c₂ = 4.18 J/(g-°C), specific heat of water
L = 947,000 J heat absorbed by the water.

Let the final temperature be T °C.
For energy balance,
m₁c₁(T₁ - T) = m₂c₂(T - T₂) + L
(1540 g)*(0.4494 J/(g-C))*(1445-T C) = (4280 g)*(4.18 J/(g-C))*(T-23.1 C) + 947000 J
692.076(1445 - T) = 17890(T - 23.1) + 947000
10⁶ - 692.076T = 17890T - 413259 + 947000
466259 = 18582.076T
T = 25.09 °C

Answer: 25.1 °C
3 0
2 years ago
The desperate contestants on a TV survival show are very hungry. The only food they can see is some fruit hanging on a branch hi
evablogger [386]

Answer:

a) v = 18.86 m / s, b)  h = 8.85 m

Explanation:

a) For this exercise we can use the conservation of energy relations.

Starting point. Like the compressed spring

          Em₀ = K_e + U = ½ k x² + m g x

the zero of the datum is placed at the point of the uncompressed spring

Final point. With the spring if compress

           Em_f = K = ½ m v²

how energy is conserved

          Em₀ = Em_f

          ½ k x² + m g x = ½ m v²

   

           v² = \frac{k}{m}  x² + 2gx

let's reduce the magnitudes to the SI system

          m = 500 g = 0.500 kg

          x = -45 cm = -0.45 m

the negative sign is because the distance in below zero of the reference frame

       

let's calculate

           v² = \frac{900}{0.500}  0.45² + 2 9.8 (- 0.45)

           v = √355.68

           v = 18.86 m / s

b) For this part we use the conservation of energy with the same initial point and as an end point at the point where the rock stops

           Em_f = U = m g h

           Em₀ = Em_f

            ½ k x²2 + m g x = m g h

            h = ½  \frac{k}{g}   x² + x

let's calculate

           h = \frac{1}{2} \ \frac{900}{9.8 } \ 0.45^2 - 0.45

           h = 8.85 m

measured from the point where the spring is uncompressed

7 0
2 years ago
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