It is called weathering. there are three different types of weathering, physical weathering, chemical weathering, and biological weathering. Your particular question is caused by chemical weathering.
Elemental hydrogen (H, element 1), nitrogen (N, element 7), oxygen (O, element 8), fluorine (F, element 9), and chlorine (Cl, element 17) are all gases at room temperature, and are found as diatomic molecules.
Titanium, Lead, Potassium, and Silicon are solids at normal temperature.
Mercury and bromine (Br) are the only liquid elements at standard pressure and temperature (Hg).
<h3>What are chemical elements?</h3>
A chemical element is a species of atoms, including the pure substance made entirely of that species, that have a specific number of protons in their nuclei. Chemical elements, in contrast to chemical compounds, cannot be reduced by any chemical process into simpler molecules.
To learn more about chemical elements visit:
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Answer:
1.24 C
Explanation:
We know that the magnitude of the induced emf, ε = -ΔΦ/Δt where Φ = magnetic flux and t = time. Now ΔΦ = Δ(AB) = AΔB where A = area of coil and change in magnetic flux = Now ΔB = 0 - 0.750 T = -0.750 T, since the magnetic field changes from 0.750 T to 0 T.
The are , A of the circular loop is πD²/4 where D = diameter of circular loop = 16.7 cm = 16.7 × 10⁻²m
So, ε = -ΔΦ/Δt = -AΔB/Δt= -πD²/4 × -0.750 T/Δt = 0.750πD²/4Δt.
Also, the induced emf ε = iR where i = current in the coil and R = resistance of wire = ρl/A where ρ = resistivity of copper wire =1.68 × 10⁻⁸ Ωm, l = length of wire = πD and A = cross-sectional area of wire = πd²/4 where d = diameter of wire = 2.25 mm = 2.25 × 10⁻³ m.
So, ε = iR = iρl/A = iρπD/πd²/4 = 4iρD/d²
So, 4iρD/d² = 0.750πD²/4Δt.
iΔt = 0.750πD²/4 ÷ 4iρD/d²
iΔt = 0.750πD²d²/16ρ.
So the charge Q = iΔt
= 0.750π(Dd)²/16ρ
= 0.750π(16.7 × 10⁻²m 2.25 × 10⁻³ m)²/16(1.68 × 10⁻⁸ Ωm)
= 123.76 × 10⁻² C
= 1.2376 C
≅ 1.24 C
Answer:
Explanation:
Calories to be burnt = 3500 - 2500 = 1000 Cals .
Efficiency of conversion to mechanical work is 25 % .
Work needed to burn this much of Cals = 1000 x 100 / 25 = 4000 Cals.
4000 Cals = 4.2 x 4000 = 16800 J .
Work done in one jump = kinetic energy while jumping
= 1/2 m v²
= .5 x 70 x 3.3²
= 381.15 J .
Number of jumps required = 16800 / 381.15
= 44 .
Answer:
C. It is the result of diffraction
Explanation:
I just took the test
