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Ilia_Sergeevich [38]
2 years ago
14

During a single-replacement reaction, a more-active metal will replace a less-active metal in a compound. Which single-replaceme

nt reaction will occur? Ni will replace Fe in a compound. Al will replace K in a compound. Mg will replace Ag in a compound. Ca will replace Li in a compound
Physics
1 answer:
son4ous [18]2 years ago
8 0

Answer:

Mg will replace Ag in a compound

Explanation:

A single replacement reaction is driven by the position of ions on the activity series.

As a rule of thumb, the position of metal ions on the activity series determines their reactivity.

Metal ions that are above another are more reactive and they will displace those that are lower.

Generally, activity increases as we go up the group.

Mg ions are higher than Ag ions on the series so, Mg will displace Ag from a solution.

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Name the type of particle formed by covalent bonds
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3 years ago
A 22-turn circular coil of wire has diameter 1.02 m. It is placed with its axis along the direction of the Earth's magnetic fiel
ollegr [7]

Answer:

ξ = 0.00845020162 V or 8.4 mV

Explanation:

Magnetic flux measures the total magnetic field that passes through a known area. Magnetic flux describe the effect of magnetic field in a given area. Mathematically,

magnetic flux (Ф) = BA cos ∅

where

A = test area

B = magnetic field

before the flip

Ф = Bπr²N

N = number of turn

magnitude of induced emf = N |ΔФ/Δt|

ξ  = 2Nπr²B/dt

ξ  = 2 × 22 × π × (1.02/2)² × 0.000047/0.2

ξ = 44 × π × 0.51² × 0.000047/0.2

ξ = 44 × π × 0.2601  × 0.000047/0.2

ξ = 0.0005378868  × 3.142/0.2

ξ = 0.00169004032/0.2

ξ = 0.00845020162 V or 8.4 mV

8 0
3 years ago
For a damped simple harmonic oscillator, the block has a mass of 1.2 kg and the spring constant is 9.8 N/m. The damping force is
ArbitrLikvidat [17]

Answer:

a) t=24s

b) number of oscillations= 11

Explanation:

In case of a damped simple harmonic oscillator the equation of motion is

m(d²x/dt²)+b(dx/dt)+kx=0

Therefore on solving the above differential equation we get,

x(t)=A₀e^{\frac{-bt}{2m}}cos(w't+\phi)=A(t)cos(w't+\phi)

where A(t)=A₀e^{\frac{-bt}{2m}}

 A₀ is the amplitude at t=0 and

w' is the angular frequency of damped SHM, which is given by,

w'=\sqrt{\frac{k}{m}-\frac{b^{2}}{4m^{2}} }

Now coming to the problem,

Given: m=1.2 kg

           k=9.8 N/m

           b=210 g/s= 0.21 kg/s

           A₀=13 cm

a) A(t)=A₀/8

⇒A₀e^{\frac{-bt}{2m}} =A₀/8

⇒e^{\frac{bt}{2m}}=8

applying logarithm on both sides

⇒\frac{bt}{2m}=ln(8)

⇒t=\frac{2m*ln(8)}{b}

substituting the values

t=\frac{2*1.2*ln(8)}{0.21}=24s(approx)

b) w'=\sqrt{\frac{k}{m}-\frac{b^{2}}{4m^{2}} }

w'=\sqrt{\frac{9.8}{1.2}-\frac{0.21^{2}}{4*1.2^{2}}}=2.86s^{-1}

T'=\frac{2\pi}{w'}, where T' is time period of damped SHM

⇒T'=\frac{2\pi}{2.86}=2.2s

let n be number of oscillations made

then, nT'=t

⇒n=\frac{24}{2.2}=11(approx)

8 0
3 years ago
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