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2 years ago

Calculate the speed of a car that traveled 200 kilometers in 3 hours.

Physics

2 answers:

nignag [31]2 years ago

8 0

Answer:

The units (km/h) tell you how to do this! 200km/3h = 66.66666666…. BUT technically you only have ONE significant digit: 3 so 66.666… rounded to ONE digit is 70km/h but that is probably not important in this intro class so V = 66.67 or 67 km/h

Kamila [148]2 years ago

7 0

Answer:

66.7km/h

Explanation:

s=d/t

200/3

=66.6666666

round off: 66.7km/h

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A quantum well is a region of semiconductor or metal film that confines particles to a small region, resulting in quantized ener

murzikaleks [220]

Answer:

226.2 m/sec

Explanation:

We have given $\Delta x=510nm=510\times 106{-9}m$

The plank's constant $h=6.67\times 10^{-34}$

Mass of electron $m=9.11\times 10^{-31}kg$

Now according to Heisenberg uncertainty principle $\Delta v\geq \frac{h}{2\pi m\Delta x}$

So $\Delta v\geq \frac{6.6\times 10^{-34}}{2\times 3.14\times 9.11\times 10^{-31}\times 510\times 10^{-9}}=226.2m/sec$

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2 years ago

Venus is often called Earth's "twin." Describe TWO ways in which Earth and Venus are similar and TWO ways in which Earth and Ven

insens350 [35]

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8 0

3 years ago

The magnetic field perpendicular to a single 16.7-cm-diameter circular loop of copper wire decreases uniformly from 0.750 T to z

sammy [17]

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

4 0

3 years ago