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creativ13 [48]
3 years ago
14

The lanthanides are all radioactive true or false

Physics
1 answer:
Oksi-84 [34.3K]3 years ago
3 0

Answer:

False

Explanation:

All the lanthanides are not radioactive in nature. Just one of the lanthanides are radioactive.

  • The lanthanides belong to the f-block on the periodic table
  • Only Promethium in this group are radioactive in nature.
  • Other elements in the lanthanide series might have radioactive isotopes.
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James E. Hansen studied climate change
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A substance feels warm
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An isolated conducting sphere has a 17 cm radius. One wire carries a current of 1.0000020 A into it. Another wire carries a curr
notsponge [240]

14 ms is required to reach the potential of 1500 V.

<u>Explanation:</u>

The current is measured as the amount of charge traveling per unit time. So the charge of electrons required for each current is determined as the product of current with time.

       Charge = Current \times Time

As two different current is passing at two different times, the net charge will be the different in current.  So,

        \text { Charge }=(1.0000020-1.0000000) \times t=2 \times 10^{-6} \times t

The electric voltage on the surface of cylinder can be obtained as the ratio of charge to the radius of the cylinder.

        V=\frac{k q}{R}

Here k = 9 * 10^9, q is the charge and R is the radius. As q=2 \times 10^{-6} \times t and R =17 cm = 0.17 m, then the voltage will be

        V=\frac{9 \times 10^{9} \times 2 \times 10^{-6} \times t}{0.17}

The time is required to find to reach the voltage of 1500 V, so

1500 =\frac{9 \times 10^{9} \times 2 \times 10^{-6} \times t}{0.17}

\begin{aligned}&t=\frac{1500 \times 0.17}{\left(9 \times 10^{9} \times 2 \times 10^{-6}\right)}\\&t=14.1666 \times 10^{-3} s=14\ \mathrm{ms}\end{aligned}

So, 14 ms is required to reach the potential of 1500 V.

3 0
3 years ago
Can someone pls help ASAP:(
diamong [38]
I’m not sure but I think it’s
△ m=5 and △= -3 and so

Answer: 5/△-3 m/s

So sorry if it’s wrong
6 0
3 years ago
The plates of a spherical capacitor have radii 6.25 cm and 15.0 crn. The space between the two spheres is filled with a material
aleksklad [387]

Answer:

Capacitance is 0.572×10⁻¹⁰ Farad

Explanation:

Radius = R₁ = 6.25 cm = 6.25×10⁻² m

Radius = R₂ = 15 cm = 15×10⁻² m

Dielectric constant = k = 4.8

Electric constant = ε₀ = 8.854×10⁻¹² F/m

ε/ε₀=k

ε=kε₀

Capacitance\ (C)=\frac{4\pi k\epsilon_0 R_1\times R_2}{R_2-R_1}\\\Rightarrow C=\frac{4\pi 4.8\times 8.854\times 10^{-12}\times 15\times 10^{-2}\times 6.25\times 10^{-2}}{15\times 10^{-2}-6.25\times 10^{-2}}\\\Rightarrow C=0.572\times 10^{-10}\ Farad

∴ Capacitance is 0.572×10⁻¹⁰ Farad

3 0
3 years ago
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