force is mass multiply by acceleration so it will be 150 multiply by 10 is 1500N
Answer:
480
Explanation:
resistance equals to potential difference divide by electric current
120÷0.25
=480
The presence of helium gas indicates the radioactive sample is most likely decaying by α-decay, or alpha decay. α-decay is the type of radioactive decay in which an atomic nucleus emits α particles. α particles are Helium nuclei. So the correct answer would be α-decay.
Answer:
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Explanation:
fevwygafjdbhsaukchavbgaekh
Answer:
Therefore the ratio of diameter of the copper to that of the tungsten is

Explanation:
Resistance: Resistance is defined to the ratio of voltage to the electricity.
The resistance of a wire is
- directly proportional to its length i.e

- inversely proportional to its cross section area i.e

Therefore

ρ is the resistivity.
The unit of resistance is ohm (Ω).
The resistivity of copper(ρ₁) is 1.68×10⁻⁸ ohm-m
The resistivity of tungsten(ρ₂) is 5.6×10⁻⁸ ohm-m
For copper:


......(1)
Again for tungsten:

........(2)
Given that
and 
Dividing the equation (1) and (2)

[since
and
]



Therefore the ratio of diameter of the copper to that of the tungsten is
