\(v=\sqrt{\frac{T}{μ}}\)
So,\( T=60^2×\frac{10×10^{−3}}{0.5}\)
=72 N
\(Δℓ=\frac{Tℓ}{YA}=\frac{72×0.5}{1.2×10^{−11}×2×10^{−6}}\)
=15×10−5 m
Given that, the extension of the wire over its natural length due to its tension is x × 10–5 m.
On comparing, x = 15
So, the answer is 15.
For the thermal decomposition of \( N_2O_5(g) \) at constant volume, the following table can be formed, for the reaction mentioned below: \[ 2 N_2O_5(g) \rightarrow 2 N_2O_4(g) + O_2(g) \] Given: Rate constant for the reaction is \( 4.606 \times 10^{-2} \text{ s}^{-1} \).
Let \( T_r \) be the \( r^{\text{th}} \) term of an A.P. If for some \( m \), \( T_m = \dfrac{1}{25} \), \( T_{25} = \dfrac{1}{20} \), and \( \displaystyle\sum_{r=1}^{25} T_r = 13 \), then \( 5m \displaystyle\sum_{r=m}^{2m} T_r \) is equal to:
The rate at which an object covers a certain distance is commonly known as speed.
The rate at which an object changes position in a certain direction is called velocity.
Read More: Difference Between Speed and Velocity