91影视

The given data is listed below as-

• The orbital speed of ground state electron in Hydrogen is $V=2190\mathrm{km}/\mathrm{s}=2190脳{10}^3{\mathrm{ms}}^{鈥�1}$.
• The mass of the child is, $m=30.0\mathrm{kg}$
• Kinetic energy, $K=100\mathrm{J}$

The kinetic energy of a particle equals the amount of work required to accelerate the particle from rest to speed V. Therefore, kinetic energy on the particle is given by-

$\mathbf{K}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{mV}}^{\mathbf{2}}$

The kinetic energy is a scalar and it is always positive or zero.

(a)

The mass of the electron is $m_e=9.1脳{10}^{-31}\mathrm{kg}$

The kinetic energy of an object is given by-

$K=\frac{1}{2}{m}_e{V}^2$

Here, m is the mass of the electron, and Vis the speed of the electron.

For, $m_e=9.1脳{10}^{-31}\mathrm{kg}$and $V=2190\mathrm{km}/\mathrm{s}=2190脳{10}^3{\mathrm{ms}}^{鈥�1}$

Therefore, the Kinetic energy of the electron is given by-

$$\begin{gathered} K=\frac{1}{2}{m}_e{V}^2 \\ K=\frac{1}{2}脳\left(9.1脳{10}^{-31}\mathrm{kg}\right)脳{\left(2190脳{10}^3\mathrm{m}/\mathrm{s}\right)}^2 \\ K=2.18脳{10}^{-18}\mathrm{J} \end{gathered}$$

Thus, the Kinetic energy of the electron is $2.18脳{10}^{-18}\mathrm{J}$

(b)

The kinetic energy under free fall along the y-axis is given by-

$K=\frac{1}{2}m{V_y}^2$

The velocity at time t of a particle is given by-

$V^2={V_0}^2+2as$鈥︹赌︹赌�..(1)

Now, the object is falling along the y-axis with constant gravitational acceleration and $S=y_2-y_1$

Therefore, equation (1) becomes

${V_y}^2={V_{0y}}^2-2g\left(y_2-y_1\right)$

For, $V_{0y}=0,g=9.8{\mathrm{ms}}^{鈥�2}$and $S=1m$

${V_y}^2={V_{0y}}^2-2g\left(y_2-y_1\right)$will become

$$\begin{gathered} {V_y}^2=0-2脳9.8{\mathrm{ms}}^{鈥�2}脳1\mathrm{m} \\ =19.6{\mathrm{m}}^2{\mathrm{s}}^{鈥�2} \end{gathered}$$

Therefore, kinetic energy is given by

$$\begin{gathered} K=\frac{1}{2}脳1\mathrm{kg}脳19.6{\mathrm{m}}^2{\mathrm{s}}^{鈥�2} \\ =9.8\mathrm{J} \end{gathered}$$

Thus, kinetic energy when a weight of 1 kg is dropped from a height of 1 mis $9.8\mathrm{J}$.

(c)

The mass of the child is $m=30\mathrm{kg}$and

Kinetic energy is $K=100\mathrm{J}$

$$\begin{gathered} K=\frac{1}{2}m{V_y}^2 \\ 100J=\frac{1}{2}脳30\mathrm{kg}脳{{\mathrm{V}}_{\mathrm{y}}}^2 \\ {V}_y=2.58{\mathrm{ms}}^{-1} \end{gathered}$$

Thus, it is reasonable that a 30.0-kg child could run fast enough to have 100 J of kinetic energy.