Answer :
We are given that the angular speed is
$$\omega = \frac{7\pi}{10} \text{ radians per hour}$$
and that the radius of the equator is
$$r = 4000 \text{ miles}.$$
The relationship between linear velocity $v$, radius $r$, and angular speed $\omega$ is given by the formula
$$v = r \omega.$$
Substitute the provided values into the formula:
$$
v = 4000 \times \frac{7\pi}{10}.
$$
We can simplify this by multiplying the constants:
$$
v = \frac{4000 \times 7\pi}{10} = \frac{28000\pi}{10} = 2800\pi.
$$
To express $v$ as a numerical value, we approximate $\pi \approx 3.141592653589793$:
$$
v \approx 2800 \times 3.141592653589793 \approx 8796.45943 \text{ miles per hour}.
$$
Thus, the linear velocity of a point on the equator is approximately
$$8796.45943 \text{ miles per hour}.$$
$$\omega = \frac{7\pi}{10} \text{ radians per hour}$$
and that the radius of the equator is
$$r = 4000 \text{ miles}.$$
The relationship between linear velocity $v$, radius $r$, and angular speed $\omega$ is given by the formula
$$v = r \omega.$$
Substitute the provided values into the formula:
$$
v = 4000 \times \frac{7\pi}{10}.
$$
We can simplify this by multiplying the constants:
$$
v = \frac{4000 \times 7\pi}{10} = \frac{28000\pi}{10} = 2800\pi.
$$
To express $v$ as a numerical value, we approximate $\pi \approx 3.141592653589793$:
$$
v \approx 2800 \times 3.141592653589793 \approx 8796.45943 \text{ miles per hour}.
$$
Thus, the linear velocity of a point on the equator is approximately
$$8796.45943 \text{ miles per hour}.$$
