Answer:
<em>The balloon is 66.62 m high</em>
Explanation:
<u>Combined Motion
</u>
The problem has a combination of constant-speed motion and vertical launch. The hot-air balloon is rising at a constant speed of 14 m/s. When the camera is dropped, it initially has the same speed as the balloon (vo=14 m/s). The camera has an upward movement for some time until it runs out of speed. Then, it falls to the ground. The height of an object that was launched from an initial height yo and speed vo is

The values are


We must find the values of t such that the height of the camera is 0 (when it hits the ground)


Multiplying by 2

Clearing the coefficient of 

Plugging in the given values, we reach to a second-degree equation

The equation has two roots, but we only keep the positive root

Once we know the time of flight of the camera, we use it to know the height of the balloon. The balloon has a constant speed vr and it already was 15 m high, thus the new height is



Because if your putting tension on something tensions obviously going to increase with more pressure and weight on it
Answer:
1 rev = 2(pi) rad pi(rad) = 180 degrees
so 33 rev/min * 1 min/60s * (2*pi)rad/1 rev *180 deg/ pi rad * .32 s = 63.36 degrees
Explanation:
63.36 estimated to 63 so 63
Answer:
Less than Mercury's
Explanation:
According to third Kepler's law, the square of the planet's orbital period is proportional to the cube of the average orbital radius of the planet's orbit. The constant of proportionality depends only on the mass of the star, recall that 51 Peg has the same mass as the Sun. Since the orbital period of this planet is less than Mercury's, its average orbital radius is less than Mercury's.