A magnetic field line: U
1 answer:
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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
The tension in the swing's chain at the bottom of the swing is 178.35 N.
The given parameters:
- Mass of the girl, m = 27 kg
- Speed of the girl, v = 3 m/s
- Radius of the circle, r = 4 m
The tension in the swing's chain at the bottom of the swing is calculated as follows;
Thus, the tension in the swing's chain at the bottom of the swing is 178.35 N.
Learn more about tension in vertical circle here: brainly.com/question/19904705
|V| = 10.33 units and the direction θ = -47.35° or 312.65°.
Given the x and y components of a vector, we can calculate the magnitude and direction from these components.
Applying the Pythagorean theorem we have that the magnitude of the vector is:
|V| =
|V| =
The expression for the direction of a vector comes from the definition of the tangent of an angle:
tan θ = ------> θ = arc tan
θ = arc tan
θ = -47.35° or 312.65°