A free-falling object is an object moving under the effect of gravitational forces alone
The correct option to select for the True or False question is False
The reason the above selected option is correct is as follows:
According to Newton's second law of motion, we have;
Force = Mass × Acceleration
The force of gravity is 
Where;
m = The mass of the object
∴ The force acting on an object in free fall, 
= m × g
Therefore the acceleration of an object in free fall is the constant acceleration due to gravity, and it therefore, does not change with time
The correct option for the question, acceleration of a free-falling object in a frictionless environment increases as a function of time is <u>False</u>
<u></u>
Learn more about object in free fall here:
brainly.com/question/13712424
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No they don't. Incident rays parallel to the axis of a concave mirror
reflect from the mirror's surface and converge at its focal point.
Answer:
T1 = 131.4 [N]
T2 = 261 [N]
Explanation:
To solve this problem we must make a sketch of how will be the semicircle, for this reason we conducted an internet search, to find the scheme of the problem. This scheme is attached in the first image.
Then we make a free body diagram, with this free body diagram, we raise the forces that act on the body. Since it is a problem involving static equilibrium, the sum of forces in any direction and moments must be equal to zero.
By performing a sum of forces on the Y axis equal to zero we can find an equation that relates the forces of tension T1 & T2.
The second equation can be determined by summing moments equal to zero, around the point of application of the T1 force. In this way we find the T2 force.
The value of T2, is replaced in the first equation and we can find the value for T1.
Therefore
T1 = 131.4 [N]
T2 = 261 [N]
The free body diagram and the developed equations can be seen in the second attached image.
Answer:
7.82 s
Explanation:
Given:
Δy = 300 m
v₀ = 0 m/s
a = 9.8 m/s²
Find: t
Δy = v₀ t + ½ at²
(300 m) = (0 m/s) t + ½ (9.8 m/s²) t²
t = 7.82 s
The energy stored in a capacitor is
E = (1/2) · (capacitance) · (voltage)²
E = (1/2) · (6 x 10⁻⁶ F) · (12 V)²
E = (3 x 10⁻⁶ F) · (144 V²)
<em>E = 4.32 x 10⁻⁴ Joule</em>
(That's 0.000432 of a Joule)
