If you go to high you’ll run out of oxygen and possibly be blown off due to high winds.
Answer:
B. fitness goal I believe. please like this, I need points. ty!
Explanation:
Answer:
v = 3.7 m/s
Explanation:
As the swing starts from rest, if we choose the lowest point of the trajectory to be the zero reference level for gravitational potential energy, and if we neglect air resistance, we can apply energy conservation as follows:
m. g. h = 1/2 m v²
The only unknown (let alone the speed) in the equation , is the height from which the swing is released.
At this point, the ropes make a 30⁰ angle with the vertical, so we can obtain the vertical length at this point as L cos 30⁰, appying simply cos definition.
As the height we are looking for is the difference respect from the vertical length L, we can simply write as follows:
h = L - Lcos 30⁰ = 5m -5m. 0.866 = 4.3 m
Replacing in the energy conservation equation, and solving for v, we get:
v = √2.g.(L-Lcos30⁰) = √2.9.8 m/s². 4.3 m =3.7 m/s
Answer:
Option B
Solution:
As per the question:
Heat produced at the rate of 10 W
The resistor R and 2R are in series.
Also, in series, same current, I' passes through each element in the circuit.
Therefore, current is constant in series.
Also,
Power,
When current, I' is constant, then
P' ∝ R
Thus

P' = 20 W
To solve this problem we will apply the concepts related to destructive interference from double-slit experiments. For this purpose we will define the path difference as,

Here,
= Wavelength
= Angle when occurs the interference point of destructive interference
Our values are given as,


Using the previous expression we have,





Therefore the distance between the two openings is 