Looking at decibel scale tables on internet, a "quiet whisper 1 meter away" is generally reported to have a sound intensity of about 20 dB, while a jackhammer 3 meters away produces a sound of about 100 dB of intensity, so option a) could be the correct one.
However, the other options could be considered correct as well, because without knowing the exact sound power produced by the whisper or by the jackhammer it's impossible to calculate the exact sound intensity with precision, and since the values reported in the three options are very similar, all the three of them can be considered as valid.
Answer:
269,000
Explanation:
The distance between Proxima Centauri and the Sun, d₁₁ = 4.246 light years
The distance between Earth and the Sun, d₂ = 1.58 × 10⁻⁵ light years
The ratio of the distances are;
d₁/d₂ = 4.246/(1.58 × 10⁻⁵) = 268734.177215 ≈ 269,000 times
Therefore, Proxima Centauri is approximately 269,000 times further from the Sun than the Earth is from the Sun
Answer:
Orbital speed=8102.39m/s
Time period=2935.98seconds
Explanation:
For the satellite to be in a stable orbit at a height, h, its centripetal acceleration V2R+h must equal the acceleration due to gravity at that distance from the center of the earth g(R2(R+h)2)
V2R+h=g(R2(R+h)2)
V=√g(R2R+h)
V= sqrt(9.8 × (6371000)^2/(6371000+360000)
V= sqrt(9.8× (4.059×10^13/6731000)
V=sqrt(65648789.18)
V= 8102.39m/s
Time period ,T= sqrt(4× pi×R^3)/(G× Mcentral)
T= sqrt(4×3.142×(6.47×10^6)^3/(6.673×10^-11)×(5.98×10^24)
T=sqrt(3.40×10^21)/ (3.99×10^14)
T= sqrt(0.862×10^7)
T= 2935.98seconds
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
