Answer:
¹/₃₈₇ second
Explanation:
<em>The period of a wave is the reciprocal of its frequency.</em>
So, simply, the frequency is ¹/₃₈₇ second(s), as that is the reciprocal of the frequency, 387 Hz.
The distance covered by the acorn is 3.136 m.
<u>Explanation:</u>
The time taken for the acorn to hit the ground is 0.8 s. As it is a free fall, the acorn will be completely under the influence of gravity. So the acceleration will be acceleration due to gravity.
Then using the second law of equation,
Since the initial velocity and time is zero, then the time taken to reach the ground is stated as 0.8 s, so
So the distance covered by the acorn is 3.136 m.
Answer:
1. 610,000 lb ft
2. 490 J
Explanation:
1. First, convert mi/hr to ft/s:
100 mi/hr × (5280 ft / mi) × (1 hr / 3600 s) = 146.67 ft/s
Now find the kinetic energy:
KE = ½ mv²
KE = ½ (1825 lb / 32.2 ft/s²) (146.67 ft/s)²
KE = 610,000 lb ft
2. KE = ½ mv²
KE = ½ (5 kg) (14 m/s)²
KE = 490 J
Answer:
ΔE = 37.8 x 10^9 J
Explanation:
The energy required will increased the potential energy and increase the kinetic energy.
As the altitude change is fairly small compared to the earth radius, we can ASSUME that the average gravity will be a good representative
Gravity acceleration at altitude would be 9.8(6400²/8000²) = 6.272 m/s²
G(avg) = (9.8 + 6.272)/2 = 8.036 m/s²
ΔPE = mG(avg)Δh = 1000(8.036)(8e6 - 6.4e6) = 12.857e9 J
The centripetal force at orbit must be equal to the gravity force
mv²/R = mg'
v²/8.0e6 = 6.272
v² = (6.272(8.0e6)) = 50.2e6 m²/s²
The maximum velocity when resting on earth at the equator is about 460 m/s.
The change in kinetic energy is
ΔKE = ½m(vf² - vi²)(1000)
ΔKE = ½(1000)(50.2e6 - 460²) = 25e9 J
Total energy increase is
25e9 + 12.857e9 = 37.8e9 J
Answer:
B. Outside the nucleus.
Explanation:
Electrons orbit the nucleus of the atom.
