<span>Work is required to pull a nucleon out of an atomic nucleus. It has more mass outside the nucleus.</span>
Answer:
Check the first and the third choices:
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- <u><em>a. The temperature of a gas is directly proportional to its volume</em></u>
- <u><em>b. The temperature-to-volume ratio of a gas is constant.</em></u>
Explanation:
Rewrite the table for better understanding:
Temperature of gas (K) Volume of gas (L)
298 4.55
315 4.81
325 4.96
335 ?
Calculate the ratios temperature to volume with 3 significant figures:
Then, those numbers show a <u><em>constant temperature-to-volume ratio</em></u>, which may be expressed in a formula as:
- Temperature / Volume = constant, which is a directly proportional variation (the volume increases in a constant proportion to the increase of the temperature).
Hence, the correct choices are:
- The temperature of a gas is directly proportional to its volume (first statement), and
- The emperature-to-volume ratio of a gas is constant (third statement).
Answer:
The puck moves a vertical height of 2.6 cm before stopping
Explanation:
As the puck is accelerated by the spring, the kinetic energy of the puck equals the elastic potential energy of the spring.
So, 1/2mv² = 1/2kx² where m = mass of puck = 39.2 g = 0.0392 g, v = velocity of puck, k = spring constant = 59 N/m and x = compression of spring = 1.3 cm = 0.013 cm.
Now, since the puck has an initial velocity, v before it slides up the inclined surface, its loss in kinetic energy equals its gain in potential energy before it stops. So
1/2mv² = mgh where h = vertical height puck moves and g = acceleration due to gravity = 9.8 m/s².
Substituting the kinetic energy of the puck for the potential energy of the spring, we have
1/2kx² = mgh
h = kx²/2mg
= 59 N/m × (0.013 m)²/(0.0392 kg × 9.8 m/s²)
= 0.009971 Nm/0.38416 N
= 0.0259 m
= 2.59 cm
≅ 2.6 cm
So the puck moves a vertical height of 2.6 cm before stopping
Answer:
Sediments can be carried from one place to another. The movement of sediments by wind, water, ice, or gravity is called erosion.
