Given:
F = 39 N, the force applied
t = 2 s, the time interval in which the force is applied.
By definition, the impulse is
Answer: 78 N-s
Answer:
Explanation:
a= 7.8i + 6.6j - 7.1k
b= -2.9 i+ 7.4 j+ 3.9k , and
c = 7.6i + 8.8j + 2.2k
r = a - b +c
=7.8i + 6.6j - 7.1k - ( -2.9i + 7.4j+ 3.9k )+ ( 7.6i + 8.8j + 2.2k)
= 7.8i + 6.6j - 7.1k +2.9i - 7.4j- 3.9k )+ 7.6i + 8.8j + 2.2k
= 18.3 i +18.3 j - k
the angle between r and the positive z axis.
cosθ = 18.3 / √18.3² +18.3² +1
the angle between r and the positive z axis.
= 18.3 / 25.75
cos θ= .71
45 degree
Answer:
i 5.3 cm ii. 72 cm
Explanation:
i
We know upthrust on iron = weight of mercury displaced
To balance, the weight of iron = weight of mercury displaced . So
ρ₁V₁g = ρ₂V₂g
ρ₁V₁ = ρ₂V₂ where ρ₁ = density of iron = 7.2 g/cm³ and V₁ = volume of iron = 10³ cm³ and ρ₂ = density of mercury = 13.6 g/cm³ and V₂ = volume of mercury displaced = ?
V₂ = ρ₁V₁/ρ₂ = 7.2 g/cm³ × 10³ cm³/13.6 g/cm³ = 529.4 cm³
So, the height of iron above the mercury is h = V₂/area of base iron block
= 529.4 cm³/10² cm² = 5.294 cm ≅ 5.3 cm
ρ₁V₁g = ρ₂V₂g
ii
ρ₁V₁ = ρ₃V₃ where ρ₁ = density of iron = 7.2 g/cm³ and V₁ = volume of iron = 10³ cm³ and ρ₃ = density of water = 1 g/cm³ and V₃ = volume of water displaced = ?
V₃ = ρ₁V₁/ρ₃ = 7.2 g/cm³ × 10³ cm³/1 g/cm³ = 7200 cm³
So, the height of column of water is h = V₃/area of base iron block
= 7200 cm³/10² cm² = 72 cm
Average speed=total distance traveled/time. Total distance=the sum of the two short distances=79*2+77*3=389 km. Total time used to travel is 2+3=5 hr. Average speed is 389/5=77.8 km/hr.
The Ideal Gas Law makes a few assumptions from the Kinetic-Molecular Theory. These assumptions make our work much easier but aren't true under all conditions. The assumptions are,
1) Particles of a gas have virtually no volume and are like single points.
2) Particles exhibit no attractions or repulsions between them.
3) Particles are in continuous, random motion.
4) Collisions between particles are elastic, meaning basically that when they collide, they don't lose any energy.
5) The average kinetic energy is the same for all gasses at a given temperature, regardless of the identity of the gas.
It's generally true that gasses are mostly empty space and their particles occupy very little volume. Gasses are usually far enough apart that they exhibit very little attractive or repulsive forces. When energetic, the gas particles are also in fairly continuous motion, and without other forces, the motion is basically random. Collisions absorb very little energy, and the average KE is pretty close.
Most of these assumptions are dependent on having gas particles very spread apart. When is that true? Think about the other gas laws to remember what properties are related to volume.
A gas with a low pressure and a high temperature will be spread out and therefore exhibit ideal properties.
So, in analyzing the four choices given, we look for low P and high T.
A is at absolute zero, which is pretty much impossible, and definitely does not describe a gas. We rule this out immediately.
B and D are at the same temperature (273 K, or 0 °C), but C is at 100 K, or -173 K. This is very cold, so we rule that out.
We move on to comparing the pressures of B and D. Remember, a low pressure means the particles are more spread out. B has P = 1 Pa, but D has 100 kPa. We need the same units to confirm. Based on our metric prefixes, we know that kPa is kilopascals, and is thus 1000 pascals. So, the pressure of D is five orders of magnitude greater! Thus, the answer is B.
