Question: A 2.0 kg sphere with a velocity of 6.0 m/s collides head-on and elastically with a stationary 10 kg sphere, What is thier velocities after collision.
Answer:
v = 6 m/s, v' = 0 m/s
Explanation:
From the question,
For Elastic collision,
mu+m'u' = mv+m'v'......................... Equation 1
Where m = mass of the first sphere, m' = mass of the second sphere, u = initial velocity of the first sphere, u' = initial velocity of the second sphere, v = final veolocity of the first sphere, v' = final velocity of the second sphere.
Also,
The relative velocity before collision = relative velocity after collision
u-u' = v-v'............................ Equation 2
Given: m = 2 kg, m' = 10 kg, u = 6 m/s, u' = 0 m/s
Substitute into equation 1 and 2
2(6)+10(0) = 2v+10v'
2v+10v' = 12.............. Equation 3
6-0 = v-v'
v-v' = 6 ................... Equation 4
Solve equation 3 and 4 simultaneously.
v = 6+v'............. Equation 5
Substitute equation 5 into equation 3
2(6+v')+10v' = 12
12+2v'+10v' = 12
12v' = 12-12
v' = 0/12
v' = 0 m/s.
Also substitute the value of v' into equation 5
v = 6+0
v = 6 m/s
Answer:
0.9
Explanation:
h = 400 mm, h' = 325 mm
Let the coefficient of restitution be e.
h' = e^2 x h
325 = e^2 x 400
e^2 = 0.8125
e = 0.9
Answer:
w = √[g /L (½ r²/L2 + 2/3 ) ]
When the mass of the cylinder changes if its external dimensions do not change the angular velocity DOES NOT CHANGE
Explanation:
We can simulate this system as a physical pendulum, which is a pendulum with a distributed mass, in this case the angular velocity is
w² = mg d / I
In this case, the distance d to the pivot point of half the length (L) of the cylinder, which we consider long and narrow
d = L / 2
The moment of inertia of a cylinder with respect to an axis at the end we can use the parallel axes theorem, it is approximately equal to that of a long bar plus the moment of inertia of the center of mass of the cylinder, this is tabulated
I = ¼ m r2 + ⅓ m L2
I = m (¼ r2 + ⅓ L2)
now let's use the concept of density to calculate the mass of the system
ρ = m / V
m = ρ V
the volume of a cylinder is
V = π r² L
m = ρ π r² L
let's substitute
w² = m g (L / 2) / m (¼ r² + ⅓ L²)
w² = g L / (½ r² + 2/3 L²)
L >> r
w = √[g /L (½ r²/L2 + 2/3 ) ]
When the mass of the cylinder changes if its external dimensions do not change the angular velocity DOES NOT CHANGE
Answer:
<u>B. the stars of spectral type A and F are considered reasonably to have habitable planets but much less likely to have planets with complex plant - or animal - like life.</u>
Explanation:
The appropriate spectral range for habitable stars is considered to be "late F" or "G", to "mid-K" or even late "A". <em>This corresponds to temperatures of a little more than 7,000 K down to a little less than 4,000 K</em> (6,700 °C to 3,700 °C); the Sun, a G2 star at 5,777 K, is well within these bounds. "Middle-class" stars (late A, late F, G , mid K )of this sort have a number of characteristics considered important to planetary habitability:
• They live at least a few billion years, allowing life a chance to evolve. <em>More luminous main-sequence stars of the "O", "B", and "A" classes usually live less than a billion years and in exceptional cases less than 10 million.</em>
• They emit enough high-frequency ultraviolet radiation to trigger important atmospheric dynamics such as ozone formation, but not so much that ionisation destroys incipient life.
• They emit sufficient radiation at wavelengths conducive to photosynthesis.
• Liquid water may exist on the surface of planets orbiting them at a distance that does not induce tidal locking.
<u><em>Thus , the stars of spectral type A and F are considered reasonably to have habitable planets but much less likely to have planets with complex plant - or animak - like life.</em></u>
