1) <span>A solar eclipse that occurs when the new moon is too far from earth to completely cover the sun can be either a partial solar eclipse or an -->
Answer: ANULAR ECLIPSE. Since the moon is too far, it will cover only a part of the sun, and only the external ring of the moon will be visible; this is called anular eclipse.
2) </span><span>anyone looking from the night side of earth can, in principle, see a -->
Answer: LUNAR ECLIPSE. If the moon is the right position, and the Earth's shadow covers partially or totally the moon, then a lunar eclipse occurs.
3) </span><span>during some lunar eclipses, the moon's appearance changes only slightly, because it passes only through the part of earth's shadow called the -->
Answer: PENUMBRA.
4) </span><span>a ... can occur only when the moon is new and has an angular size larger than the sun in the sky -->
Answer: TOTAL SOLAR ECLIPSE. When the moon is new, it means it is between the sun and the Earth, and its dark side faces the Earth. If the moon's angular size is also larger than the sun angular size, than it will completely cover the sun, and a total solar eclipse occurs.
5) </span><span>a partial lunar eclipse begins when the moon first touches earth's -->
Answer: SHADOW. The Earth's shadow will start to cover the moon, and partial lunar eclipse will start.
6) </span><span> a point at which the moon crosses earth's orbital plane is called a(n) -->
Answer: NODE. Eclipses occur only when the Moon is at or close to a node, otherwise sun, earth and moon are not "aligned".</span>
After the collision the magnitude of the momentum of the system is Mv
Given:
mass of 1st object = M
speed of 1st object = v
mass of 2nd object = M
speed of 2nd object = 0
To Find:
magnitude of the momentum after collision
Solution: Product of the mass of a particle and its velocity. Momentum is a vector quantity; i.e., it has both magnitude and direction. Isaac Newton's second law of motion states that the time rate of change of momentum is equal to the force acting on the particle.
Applying conservation of linear momentum
Mv + M(0) = 2MV
Mv = 2MV
V = v/2
So, after collision momentum is
p = 2MV = 2xMxv/2 = Mv
So, after collision momentum is Mv
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Answer:
2.86 m
Explanation:
Given:
M₁ = 10 kg
M₂ = 5 kg
= 0.5
height, h = 5 m
distance traveled, s = 2 m
spring constant, k = 250 N/m
now,
the initial velocity of the first block as it approaches the second block
u₁ = √(2 × g × h)
or
u₁ = √(2 × 9.8 × 5)
or
u₁ = 9.89 m/s
let the velocity of second ball be v₂
now from the conservation of momentum, we have
M₁ × u₁ = M₂ × v₂
on substituting the values, we get
10 × 9.89 = 5 × v₂
or
v₂ = 19.79 m/s
now,
let the velocity of mass 2 when it reaches the spring be v₃
from the work energy theorem, we have
Work done by the friction force = change in kinetic energy of the mass 2
or
or
v₃ = 20.27 m/s
now, let the spring is compressed by the distance 'x'
therefore, from the conservation of energy
we have
Energy of the spring = Kinetic energy of the mass 2
or
on substituting the values, we get
or
x = 2.86 m
The rms speed can be calculated using the following rule:
rms = sqrt ((3RT) / (M)) where:
R is the gas constant = 8.314 J/mol-K
T is the temperature = 31.5 + 273 = 304.5 degrees kelvin
M is the molar mass = 2*14 = 28 grams = 0.028 kg
Substitute with the givens to get the rms speed as follows:
rms speed = sqrt [(3*8.314*304.5) / (0.028)] = 520.811 m/sec
