When a cold air <span>mass replaces a warm air mass, this is called a cold front. Some characteristics of a called front before passing are winds coming from south or southwest area, warm temperature, falling pressure, and drizzles. When it passes, the winds are shifting, there is a sudden drop of temperature, minimum pressure followed by a sharp rise. After passing, the winds head to the west or northwest area, temperature is steadily dropping and the pressure is rising steadily.</span>
Answer:
The correct option is;
d) F, because the net force is equal to the mutual contact force between the blocks
Explanation:
The given information are
The mass of block A = m
The velocity of block A = +v
The mass of block B = 2·m
The velocity of block B = -v
Given that the two blocks collide, we have;
Initial total momentum = m × v + 2·m×(-v) = m·(v - 2·v) = -m·v
Final total momentum = m × v₁ + 2·m×v₂ = m·(v₁ + 2·v₂)
From the law of conservation of linear momentum, we have;
m·(v₁ + 2·v₂) = -m·v
v₁ + 2·v₂ = -v
Therefore, the resultant velocity of the two blocks is -v, and the direction of the block A is reversed and the resultant inertia is equivalent to the inertia of block A
Therefore;
The force exerted on block B = The force exerted on block A = The rate of change of momentum experienced by the two blocks = The mutual contact force experienced between the blocks.
Answer:
3.87 or 3.9 mg
Explanation:
2.7h * 60mins = 162mins
number of 1/2 lives = 162/22= 7.364
amount remaining = 639(1/2)^7.364 = 3.8796
answer = 3.87 or 3.9
Answer:
15.4 kg.
Explanation:
From the law of conservation of momentum,
Total momentum before collision = Total momentum after collision
mu+m'u' = V(m+m').................... 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 = common velocity of both sphere.
Given: m = 7.7 kg, u' = 0 m/s (at rest)
Let: u = x m/s, and V = 1/3x m/s
Substitute into equation 1
7.7(x)+m'(0) = 1/3x(7.7+m')
7.7x = 1/3x(7.7+m')
7.7 = 1/3(7.7+m')
23.1 = 7.7+m'
m' = 23.1-7.7
m' = 15.4 kg.
Hence the mass of the second sphere = 15.4 kg
