Which kind of air front forms when a warm air mass meets the area of a cooler air mass?
1 answer:
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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
Answer:
A= 148.92 m/s²
Explanation:
Given that
U(x,y) = (6.00 )x² - (3.75 )y ³
m= 0.04 kg
Now force in the x-direction
Fx= - dU/dx
U(x,y) = (6.00 )x² - (3.75 )y ³
dU/dx= 12 x
When x=0.4 m
dU/dx= 12 x 0.4 = 4.8
So we can say that
Fx= - 4.8 N
From Newtons law
F= m a
- 4.8 = 0.04 x a
a = -120 m/s²
Acceleration in x direction ,a = -120 m/s²
In y -direction
F= - dU/dy
U(x,y) = (6.00 )x² - (3.75 )y ³
dU/dy = 0 - 3.75 x 3 y²
When y = 0.56 m
dU/dy = - 3.75 x 3 x 0.56 x 0.56
dU/dy = - 3.52
So we can say that force in y -direction
F= 3.52 N
F= m a'
3.52 = 0.04 x a'
a'=88.2 m/s²
acceleration in y direction is 88.2 m/s²
The resultant acceleration
A= 148.92 m/s²