Answer:
3.71 m/s in the negative direction
Explanation:
From collisions in momentum, we can establish the formula required here which is;
m1•u1 + m2•v2 = m1•v1 + m2•v2
Now, we are given;
m1 = 1.5 kg
m2 = 14 kg
u1 = 11 m/s
v1 = -1 m/s (negative due to the negative direction it is approaching)
u2 = -5 m/s (negative due to the negative direction it is moving)
Thus;
(1.5 × 11) + (14 × -5) = (1.5 × -1) + (14 × v2)
This gives;
16.5 - 70 = -1.5 + 14v2
Rearranging, we have;
16.5 + 1.5 - 70 = 14v2
-52 = 14v2
v2 = - 52/14
v2 = 3.71 m/s in the negative direction
Answer:
true is the answer of the question
A globe sitting on the desk can't demonstrate the speed of axial rotation
or the speed of orbital revolution.
The complete queston is The amount of a radioactive element A at time t is given by the formula
A(t) = A₀e^kt
Answer: A(t) =N e^( -1.2 X 10^-4t)
Explanation:
Given
Half life = 5730 years.
A(t) =A₀e ^kt
such that
A₀/ 2 =A₀e ^kt
Dividing both sides by A₀
1/2 = e ^kt
1/2 = e ^k(5730)
1/2 = e^5730K
In 1/2 = 5730K
k = 1n1/2 / 5730
k = 1n0.5 / 5730
K= -0.00012 = 1.2 X 10^-4
So that expressing N in terms of t, we have
A(t) =A₀e ^kt
A₀ = N
A(t) =N e^ -1.2 X 10^-4t
F=K*X,
F=M*a
M*a=K*X
2.5*9.81=K*0.0276
24.525=K*0.0276
24.525/0.0276=K
K= 888.6 N/m ---- force constant
assuming 2.5 refers to the new extension, just divide F/ 0.025
to get
981N/m
