<span>First, we use the kinetic energy equation to create a formula:
Ka = 2Kb
1/2(ma*Va^2) = 2(1/2(mb*Vb^2))
The 1/2 of the right gets cancelled by the 2 left of the bracket so:
1/2(ma*Va^2) = mb*Vb^2 (1)
By the definiton of momentum we can say:
ma*Va = mb*Vb
And with some algebra:
Vb = (ma*Va)/mb (2)
Substituting (2) into (1), we have:
1/2(ma*Va^2) = mb*((ma*Va)/mb)^2
Then:
1/2(ma*Va^2) = mb*(ma^2*Va^2)/mb^2
We cancel the Va^2 in both sides and cancel the mb at the numerator, leving the denominator of the right side with exponent 1:
1/2(ma) = (ma^2)/mb
Cancel the ma of the left, leaving the right one with exponent 1:
1/2 = ma/mb
And finally we have that:
mb/2 = ma
mb = 2ma</span>
Answer:
yes independent of the sign or valve of Q
Explanation:
Impulse = Integral of F(t) dt from 0.012s to 0.062 s
Given that you do not know the function F(t) you have to make an approximation.
The integral is the area under the curve.
The problem suggest you to approximate the area to a triangle.
In this triangle the base is the time: 0.062 s - 0.012 s = 0.050 s
The height is the peak force: 35 N.
Then, the area is [1/2] (0.05s) (35N) = 0.875 N*s
Answer> 0.875 N*s
Answer:
the expected distance is 4.32 m
Explanation:
given data
half life time = 1.8 × 
s
speed = 0.8 c = 0.8 × 3 × 
to find out
expected distance over
solution
we know c is speed of light in air is 3 × m/s
we calculate expected distance by given formula that is
expected distance = half life time × speed .........1
put here all these value
expected distance = half life time × speed
expected distance = 1.8 × × 0.8 × 3 ×
expected distance = 4.32
so the expected distance is 4.32 m
