Explanation:
1) This is a synthesis reaction (two or more reactants combine to form a single product).
2) The coefficients are added to balance the reaction.
3) Adding the states of matter (solid, liquid, gas) will make the reaction more precise.
Answer:
, the minus meaning west.
Explanation:
We know that linear momentum must be conserved, so it will be the same before (
) and after (
) the explosion. We will take the east direction as positive.
Before the explosion we have 
.
After the explosion we have pieces 1 and 2, so 
.
These equations must be vectorial but since we look at the instants before and after the explosions and the bomb fragments in only 2 pieces the problem can be simplified in one dimension with direction east-west.
Since we know momentum must be conserved we have:
Which means (since we want 
and 
):
So for our values we have:
Answer:
ω = √(2T / (mL))
Explanation:
(a) Draw a free body diagram of the mass. There are two tension forces, one pulling down and left, the other pulling down and right.
The x-components of the tension forces cancel each other out, so the net force is in the y direction:
∑F = -2T sin θ, where θ is the angle from the horizontal.
For small angles, sin θ ≈ tan θ.
∑F = -2T tan θ
∑F = -2T (Δy / L)
(b) For a spring, the restoring force is F = -kx, and the frequency is ω = √(k/m). (This is derived by solving a second order differential equation.)
In this case, k = 2T/L, so the frequency is:
ω = √((2T/L) / m)
ω = √(2T / (mL))
Answer:
Explanation:
There will be conservation of momentum along horizontal plane because no force acts along horizontal plane.
momentum of first piece = .320 kg x 2 m/s
= 0.64 kg m/s along x -axis.
momentum of second piece = .355 kg x 1.5 m/s
= 0.5325 kg m/s along y- axis .
Let the velocity of third piece be v and it is making angle of θ with x -axis .
Horizontal component of its velocity = .100 kg x v cosθ = .1 v cosθ
vertical component of its velocity = .100 kg x v sinθ = .1 v sinθ
For making total momentum in the plane zero
.1 v cosθ = 0.64 kg m/s
.1 v sinθ = 0.5325 kg m/s
Dividing
Tanθ = .5325 / .64 = .83
θ = 40⁰.
The angle will be actually 180 + 40 = 220 ⁰ from positive x -axis.
Answer:
The wave in the string travels with a speed of 528.1 m/s
Explanation:
Wave speed of sound waves in a string, v, is related to the Tension in the string, T, and the mass per unit length, μ, by the relation,
v = √(T/μ)
μ = 5.20 × 10⁻³ kg/m
T = 1450N
v = √(1450/0.0052) = 528.1 m/s
Hope this Helps!!!
