Answer:
x component 60.85 m
y component 101.031 m
Explanation:
We have given distance r = 118 km
Angle which makes from ground = 58.9°
(a) X component of distance is given by 
(b) Y component of distance is given by 
These are the x and y component of position vector
Maybe you can split up the questions. I will try to answer your first question.
1. In an elastic collision, momentum is conserved. The momentum before the collision is equal to the momentum after the collision. This is a consequence of Newton's 3rd law. (Action = Reaction)
2. Momentum: p = m₁v₁ + m₂v₂
m₁ mass of ball A
v₁ velocity of ball A
m₂ mass of ball B
v₂ velocity of ball B
Momentum before the collision:
p = 2*9 + 3*(-6) = 18 - 18 = 0
Momentum after the collision:
p = 2*(-9) + 3*6 = -18 + 18 = 0
3: mv + m(-v) = m(-v) + m(v)
the velocities would reverse.
4.This question is not factual since the energy of an elastic collision must also be conserved. The final velocities should be: v₁ = -1 m/s and v₂ = 5 m/s. That said assuming the given velocities were correct:
before collision
p = 10*3 + 5*(-3) = 30 - 15 = 15
after collision:
p = 10*(-2) + 5 * v₂ = 15
v₂ = 7
5.You figure out.
Answer:
No
Explanation:
You can easily injury yourself due to the force of gravity.
The total mechanical energy of the block-spring system is given by the sum of the potential energy and the kinetic energy of the block:
where
k is the spring constant
x is the elongation/compression of the spring
m is the mass of the block
v is the speed of the block
At the point of maximum displacement of the spring, the velocity of the block is zero: v=0, so the kinetic energy is zero and the mechanical energy is just potential energy of the spring:
(1)
where we used x=A, the amplitude (which is the maximum displacement of the spring).
Since we know
A = 11.0 cm= 0.11 m
E = 1.10 J
We can re-arrange (1) to find the spring constant:
Answer:
y = 2.74 m
Explanation:
The linear thermal expansion processes are described by the expression
ΔL = α L ΔT
Where α the thermal dilation constant for concrete is 12 10⁻⁶ºC⁻¹, ΔL is the length variation and ΔT the temperature variation in this case 20ªc
If the bridge is 250 m long and is covered by two sections each of them must be L = 125 m, let's calculate the variation in length
ΔL = 12 10⁻⁶ 125 20
ΔL = 3.0 10⁻² m
Let's use trigonometry to find the height
The hypotenuse Lf = 125 + 0.03 = 125.03 m
Adjacent leg L₀ = 125 m
cos θ = L₀ / Lf
θ = cos⁻¹ (L₀ / Lf)
θ = cos⁻¹ (125 / 125.03)
θ = 1,255º
We calculate the height
tan 1,255 = y / x
y = x tan 1,255
y = 125 tan 1,255
y = 2.74 m
