Answer:
V =1/EI(-Ax^n+y/(n+1)(n+2)(n+3)(n+y)+AL^n+1 x^3/6(n+1)(n+2)+AL^n+3(6-(n+3)(n+4)x/6(n+1)(n+2)(n+3)(n+4))
Explanation:
Please look at attachment for step by step solution and guide.
A 10-meter long ramp has a mechanical advantage of 5. What is the height of the ramp?
1 answer:
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Answer:
vo = 227 m/s
Explanation:
Let's analyze the problem, we have two parts one during the crash for which we will use the moment and another when it slides where we will use the energy. As we give the data of the latter let's start here
Before starting let's reduce the unit to the SI system
m = 6.00 g (1kg / 1000g) = 6.00 10⁻³ kg
M = 1.12 kg
We write the energy at two points, one initial right after the crash and another when the body has stopped
Just after shock
The bodies are united, so the mass is the sum of the mass of the bullet and the block
Em₀ = K = ½ (m + M) v²
When the body has stopped
= 0
When in the system there is friction force the variation the mechanical energy is equal to the work of the friction force. Notice that the force of friction opposes the movement so that their work is negative
W fr = ΔEm = -Em₀
-fr d = 0 - ½ (m + M) v²
To find the force of friction let's use Newton's second law
Axis y
N-W = 0
N = W = (m + M) g
The equation for the force of friction is
fr = μ N
fr = μ (m + M) g
Let's replace in the work and energy equation
-μ (m + M) g d = 0 - ½ (m + M) v²
From here we can find the system speed. Let's calculate
v³ = 2 μ g d
v = √ (2μ gd)
v = √ (2 0.250 9.8 0.300)
v = 1.21 m / s
Now let's solve the crash, let's look for the moment before and after it
Before the crash
po = m vo
After the crash
= (m + M) v
The system is formed by the bullet and the block, therefore, the forces during the impact are internal and the amount of movement is conserved
po =
m vo = (m + M) v
vo = v (m + M) / m
vo = 1.21 (0.00600 + 1.12) /0.00600
vo = 227 m / s
Answer:
a) F_net = 30.47 N , θ = 10.6º
b) Fₓ = 29.95 N
Explanation:
For this exercise we use coulomb's law
F₁₂ = k
the direction of the force is on the line between the two charges and the sense is repulsive if the charges are equal and attractive if the charges are different.
As we have several charges, the easiest way to solve the problem is to add the components of the force in each axis, see attached for a diagram of the forces
X axis
Fₓ =
Y axis
Fy =
let's find the magnitude of each force
= 9 10⁹ 2.76 10⁻⁴ 1.02 10⁻⁴ / 3²
F_{ab} = 2.82 10¹ N
F_{ab} = 28.2 N
= 9 10⁹ 6.54 10⁻⁴ 1.02 10⁻⁴ / 4²
F_{bc} = 3.75 10¹ N
F_{bc} = 37.5 N
let's use trigonometry to decompose this force
tan θ = y / x
θ = tan⁻¹ and x
θ= tan⁻¹ ¾
θ = 37º
let's break down the force
sin 37 = F_{bcy} / F_{bc}
F_{bcy} = F_{bc} sin 37
F_{bcy} = 37.5 sin 37
F_{bcy} = 22.57 N
cos 37 = F_{bcx} /F_{bc}
F_{bcx} = F_{bc} cos 37
F_{bcx} = 37.5 cos 37
F_{bcx} = 29.95 N
let's do the sum to find the net force
X axis
Fₓ = 29.95 N
Axis y
Fy = 28.2 -22.57
Fy = 5.63 N
we can give the result in two ways
a) F_net = Fₓ i ^ + j ^
F_net = 29.95 i ^ + 5.63 j ^
b) in the form of module and angle
let's use the Pythagorean theorem
F_net =
F_net = √(29.95² + 5.63²)
F_net = 30.47 N
we use trigonometry for the direction
tan θ=
θ = tan⁻¹ \frac{ F_{y} }{ F_{x} }
θ = tan⁻¹ (5.63 / 29.95)
θ = 10.6º
