Answer:
a) vboat = 5.95 m/s b) vriver= 1.05 m/s
Explanation:
a) As observed from the shore, the speed of the boats can be expressed as the vector sum, of the boat speed relative to the water and the river speed relative to the shore, as follows:
vb₁s = vb₁w + vrs
In one case, the boat is moving in the same direction as the water:
vb₁s = vb₁w + vrs = 7.0 m/s (1)
For the other boat, it is clear that is moving in an opposite direction:
vb₂s = vb₂w - vrs = 4.9 m/s (2)
As we know that vb₁w = vb₂w, adding both sides, we can remove the river speed from the equation, as follows:
vb₁w = vb₂w = 
b) Replacing this value in (1) and solving for vriver, we have:
vriver = 7.0 m/s - 5.95 m/s = 1.05 m/s
(we could have arrived to the same result subtracting both sides in (1), and (2))
Answer:
The tension in the rope is 281.60 N.
Explanation:
Given that,
Length = 3.0 m
Weight = 600 N
Distance = 1.0 m
Angle = 60°
Consider half of the ladder,
let tension be T, normal reaction force at ground be F, vertical reaction at top hinge be Y and horizontal reaction force be X.
....(I)
.....(II)
On taking moment about base
Put the value into the formula
....(III)
We need to calculate the force for ladder
We need to calculate the tension in the rope
From equation (3)
Hence, The tension in the rope is 281.60 N.
Answer:
When reviewing the results, the correct one is C
Explanation:
The right hand rule is widely useful in knowing the direction of force in a maganto field,
The ruler sets the thumb in the direction of the positive particle, the fingers extended in the direction of the magnetic field, and the palm in the direction of the force.
Let's apply this to our exercise.
The thumb that is the speed goes in the negative direction of the axis,
The two extended that the magnetic field look negative x,
The span points entered the dear sheet the negative the Z axis
When reviewing the results, the correct one is C
Answer:
y = 54.9 m
Explanation:
For this exercise we can use the relationship between the work of the friction force and mechanical energy.
Let's look for work
W = -fr d
The negative sign is because Lafourcade rubs always opposes the movement
On the inclined part, of Newton's second law
Y Axis
N - W cos θ = 0
The equation for the force of friction is
fr = μ N
fr = μ mg cos θ
We replace at work
W = - μ m g cos θ d
Mechanical energy in the lower part of the embankment
Em₀ = K = ½ m v²
The mechanical energy in the highest part, where it stopped
= U = m g y
W = ΔEm = - Em₀
- μ m g d cos θ = m g y - ½ m v²
Distance d and height (y) are related by trigonometry
sin θ = y / d
y = d sin θ
- μ m g d cos θ = m g d sin θ - ½ m v²
We calculate the distance traveled
d (g syn θ + μ g cos θ) = ½ v²
d = v²/2 g (sintea + myy cos tee)
d = 9.8 12.6 2/2 9.8 (sin16 + 0.128 cos 16)
d = 1555.85 /7.8145
d = 199.1 m
Let's use trigonometry to find the height
sin 16 = y / d
y = d sin 16
y = 199.1 sin 16
y = 54.9 m
