Answer:
Explanation:
Weight of the ladder is 355N
WL = 355N
The weight of the ladder acts at center of the ladder I.e at 4m from the bottom
Weight of firefighters is 870N
Wf = 870N
The fire fighter is at 6.3m from the bottom of the ladder.
N1 is the normal force exerted by the wall
N2 is the normal force exerted by the ground
Using Newton law
Check attachment
ΣFy = 0, since body is in equilibrium
N₂ - WL - Wf = 0
N₂ = WL + Wf
N₂ = 870 + 355
N₂ = 1225 N
This the normal force exerted by the ground on the wall.
Now to get N₁, let take moment about point A
So, before we take the moment we need to make sure that the forces are perpendicular to the plane(ladder), we need to resolve the weight of the ladder, firefighter and the normal of the wall to be perpendicular to the plane.
ΣMa = 0
Clockwise moment is equal to anti-clockwise moment
Moment Is the produce of force and perpendicular distance.
M = F×r
So,
WL•Cos50 × 4 + Wf•Cos50 × 6.3 —N₁•Sin50 × 8 = 0
355•Cos50 × 4 + 870•Cos50 × 6.3 =
N₁•Sin50 × 8
1825.52 + 3523.12 = 6.13N₁
6.13N₁ = 5348.64
N₁ = 5348.64/6.13
N₁ = 872.54 N.
The normal force exerted by the wall is 872.54N
The final velocity of the passenger is zero as he is brought to rest by the inflated bag.
Apply the equation of motion
Replacing with our values,
Calculate the force using the force equation,
Therefore the magnitude of force acts on the passenger's upper torso is 34.923kN
66.7 Watts
Explanation:
Let ohms,
ohms and
ohms. Since
and
are in parallel, their combined resistance
is given by
or
The total current flowing through the circuit <em>I</em> is given
where
Therefore, the total current through the circuit is
In order to find the voltage drop across the 6-ohm resistor, we first need to find the voltage drop across the 1-ohm resistor :
This means that voltage drop across the 6-ohm resistor is 20 V. The power dissipated <em>P</em> by the 6-ohm resitor is given by
Answer:
You must add 8cm of water to the tank
Explanation:
In order to find how much the height is we will use the Snell Refraction law
.
This law relates the index of refraction of the water (n2), the index of refraction of the air (n1), the incidence angle relative to the vertical (theta1) and the refraction angle relative to the vertical (theta2) by using the next equation:
Then we will find the refraction angle relative to the vertical this way:
Then, theta2=32.12°
Now that we have this information we can imagine a triangle with a 30cm height and a 32.12° angle. This way we can find how much X is, this X will be the distance between the vertical line and the spot the beam hits the bottom, so we can use some trigonometry to find it, this way:
Then, X=18.8cm, we can approximate it to 19cm
Once we have X we will add 5cm to it which is how much the beam needs to be moved, then the new X will be 24cm
Now, with the new horizontal distance we will find the new vertical distance, let´s call it Y, this way we will know how much water we must add to move the beam, then we will have a triangle with a vertical distance called Y, the same 32.12° angle will be used as we are still working with the air-water interface and a 19cm horizontal distance, then:
Then, Y=38cm
In this case, you must add 8cm of water to the tank to move the beam on the bottom 5cm