Write a balanced molecular equation for the reaction that takes place when iron metal combines with aqueous hydrochloric acid, HCl, to form the iron(II) salt.
It is not necessary to include states such as (aq) or (g).
Fe(s) + 2HCl(aq) → FeCl2(aq) + H2(g)
a plane weighing 242 kN(24.7) lands on an aircraft carrier.the plane is moving horizontally at 65.3 m/s (146.1 mi/h) when its tailhook grabs hold of the arresting cables.The cables bring the plane to a stop in a distance of 84.0m.How much work is done on the plane by the arresting cables
We can approach this problem using the work-energy principle:
Work done on the plane = Change in the plane's kinetic energy
The initial kinetic energy of the plane is:
K1 = (1/2)mv1^2
= (1/2)(242,000 kg)(65.3 m/s)^2
= 502,382,300 J
The final kinetic energy of the plane is zero, since it comes to a stop. Therefore, the change in kinetic energy is:
ΔK = K2 - K1 = -K1
The negative sign indicates that the kinetic energy of the plane decreases (since it stops). Therefore, the work done on the plane by the arresting cables is:
W = -ΔK = K1 = 502,382,300 J
The work done on the plane by the arresting cables is 502,382,300 J.
the mechanical power output of a cyclist moving at a constant speed of 5.00 m/s on a level road is 115W.What is the force exerted on the cyclist and the bicycle by the air
We can use the equation for mechanical power:
P = Fv
where P is power, F is force, and v is velocity. Since the cyclist is moving at a constant speed, there is no acceleration and no net force acting on the cyclist. Therefore, the force exerted by the air must be equal in magnitude and opposite in direction to the force exerted by the cyclist on the air. We can write:
P = -F_air*v
where F_air is the force exerted by the air on the cyclist. Solving for F_air, we get:
F_air = -P/v = -(115 W)/(5.00 m/s) = -23.0 N
The negative sign indicates that the force exerted by the air is in the opposite direction to the cyclist's motion. Therefore, the force exerted on the cyclist and the bicycle by the air is 23.0 N.
the mechanical power output of a cyclist moving at a constant speed of 5.00m/s on a level road is 115W.by bending low over the handlebars ,the cyclist reduces the air resistance to 18.0N .If she maintains a power output of 115W what will her speed be
We can use the equation for mechanical power and the work-energy principle to solve this problem. Firstly, we can find the initial force exerted by the air on the cyclist using:
P = -F_air*v1
where v1 is the initial velocity with full air resistance. Solving for F_air, we get:
F_air = -P/v1 = -(115 W)/(5.00 m/s) = -23.0 N
The force exerted by the cyclist on the pedals is:
F_pedals = F_friction + F_air = m*a
where F_friction is the force of rolling friction and m*a is the net force on the cyclist. Since the cyclist is moving at a constant speed, the acceleration is zero and the net force is zero. Therefore:
F_pedals = F_friction + F_air = 0
Solving for F_friction, we get:
F_friction = -F_air = 18.0 N
Now we can use the work-energy principle to find the cyclist's new speed:
W = ΔK = (1/2)m*v2^2 - (1/2)m*v1^2
where W is the work done by the cyclist's legs, ΔK is the change in kinetic energy, and m is the mass of the cyclist and the bicycle (which we assume is constant). Rearranging and substituting in the values we know, we get:
v2 = sqrt(v1^2 + 2*W/m)
where we can find W using:
W = P*t
where t is the time the cyclist maintains a power output of 115 W. Assuming the time is 1 hour (or 3600 seconds), we get:
W = P*t = (115 W)*(3600 s) = 414000 J
Substituting in the appropriate values, we get:
v2 = sqrt((5.00 m/s)^2 + 2*(414000 J)/(m*))
We can simplify this expression by assuming the mass of the cyclist and the bicycle is around 100 kg (a reasonable assumption for most road bikes). Plugging in this value, we get:
v2 = sqrt((5.00 m/s)^2 + 8280 m^2/s^2)
v2 = 29.2 m/s
Therefore, the cyclist's new speed is approximately 29.2 m/s (or 104.9 km/h).
a child playground swing is supported by chains are vertical,what is the maximum height of the swing ?Assume the masses of the chains are negligible
The maximum height of the swing is when all of the potential energy at the highest point is converted into kinetic energy at the lowest point. We can use the conservation of energy to find the maximum height.
At the highest point, the swing is momentarily at rest and has no kinetic energy. Therefore, all of its energy is potential energy due to its height above the lowest point. At the lowest point, the swing is at its maximum speed and has no potential energy. Therefore, all of its energy is kinetic energy. By conservation of energy, we can equate the two energies at the highest and lowest points:
Potential energy at highest point = Kinetic energy at lowest point
mgh = (1/2)mv^2
where m is the mass of the child and the swing, g is the acceleration due to gravity, h is the height above the lowest point, and v is the maximum speed of the swing at the lowest point.
Simplifying and solving for h, we get:
h = (1/2)v^2/g
We can find the maximum speed of the swing using:
v = sqrt(2gL)
where L is the length of the chains from the pivot point to the seat of the swing. Substituting this expression into the equation for h, we get:
h = (1/2)*(sqrt(2gL))^2/g = L
Therefore, the maximum height of the swing is equal to its length (assuming the chains are vertical and have no slack).