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).
a child playground swing is supported by chains that are 4.00m long.If a child in the swing is 0.36m above the ground and moving at 5.90m/s when the chains are vertical ,what is the maximum height of the swing ?Assume the masses of the chains are negligible
We can use the conservation of energy principle to find the maximum height of the swing if we know the initial speed of the swing and the height of the child above the ground. At the highest point of the swing, the child is momentarily at rest before it starts to move downwards, so its kinetic energy is zero. At this highest point, all the energy is potential energy due to the height of the child above the ground. At the lowest point of the swing, the child has maximum speed but zero potential energy. Therefore, the total energy is conserved, and we can express the conservation of energy as:
PE_max = KE_max
where PE_max is the potential energy at the highest point and KE_max is the kinetic energy at the lowest point.
We can express the potential energy in terms of the height h above the equilibrium point of the swing and the mass m of the child and the swing as:
PE_max = mgh
The kinetic energy can be expressed in terms of the mass and speed at the point where the chains are vertical as:
KE_max = (1/2)mv^2
We can equate the two energies using conservation of energy:
mgh = (1/2)mv^2
Solving for h, we get:
h = (1/2)v^2/g
where g is the acceleration due to gravity. We can substitute v and g with the values given in the problem and calculate h as:
h = (1/2)*(5.90 m/s)^2/9.81 m/s^2 = 1.76 m
Therefore, the maximum height of the swing is 1.76 m above the equilibrium point.
a gymnast of mass 52.0 kg is jumping on a trampoline.She jumps so that her feet stretch the trampoline 68.0cm down.How far does the trampoline stretch when she stands on it at rest?Assume that the trampoline is described by hooke law when it is stretched
We can use Hooke's Law to solve this problem, which states that the force required to stretch a spring (or a trampoline) is directly proportional to the amount it is stretched. We can write this as:
F = -kx
where F is the force applied to the trampoline, x is the displacement of the trampoline from its rest position, and k is the spring constant (a measure of the stiffness of the trampoline).
We can first determine the spring constant by using the given information that the gymnast's feet stretch the trampoline 68.0 cm down. At this point, the trampoline is at its maximum displacement, so the force acting on the trampoline is equal to the weight of the gymnast. Therefore, we can write:
F = mg
where m is the mass of the gymnast and g is the acceleration due to gravity. At the maximum displacement, x = 0.68 m. Substituting these values into Hooke's Law, we get:
mg = -k(0.68 m)
Solving for k, we get:
k = -(mg/0.68 m) = -(52.0 kg)(9.81 m/s^2)/0.68 m = -755 N/m
Now we can use this spring constant to find the displacement of the trampoline when the gymnast stands on it at rest (i.e., when there is no net force acting on the trampoline). In this case, the force applied to the trampoline is just the weight of the gymnast, so we can write:
F = mg = -kx
Solving for x, we get:
x = -(mg)/k = -(52.0 kg)(9.81 m/s^2)/(-755 N/m) = 0.675 m
Therefore, the trampoline stretches by 0.675 m (67.5 cm) when the gymnast stands on it at rest. Note that the sign of x is negative because the trampoline is displaced downwards from its rest position.
a gymnast of mass 52.0 kg is jumping on a trampoline . she jumps so that her feet reach a maximum height of 3.41 m above the trampoline and when she lands her feet stretch thr trampoline 68.0 cm down.How far does the trampoline stretch when she stands on it at rest?Assume that the trampoline is described by Hooke law when it is stretched.
We can use the law of conservation of energy to solve this problem. At the maximum height, the gymnast has zero kinetic energy and has potential energy due to her height above the trampoline. At the lowest point (when she lands), she has maximum kinetic energy and zero potential energy. Therefore, we can write:
mgh = (1/2)mv^2
where m is the mass of the gymnast, g is the acceleration due to gravity, h is the maximum height, and v is the velocity of the gymnast when she lands.
We can solve this equation for v:
v = sqrt(2gh)
Substituting the given values, we get:
v = sqrt(2 * 9.81 m/s^2 * 3.41 m) = 8.24 m/s
When the gymnast lands, the force on the trampoline is equal to the weight of the gymnast plus the force due to her velocity (which is downwards). Therefore, we can write:
F = mg + ma
where a is the acceleration of the gymnast as she comes to a stop on the trampoline. We can assume that this acceleration is constant and use the relationship between acceleration and displacement for an object under constant acceleration:
a = -kx/m
where k is the spring constant of the trampoline, and x is the displacement of the trampoline from its rest position. At the maximum displacement, x = 0.68 m. Rearranging this equation, we get:
x = -(ma)/k
Substituting for F and a, we get:
x = -(mg + ma)/k
Simplifying and substituting the given values, we get:
x = -[(52 kg)(9.81 m/s^2) + (52 kg)(8.24 m/s)] / (-755 N/m) = 0.255 m
Therefore, the trampoline stretches 0.255 m (25.5 cm) when the gymnast stands on it at rest.
Ugonna stands at the top of an incline and pushes a 115 kg crate to get it started sliding down the incline .The crate slows to a halt after traveling 1.33 m along the incline.
We can use the work-energy principle to find the work done by Ugonna as he pushes the crate and the potential energy change of the crate as it moves down the incline. By the work-energy principle, the work done on the crate by Ugonna must be equal to the change in the crate's kinetic and potential energy:
Work done by Ugonna = Potential energy change + Kinetic energy change
The kinetic energy of the crate is initially zero, since it is at rest. At the end of the incline, the crate is also momentarily at rest, so its kinetic energy is zero. Therefore, the change in kinetic energy is zero, and we can simplify the equation to:
Work done by Ugonna = Potential energy change
The potential energy of the crate at the top of the incline is:
PE_1 = mgh
where m is the mass of the crate, g is the acceleration due to gravity, and h is the height of the incline. The potential energy of the crate at the bottom of the incline is:
PE_2 = mgh
where h is now the distance between the top of the incline and the bottom of the incline, which we need to find. The work done by Ugonna is the force he applies to the crate multiplied by the distance over which he applies the force:
Work done by Ugonna = F * d
where F is the force applied by Ugonna and d is the distance he pushes the crate before releasing it. The force of friction opposes the motion of the crate down the incline, so we have:
F = f_friction + mgsinθ
where f_friction is the force of friction, θ is the angle of the incline, and mgsinθ is the component of the crate's weight parallel to the incline.
Therefore, we can write:
Work done by Ugonna = (f_friction + mgsinθ)*d
Solving for h, we get:
h = (d*f_friction + d*mgsinθ - Work done by Ugonna) / mg
We need to find the force of friction in order to find the height h. The force of friction is equal to the normal force times the coefficient of kinetic friction, which we assume is 0.35 for wood on wood. The normal force on the crate is:
N = mgcosθ
where θ is the angle of the incline. Therefore, we have:
f_friction = μ_kinetic*N = μ_kinetic*mgcosθ
Substituting this expression into our equation for h, we get:
h = (d*μ_kinetic*mgcosθ + d*mgsinθ - Work done by Ugonna) / mg
Substituting in the given values, we get:
h = (1.33 m * 0.35 * 115 kg * 9.81 m/s^2 * cos(θ) + 1.33 m * 115 kg * 9.81 m/s^2 * sin(θ) - Work done by Ugonna) / (115 kg * 9.81 m/s^2)
We need to know the work done by Ugonna in order to find h. We assume that Ugonna applies a constant force over the distance he pushes the crate. The force of Ugonna can be found by writing the equation of motion for the crate on the incline, assuming that the acceleration is constant and equal to the acceleration due to gravity times the sin of the angle of the incline:
F_net = ma = mgsinθ - f_friction
where f_friction is the force of kinetic friction, and we assume that the crate is moving with constant velocity down the incline. Therefore:
f_friction = μ_kinetic*N = μ_kinetic*mgcosθ
Substituting this into the equation of motion, we get:
F_net = mgsinθ - μ_kinetic*mgcosθ = ma
Solving for a, we get:
a = g*(sinθ - μ_kinetic*cosθ)
Substituting the given values, we get:
a = 9.81 m/s^2 * (sin(θ) - 0.35*cos(θ))
We can use this acceleration to find the force exerted by Ugonna using Newton's second law:
F_net = ma = Ugonna's force - f_friction = Ugonna's force - μ_kinetic*mgcosθ
Therefore, we get:
Ugonna's force = mgsinθ + μ_kinetic*mgcosθ
We can substitute this expression for Ugonna's force into our equation for h, along with the given values:
h = (1.33 m * 0.35 * 115 kg * 9.81 m/s^2 * cos(θ) + 1.33 m * 115 kg * 9.81 m/s^2 * sin(θ) - Ugonna's force * 1.33 m) / (115 kg * 9.81 m/s^2)
We can now solve this expression for h for different values of θ (the angle of the incline) to find the height h corresponding to the known distance, force, and mass. For