Question 2 Ammonium phosphate NH43PO4 is an important ingredient in many fertilizers. It can be made by reacting phosphoric acid
1 answer:
Answer:
The answer is 12.27 g (NH4)3PO4
Explanation:
Step 1: balance the chemical equation:
H3PO4 + 3NH3 → (NH4)3PO4
step 2: the molar masses of each of the reagents and products are calculated using the periodic table:
H3PO4:
3 atoms of H: 3x1=3 g/mol
1 atom of P: 1x30.97=30.97 g/mol
4 atoms of O: 4x16=64 g/mol
Molar mass=3+30.97+64=97.97 g/mol
3NH3:
3 atoms of N: 3x14=42 g/mol
9 atoms of H: 9x1=9 g/mol
Molar mass=42+9=51 g/mol
(NH4)3PO4:
3 atoms of N: 3x14=42 g/mol
12 atoms of H: 12x1=12 g/mol
1 atom of P: 1x30.97=30.97 g/mol
4 atoms of O: 4x16=64 g/mol
Molar mass=42+12+30.97+64=148.97 g/mol
we make a rule of three to calculate the amount of ammonium phosphate:
51 g NH3------------------148.97 g (NH4)3PO4
4.2 g NH3-----------------x g (NH4)3PO4
Clearing the x, we have:
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Answer:
The magnitude of the acceleration of the box is 2.01 m/s².
Explanation:
Hi there!
Please, see the attached figure for a graphical description of the problem.
We have the following horizontal forces:
Fr = friction force.
Fx = Horizontal component of the applied force, F.
And we have the following vertical forces:
Fy = vertical component of the applied force.
N = normal force exerted on the box.
W = weight of the box.
According to Newton´s second law:
∑F = m · a
Then, in the horizontal direction:
Fx - Fr = m · a
Where "m" is the mass of the box and "a" its acceleration.
Fx can be obtained by trigonometry (see figure):
Fx = F · cos 30°
Fx = 90.0 N · cos 30°
Fr is calculated as follows:
Fr = μ · N
Where μ is the coefficient of friction and N the normal force.
So, we have to find the magnitude of the normal force.
Using Newton´s second law in the vertical direction:
∑F = N + Fy - W = m · a
Notice that the box has no vertical acceleration, then:
N + Fy - W = 0
Solving for N:
N = W - Fy
The weight is calculated as follows:
W = m · g
Where g is the acceleration due to gravity:
W = 20.0 kg · 9.8 m/s² = 196 N
And the vertical component of the applied force can be obtained by trigonometry:
Fy = F · sin 30°
Fy = 90.0 N · sin 30°
The normal force will be:
N = W - Fy = 196 N - 90.0 N · sin 30°
N = 151 N
Now, we can calculate the friction force:
Fr = μ · N
Fr = 0.250 · 151 N
Fr = 37.8 N
And now, we can obtain the acceleration of the box:
Fx - Fr = m · a
(Fx - Fr) / m = a
(90.0 N · cos 30° - 37.8 N ) / 20.0 kg = a
a = 2.01 m/s²
The magnitude of the acceleration of the box is 2.01 m/s².
Answer:C
Explanation:
When a constant horizontal force is applied to the box, box started moving in the horizontal direction such that it moves with constant velocity
Constant velocity implies that net force on the box is zero
i.e. there must be an opposing force which is equal to the applied force and friction force can serve that purpose.
So option c is the correct choice.
The first: alright, first: you draw the person in the elevator, then draw a red arrow, pointing downwards, beginning from his center of mass. This arrow is representing the gravitational force, Fg.
You can always calculate this right away, if you know his mass, by multiplying his weight in kg by the gravitational constant
let's do it for this case:
the unit of your fg will be in Newton [N]
so, first step solved, Fg is 637.65N
Fg is a field force by the way, and at the same time, the elevator is pushing up on him with 637.65N, so you draw another arrow pointing upwards, ending at the tip of the downwards arrow.
now let's calculate the force of the elevator
so you draw another arrow which is pointing downwards on him, because the elevator is accelating him upwards, making him heavier
the elevator force in this case is a contact force, because it only comes to existence while the two are touching, while Fg is the same everywhere
You can always calculate this right away, if you know his mass, by multiplying his weight in kg by the gravitational constant
let's do it for this case:
the unit of your fg will be in Newton [N]
so, first step solved, Fg is 637.65N
Fg is a field force by the way, and at the same time, the elevator is pushing up on him with 637.65N, so you draw another arrow pointing upwards, ending at the tip of the downwards arrow.
now let's calculate the force of the elevator
so you draw another arrow which is pointing downwards on him, because the elevator is accelating him upwards, making him heavier
the elevator force in this case is a contact force, because it only comes to existence while the two are touching, while Fg is the same everywhere