Which statement best describes the force of gravity
1 answer:
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First, you need to make certain assumptions before solving this question. Why? Because there are no information given about the direction of these forces. In such questions as above, ALWAYS make the following assumptions:
1) Take first force, say
, and assume that it is pointing towards the x-direction.
Let us take the 7N force! By keeping the above assumption in our minds, the force vector would be like:
= 
, where 
= Unit vector in the x-direction.
2) Take the second force, say
, and assume that it is making an angle 
with the first force .
Let us take the 15N force! By keeping the above assumption in our minds, the forces vector would be like:
Now from simple vector addition, we know that,
--- (A)
Where
= Resultant vector.
NOTE: In equation (A), all forces are in vector notation. Assume that there is an arrow head on top of them.
Let us find
first!
= 
=> = 
Now the magnitude of is,
|| = 
=> || = 
Since
, therefore,
=> || = 
Since || = ||, and the magnitude of the resultant force is 20N, therefore,
|| = | |
20 =
Take square on both sides,
400 =
Ans: Angle formed by the two forces, 7N and 15N, is: 53.13°
-israr
1) Take first force, say
Let us take the 7N force! By keeping the above assumption in our minds, the force vector would be like:
2) Take the second force, say
Let us take the 15N force! By keeping the above assumption in our minds, the forces vector would be like:
Now from simple vector addition, we know that,
Where
NOTE: In equation (A), all forces are in vector notation. Assume that there is an arrow head on top of them.
Let us find
=>
Now the magnitude of
|
=> |
Since
=> |
Since |
|
20 =
Take square on both sides,
400 =
Ans: Angle formed by the two forces, 7N and 15N, is: 53.13°
-israr
The kinematic relationships we can find the position, acceleration and launch angle of the body on the planet Exidor.
a) the position are
time (s) x (m) y(m)
0 0 0
2.0 3.6 1.2
3.0 5.4 0.9
b) The aceleration is g = 0.6 m / s²
c) The launch angle θ = 33.7º
given parameters
- the initial velocity of the body vₓ = 1.8m / s and v_y = 1.2 m / s
- the movement times t = 1.0s, 2.0s and 3.0 s
to find
a) position
b) acceleration
c) launch angle
Projectile launch is an application of kinematics to the movement of the body in two dimensions where there is no acceleration on the x axis and the y axis has the planet's gravity acceleration
b) To calculate the acceleration of the plant acting on the y-axis, we use that the vertical velocity of the body at the highest point is zero.
v_y = v_{oy} - g t
where v and v({oy} are the velocities of the body, g the acceleration of the planet's gravity and t the time
0 = v_{oy} - gt
g = v_{oy} / t
from the graph we observe that the highest point occurs for t = 2.0 s
g = 1.2 / 2.0
g = 0.6 m / s²
a) The position is requested for several times
X axis
in this axis there is no acceleration so we can use the uniform motion relationships
vₓ = x / t
x = vₓ t
where x is the position, vx is the velocity and t is the time
we calculate for the time
t = 0.0 s
x₀ = 0
t = 2.0 s
x₂ = 1.8 2
x₂ = 3.6 m
t = 3.0 s
x₃ = 1.8 3
x₃ = 5.4 m
Y axis
In this axis there is the acceleration of the planet, let us use for the position the relation
y = v_{oy} t - ½ g t²
t = 0.0 s
y₀ = 0
y₀ = 0 m
t = 2.0 s
y₂ = 1.2 2 - ½ 0.6 2²
y₂ = 1.2 m
t = 3.0 s
y₃ = 1.2 3 - ½ 0.6 3²
y₃ = 0.9 m
c) the launch angle use the trigonometry relation
tan θ =
θ = tan⁻¹
θ = tan⁻¹
θ = 33.7º
measured counterclockwise from the positive side of the x-axis
With the kinematic relationships we can find the position, acceleration and launch angle of the body on the planet Exidor.
a) the position are
time (s) x (m) y(m)
0 0 0
2.0 3.6 1.2
3.0 5.4 0.9
b) The aceleration is g = 0.6 m / s²
c) The launch angle θ = 33.7ºto)
learn more about projectile launch here:
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: