This is the equation for the formation of magnesium chloride. Mg(s) + 2HCl(l) → MgCl2(aq) + H2(g) Which are the reactants and th
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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:
Explanation:
Given that
Constant rate of leak =R
Mass at time T ,m=RT
At any time t
The mass = Rt
So the total mass in downward direction=(M+Rt)
Now force equation
(M+Rt) a =P- (M+Rt) g
We know that
This is the velocity of bucket at the instance when it become empty.
Answer:
<em>a) 3.56 x 10^22 N</em>
<em>b) 3.56 x 10^22 N</em>
<em></em>
Explanation:
Mass of the sun M = 2 x 10^30 kg
mass of the Earth m = 6 x 10^24 kg
Distance between the sun and the Earth R = 1.5 x 10^11 m
From Newton's law,
F =
where F is the gravitational force between the sun and the Earth
G is the gravitational constant = 6.67 × 10^-11 m^3 kg^-1 s^-2
m is the mass of the Earth
M is the mass of the sun
R is the distance between the sun and the Earth.
Substituting values, we have
F = = <em>3.56 x 10^22 N</em>
<em></em>
A) The force exerted by the sun on the Earth is equal to the force exerted by the Earth on the Sun also, and the force is equal to <em>3.56 x 10^22 N</em>
b) The force exerted by the Earth on the Sun = <em>3.56 x 10^22 N</em>