Latitude and longitude picture.
1 answer:
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Answer:
W = 19.845 J
Explanation:
Work is defined as W = Fdcos, where F is the force exerted and d is the distance. Because the direction the ball is falling is the same direction as the force itself,
= 0 deg, and since cos(0) = 1, this equation is equivalent to W = Fd. In this case, the force exerted is the weight force, which is equivalent to m * g. Substituting you get:
W = mgd = 0.810 kg * 9.8 m/s^2 * 2.5m
W = 19.845 J
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:
The image mentioned is in the attachment
Answer: a) P = 2450 Pa;
b) P = 2940 Pa;
c) F = 4.9 N
Explanation:
a) Pressure is a force applied to a surface of an object or fluid per unit area.
The image shows a block applying pressure on the large side of the piston. The force applied is due to gravitation, so:
P =
P =
P =
P = 2450 Pa
The pressure generated by the block is P = 2450 Pa.
b) A static liquid can also exert pressure and can be calculated as:
ρ.g.h
where
ρ is the density of the fluid
h is the depth of the fluid
g is acceleration of gravity
600.9.8.0.5
2940 Pa
The pressure in the fluid at 50 cm deep is 2940 Pa.
c) For the system to be in equilibrium both pressures, pressure on the left side and pressure on the right side, have to be the same:
=
F =
Adjusting the units, = 0.002 m².
F =
F = 4.9 N
The force necessary to be equilibrium is F = 4.9 N.
