Part CPart complete Determine the maximum force P that can be applied without causing the two 43-kg crates to move. The coeffici
1 answer:
Answer:
the maximum force will be equal to 134.84 N
Explanation:
We have given mass m = 43 kg
Coefficient of static friction
Acceleration due to gravity
We have to find the maximum force which , when applied there is no movement of crates
This maximum force will be equal to frictional force
Frictional force is given by
So the maximum force will be equal to 134.84 N
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Answer:
Amplitude = 8 Volts
Frequency = 0.067 kHz
Explanation:
Note: The missing picture in question is attached for your review.
Given:
Volts/Div = 2 V/div
Time/Div = 5 msec/div
Finding Amplitude:
Now, as you can see in the attached picture, there are 4 division between two peaks of the waveform, so,
(Multiplying by 2 V/div because oscilloscope dial is set at 2 V/div)
Finding Frequency:
As can be seen in attached picture, 3 division are there for one complete cycle of waveform,so,
Since,
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: