"1.<span> characterized by order and planning</span>
2.<span> not haphazard"
is the answers i found in the dictionary </span>
The satellite exerts a force of 1900 N on the Earth
Explanation:
To answer this question, let's remind Newton's third law:
"When an object A exerts a force (called action) on an object B, then object B exerts an equal and opposite force (called reaction) on object A"
In this problem, we can identify:
The Earth as object A
The satellite as object B
This means that if we apply the law, the force that the Earth exerts on the satellite must be equal (in magnitude) and opposite (in direction) to the force the satellite exerts on the Earth.
Since the Earth exerts a force of 1900 N on the satellite, this means that the satellite also exerts a force of 1900 N on the Earth.
Learn more about Newton's third law:
brainly.com/question/11411375
#LearnwithBrainly
The kinetic energy of an object is directly proportional to its mass and the square of its velocity
KE = 1/2 (mv²)
KE = Kinetic Energy
m = mass in kg
v = velocity in m/s
Given:
m = .8 kg
v = 11.2 m/s
Substitute:
KE = 1/2 (.8)(11.2²)
KE = 50.18 J
Answer:
2.5 m
Explanation:
Weight of billboard worker = 800 N
Number of ropes = 2
Length of scaffold = 4 m
Weight of scaffold = 500 N
Tension in rope = 550 N
The sum of the torques will be

The position of the person will be 2.5 m
Complete Question
An aluminum "12 gauge" wire has a diameter d of 0.205 centimeters. The resistivity ρ of aluminum is 2.75×10−8 ohm-meters. The electric field in the wire changes with time as E(t)=0.0004t2−0.0001t+0.0004 newtons per coulomb, where time is measured in seconds.
I = 1.2 A at time 5 secs.
Find the charge Q passing through a cross-section of the conductor between time 0 seconds and time 5 seconds.
Answer:
The charge is 
Explanation:
From the question we are told that
The diameter of the wire is 
The radius of the wire is 
The resistivity of aluminum is 
The electric field change is mathematically defied as

Generally the charge is mathematically represented as

Where A is the area which is mathematically represented as

So

Therefore

substituting values
![Q = 120 \int\limits^{t}_{0} { [ 0.0004t^2 - 0.0001t +0.0004] } \, dt](https://tex.z-dn.net/?f=Q%20%3D%20120%20%5Cint%5Climits%5E%7Bt%7D_%7B0%7D%20%7B%20%5B%200.0004t%5E2%20-%200.0001t%20%2B0.0004%5D%20%7D%20%5C%2C%20dt)
![Q = 120 [ \frac{0.0004t^3 }{3} - \frac{0.0001 t^2}{2} +0.0004t] } \left | t} \atop {0}} \right.](https://tex.z-dn.net/?f=Q%20%3D%20120%20%5B%20%5Cfrac%7B0.0004t%5E3%20%7D%7B3%7D%20-%20%5Cfrac%7B0.0001%20t%5E2%7D%7B2%7D%20%2B0.0004t%5D%20%7D%20%20%5Cleft%20%7C%20t%7D%20%5Catop%20%7B0%7D%7D%20%5Cright.)
From the question we are told that t = 5 sec
![Q = 120 [ \frac{0.0004t^3 }{3} - \frac{0.0001 t^2}{2} +0.0004t] } \left | 5} \atop {0}} \right.](https://tex.z-dn.net/?f=Q%20%3D%20120%20%5B%20%5Cfrac%7B0.0004t%5E3%20%7D%7B3%7D%20-%20%5Cfrac%7B0.0001%20t%5E2%7D%7B2%7D%20%2B0.0004t%5D%20%7D%20%20%5Cleft%20%7C%205%7D%20%5Catop%20%7B0%7D%7D%20%5Cright.)
![Q = 120 [ \frac{0.0004(5)^3 }{3} - \frac{0.0001 (5)^2}{2} +0.0004(5)] }](https://tex.z-dn.net/?f=Q%20%3D%20120%20%5B%20%5Cfrac%7B0.0004%285%29%5E3%20%7D%7B3%7D%20-%20%5Cfrac%7B0.0001%20%285%29%5E2%7D%7B2%7D%20%2B0.0004%285%29%5D%20%7D)
