Answer:
a = 5.5 [m/s²]
Explanation:
To solve this problem we must use Newton's second law, which tells us that the force is equal to the product of mass by acceleration.
ΣF =m*a
where:
F = Force [N] (units of Newtons)
m = mass = 20 [kg]
a = acceleration [m/s²]
Let's take the force of 230 [N] as positive, in this way the other force will be negative, by pointing in the opposite direction.
![230 - 120 = 20*a\\110 = 20*a\\a=5.5 [m/s^{2} ]](https://tex.z-dn.net/?f=230%20-%20120%20%3D%2020%2Aa%5C%5C110%20%3D%2020%2Aa%5C%5Ca%3D5.5%20%5Bm%2Fs%5E%7B2%7D%20%5D)
Answer:
The planets and moons that orbit in the solar system.
Explanation:
For example the earth moves at 67,000 mph (107,000 km/h), and is constant from the gravitational pull of the sun. The moon orbits at about 2,288 mph (3,683 km/h). these are both traveling at different velocities but at a constant speed.
Answer: D. the wave has traveled 97.2 cm in 1 second.
Explanation:
Got it right on edge, hope this helps :)
Answer:
66.4 m
Explanation:
To solve the problem, we can use the length contraction formula, which states that the length observed in the reference frame moving with the object (the rocket) is given by

where
is the proper length (the length measured from an observer at rest)
v is the speed of the object (the rocket)
c is the speed of light
Here we know
v = 0.85c
L = 35.0 m
So we can re-arrange the equation to find the length of the rocket at rest:

The magnitude and direction of the electric field in the wire are mathematically given as
![L &=[(v / L) v / m] \hat{i}](https://tex.z-dn.net/?f=L%20%26%3D%5B%28v%20%2F%20L%29%20v%20%2F%20m%5D%20%5Chat%7Bi%7D)
<h3>What is the magnitude and direction of the electric field in the wire?</h3>
Generally, the equation for is mathematically given as
A cylindrical wire that is straight and parallel to the x-axis has the following dimensions: length L, diameter d, resistivity p, diameter d, potential v, and z length. combining elements from both sides
E d 
![\begin{aligned}&-E \int_0^L d x=\int_v^0 d v \\\therefore E \cdot L &=v \\L &=[(v / L) v / m] \hat{i}\end{aligned}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7D%26-E%20%5Cint_0%5EL%20d%20x%3D%5Cint_v%5E0%20d%20v%20%5C%5C%5Ctherefore%20E%20%5Ccdot%20L%20%26%3Dv%20%5C%5CL%20%26%3D%5B%28v%20%2F%20L%29%20v%20%2F%20m%5D%20%5Chat%7Bi%7D%5Cend%7Baligned%7D)
In conclusion, the magnitude and direction of the electric field in the wire are given as
![L &=[(v / L) v / m]](https://tex.z-dn.net/?f=L%20%26%3D%5B%28v%20%2F%20L%29%20v%20%2F%20m%5D)
Read more about electric fields
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