Radiation present in the environment but not produced by humans is called ______.
1 answer:
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Answer:
E = 31.329 N/C.
Explanation:
The differential electric field at the center of curvature of the arc is
<em>(we have a cosine because vertical components cancel, leaving only horizontal cosine components of E. )</em>
where is the radius of curvature.
Now
,
where is the charge per unit length, and it has the value
Thus, the electric field at the center of the curvature of the arc is:
Now, we find and
. To do this we ask ourselves what fraction is the arc length 3.0 of the circumference of the circle:
and this is
radians.
Therefore,
evaluating the integral, and putting in the numerical values we get:
The center of mass of the two objects is the average position of the parts of the two object system
The center of mass of block <em>A</em>, and block <em>B</em> after displacement of block <em>B</em> is at <u>20 m from block </u><u><em>A</em></u>
<em />
Reason:
The given parameters are;
The position of block A and block B at t = 0 is x = 0
The mass of block A, m₁ = 8 kg
Mass of block B, m₂ = 16 kg
Speed of block <em>A</em> = 0 m/s
Speed of block <em>B</em>, v₂ = 10 m/s
Location of the center of mass of the two object at t = 3 s; Required
Solution;
The location of block <em>A</em>, after 3 s is x₁ = 0 (block A is at rest)
The location of block <em>B</em>, = v₂ × t
The location of block <em>B</em>, after 3 s is x₂ = 10 m/s × 3 s = 30 m
The center of mass of two masses are given as follows;
The center of mass of the two objects is at at the position x = <u>20 m</u> (from block <em>A</em>)
Learn more about the center of mass here:
Answer:
a) see attached, a = g sin θ
b)
c) v = √(2gL (1-cos θ))
Explanation:
In the attached we can see the forces on the sphere, which are the attention of the bar that is perpendicular to the movement and the weight of the sphere that is vertical at all times. To solve this problem, a reference system is created with one axis parallel to the bar and the other perpendicular to the rod, the weight of decomposing in this reference system and the linear acceleration is given by
Wₓ = m a
W sin θ = m a
a = g sin θ
b) The diagram is the same, the only thing that changes is the angle that is less
θ' = 9/2 θ
c) At this point the weight and the force of the bar are in the same line of action, so that at linear acceleration it is zero, even when the pendulum has velocity v, so it follows its path.
The easiest way to find linear speed is to use conservation of energy
Highest point
Em₀ = mg h = mg L (1-cos tea)
Lowest point
Emf = K = ½ m v²
Em₀ = Emf
g L (1-cos θ) = v² / 2
v = √(2gL (1-cos θ))
