Nuclear sizes are expressed in a unit named
1 answer:
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To develop this problem it is necessary to apply the concepts related to Gravitational Potential Energy.
Gravitational potential energy can be defined as
As M=m, then
Where,
m = Mass
G =Gravitational Universal Constant
R = Distance /Radius
PART A) As half its initial value is u'=2u, then
Therefore replacing we have that,
Re-arrange to find v,
Therefore the velocity when the separation has decreased to one-half its initial value is 816m/s
PART B) With a final separation distance of 2r, we have that
Therefore
Therefore the velocity when they are about to collide is
Answer:
Fnet - Fg
Explanation:
When an object is in an elevator, its weight varies with respect to the direction of movement of the elevator and the elevators acceleration.
The weight, W, of an object can be expressed as;
W = mg
where m is the object's mass, and g is the acceleration due gravity.
If the object is in an elevator that speed up, an apparent weight would be felt since both mass and elevator are moving against gravitational pull of the earth.
So that,
= mg + ma
where: mg is the weight of the object, and ma is the apparent weight.
Apparent weight (ma) = - mg
Answer:
a) 2.063*10^-4
b) 1.75*10^-4
Explanation:
Given that: d= 1.628 mm = 1.628 x 10-3 I= 12 mA = 12.0 x 10-8 A The Cross-sectional area of the wire is:
a) <em>The Potential difference across a 2.00 in length of a 14-gauge copper </em>
<em> wire: </em>
L= 2.00 m
From Table Copper Resistivity = 1.72 x 10-8 S1 • m The Resistance of the Copper wire is:
=0.0165Ω
The Potential difference across the copper wire is:
V=IR
=2.063*10^-4
b) The Potential difference if the wire were made of Silver: From Table: Silver Resistivity p= 1.47 x 10-8 S1 • m
The Resistance of the Silver wire is:
=0.014Ω
The Potential difference across the Silver wire is:
V=IR
=1.75*10^-4