Answer:
v=30 m/s
Explanation:
h - height
g - acceleration due to gravity=10
t - time
v- velocity
45 = 5t²
t² = 9
t=3 seconds
v=g×t
v=10×3
v=30 m/s
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2 answers:
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Answer:
a) 156960 N/m
b) Yes the player can play on the team.
Explanation:
F = Force
a = Acceleration due to gravity = 9.81 m/s²
m = Mass
k = Spring constant
x = Deformation length of spring
F = ma
From Hooke's law
The effective spring constant is 156960 N/m
When x = 0.48 cm = 0.0048 m
The mass of the player is 76.8 kg which is less than the required mass limit of the players' which is 85 kg. So, the player is eligible to play.
Number 3.
When you have two parallel wire and the charges are moving, they create a magnetic field around the wire (right-hand rule, thumb points in the direction of current).
If you want the math, here ya go:
F: magnetic force on the wire, L: length of the wire I: current in one wire I': current in other wire r: distance of the wire.
As we can see, two wires that has current through it will generate a force on each other.
When you have two parallel wire and the charges are moving, they create a magnetic field around the wire (right-hand rule, thumb points in the direction of current).
If you want the math, here ya go:
F: magnetic force on the wire, L: length of the wire I: current in one wire I': current in other wire r: distance of the wire.
As we can see, two wires that has current through it will generate a force on each other.
To solve this problem it is necessary to apply the concepts related to intensity as a function of power and area.
Intensity is defined to be the power per unit area carried by a wave. Power is the rate at which energy is transferred by the wave. In equation form, intensity I is
The area of a sphere is given by
So replacing we have to
Since the question tells us to find the proportion when
So considering the two intensities we have to
The ratio between the two intensities would be
The power does not change therefore it remains constant, which allows summarizing the expression to
Re-arrange to find
Therefore the intensity at five times this distance from the source is