A helium nucleus is composed of two protons (positively charged) and two neutrons (charge neutral). The distance between the two
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Answer:
<h2>40 kg</h2>Explanation:
Find the diagram relating to the question for proper explanation of the question below.
Using the principle of moment
Sum of clockwise moments = Sum of anticlockwise moments
Moment = Force * perpendicular distance
For anti-clockwise moment:
Since the 30 kg moves in the anticlockwise direction according to the diagram
ACW moment = 30 * 1 = 30 kgm
For clockwise moment
If another child sits 0.75 m away from the pivot point on the opposite side, moment of the child in clockwise direction = M * 0.75 = 0.75M (M is the mass of the unknown child).
Equating both moments we have;
0.75M = 30
M = 30/0.75
M = 40 kg
The second child's mass is 40 kg
Answer:
a) x = 0.098
b) x = 2.72 m
Explanation:
(a) To find the stretch of the fire net when the same person is lying in it, you can assume that the net is like a spring with constant spring k. It is necessary to find k.
When the person is falling down he acquires a kinetic energy K, this energy is equal to the elastic potential energy of the net when it is max stretched.
Then, you have:
(1)
m: mass of the person = 62kg
k: spring constant = ?
v: velocity of the person just when he touches the fire net = ?
x: elongation of the fire net = 1.4 m
Before the calculation of the spring constant, you calculate the final velocity of the person by using the following formula:
vo: initial velocity = 0 m/s
g: gravitational acceleration = 9.8 m/s^2
y: height from the person jumps = 20.0m
With this value you can find the spring constant k from the equation (1):
When the person is lying on the fire net the weight of the person is equal to the elastic force of the fire net:
you solve the last expression for x:
When the person is lying on the fire net the elongation of the fire net is 0.098m
b) To find how much would the net stretch, If the person jumps from 38 m, you first calculate the final velocity of the person again:
Next, you calculate x from the equation (1):
The net fire is stretched 2.72 m