5. You are skateboarding on top of a small hill and take a break. You have 100 J of potential energy. If
1 answer:
Given the potential energy of 100 J, the height of the hill is 0.128 m (12.8 cm)
Explanation:
The gravitational potential energy of an object is given by:
where
m is the mass of the object
g is the acceleration of gravity
h is the heigth of the object, relative to the ground
In this problem, we have
U = 100 J is the potential energy of the person
is the acceleration of gravity
m = 80 kg is the mass of the person
Solving for h, we find at what height the person is:
Learn more about gravitational potential energy here:
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Answer:
During those 3.00 seconds before stopping, the car travels a distance of 6 m.
Explanation:
The simple rule of three is a tool that is used to quickly solve problems, where three pieces of information must be known, and one of them operates as an unknown to be known.
Two magnitudes are directly proportional if one magnitude increases the other also does it, and if the magnitude decreases the other in the same way.
Being a, b and c known data and x the unknown, the value that we want to know, the rule of three when the magnitudes are directly proportional is applied as follows:
a ⇒ b
c ⇒ x
So:
In this case, knowing that a truck travels at 2 m/s, the rule of three applies as follows: if in 1 second the truck travels 2 m, in 3 seconds how much distance does it travel?
distance= 6 m
<u><em> During those 3.00 seconds before stopping, the car travels a distance of 6 m.</em></u>
Answer:
0.03167 m
1.52 m
Explanation:
x = Compression of net
h = Height of jump
g = Acceleration due to gravity = 9.81 m/s²
The potential energy and the kinetic energy of the system is conserved
The spring constant of the net is 20130.76 N
From Hooke's Law
The net would strech 0.03167 m
If h = 35 m
From energy conservation
Solving the above equation we get
The compression of the net is 1.52 m
Given:
v = 50.0 m/s, the launch velocity
θ = 36.9°, the launch angle above the horizontal
Assume g = 9.8 m/s² and ignore air resistance.
The vertical component of the launch velocity is
Vy = (50 m/s)*sin(50°) = 30.02 m/s
The time, t, to reach maximum height is given by
(30.02 m/s) - (9.8 m/s²)*(t s) = 0
t = 3.0634 s
The time fo flight is 2*t = 6.1268 s
The horizontal velocity is
u = (50 m/s)cos(36.9°) = 39.9842 m/s
The horizontal distance traveled at time t is given in the table below.
Answer:
t, s x, m
------ --------
0 0
1 39.98
2 79.79
3 112.68
4 159.58
5 199.47
6 239.37
v = 50.0 m/s, the launch velocity
θ = 36.9°, the launch angle above the horizontal
Assume g = 9.8 m/s² and ignore air resistance.
The vertical component of the launch velocity is
Vy = (50 m/s)*sin(50°) = 30.02 m/s
The time, t, to reach maximum height is given by
(30.02 m/s) - (9.8 m/s²)*(t s) = 0
t = 3.0634 s
The time fo flight is 2*t = 6.1268 s
The horizontal velocity is
u = (50 m/s)cos(36.9°) = 39.9842 m/s
The horizontal distance traveled at time t is given in the table below.
Answer:
t, s x, m
------ --------
0 0
1 39.98
2 79.79
3 112.68
4 159.58
5 199.47
6 239.37