Answer:
The maximum potential energy of the system is 0.2 J
Explanation:
Hi there!
When the spring is stretched, it acquires potential energy. When released, the potential energy is converted into kinetic energy. If there is no friction nor any dissipative forces, all the potential energy will be converted into kinetic energy according to the energy conservation theorem.
The equation of elastic potential energy (EPE) is the following:
EPE = 1/2 · k · x²
Where:
k = spring constant.
x = stretching distance.
The elastic potential energy is maximum when the block has no kinetic energy, just before releasing it.
Then:
EPE = 1/2 · 40 N/m · (0.1 m)²
EPE = 0.2 J
The maximum potential energy of the system is 0.2 J
Answer:
diameter of largest orbit is 0.60 m
Explanation:
given data
isotopes accelerates KE = 6.5 MeV
magnetic field B = 1.2 T
to find out
diameter
solution
first we find velocity from kinetic energy equation
KE = 1/2 × m×v² ........1
6.5 × 1.6 ×
= 1/2 × 1.672 ×
×v²
v = 3.5 ×
m/s
so
radius will be
radius =
........2
radius =
radius = 0.30
so diameter = 2 × 0.30
so diameter of largest orbit is 0.60 m
Answer: True
Explanation: Because of the way this water cycle has always circulated our planet, there is indeed a chance that the water in your glass is the same water that thirsty dinosaurs were drinking about 65 million years ago
Explanation:
(a) Displacement of an object is the shortest path covered by it.
In this problem, a student is biking to school. She travels 0.7 km north, then realizes something has fallen out of her bag. She travels 0.3 km south to retrieve her item. She then travels 0.4 mi north to arrive at school.
0.4 miles = 0.64 km
displacement = 0.7-0.3+0.64 = 1.04 km
(b) Average velocity = total displacement/total time
t = 15 min = 0.25 hour

Hence, this is the required solution.
Answer:
E = 10t^2e^-10t Joules
Explanation:
Given that the current through a 0.2-H inductor is i(t) = 10te–5t A.
The energy E stored in the inductor can be expressed as
E = 1/2Ll^2
Substitutes the inductor L and the current I into the formula
E = 1/2 × 0.2 × ( 10te^-5t )^2
E = 0.1 × 100t^2e^-10t
E = 10t^2e^-10t Joules
Therefore, the energy stored in the inductor is 10t^2e^-10t Joules