2000J
Explanation:
Given parameters:
Extension = 0.5m
Spring constant = 16000N/m
Unknown:
Energy stored in the bow string = ?
Solution:
The energy stored in a bow string is an elastic potential energy.
It can be calculated using the expression below;
Elastic energy = 
K e²
Where k is the spring constant
e is the extension
Input the parameters;
Elastic energy = K e²
= x 16000 x 0.5²
= 2000J
learn more:
Potential energy brainly.com/question/10770261
#learnwithBrainly
<h3><u>Answer</u> :</h3>
◈ As per newton's second law of motion, Force is defined as the product of mass and acceleration
Mathematically,
Unit of mass : kg
Unit of acceleration : m/s²
Therefore,
Unit of force ➠ <u>kg m/s²</u>
SI unit : <u>N (newton)</u> or <u>kg m/s²</u>
Answer:
magnitude of the magnetic field 0.692 T
Explanation:
given data
rectangular dimensions = 2.80 cm by 3.20 cm
angle of 30.0°
produce a flux Ф = 3.10 × 
Wb
solution
we take here rectangular side a and b as a = 2.80 cm and b = 3.20 cm
and here angle between magnitude field and area will be ∅ = 90 - 30
∅ = 60°
and flux is express as
flux Ф = 
.................1
and Ф = BA cos∅ ............2
so B = 
and we know
A = ab
so
B = 
..............3
put here value
B = 
solve we get
B = 0.692 T
