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
For the First answer, It would be "A"
The for the next one the answer is "C"
I hope this helps. :)
Answer:
a = -0.33 m/s² k^
Direction: negative
Explanation:
From Newton's law of motion, we know that;
F = ma
Now, from magnetic fields, we know that;. F = qVB
Thus;
ma = qVB
Where;
m is mass
a is acceleration
q is charge
V is velocity
B is magnetic field
We are given;
m = 1.81 × 10^(−3) kg
q = 1.22 × 10 ^(−8) C
V = (3.00 × 10⁴ m/s) ȷ^.
B = (1.63T) ı^ + (0.980T) ȷ^
Thus, since we are looking for acceleration, from, ma = qVB; let's make a the subject;
a = qVB/m
a = [(1.22 × 10 ^(−8)) × (3.00 × 10⁴)ȷ^ × ((1.63T) ı^ + (0.980T) ȷ^)]/(1.81 × 10^(−3))
From vector multiplication, ȷ^ × ȷ^ = 0 and ȷ^ × i^ = -k^
Thus;
a = -0.33 m/s² k^
There are no appropriate units for power on the list you provided
Answer:
the height of the potential energy is 3,200 J
Explanation:
The computation of the kinetic energy is shown below:
Kinetic energy = 1 ÷ 2 × mass × velocity^2
= 1 ÷ 2 × 4 kg × 40 m/s^2
= 3,200 J
Hence the height of the potential energy is 3,200 J
