Answer:
Explanation:
Energy density in magnetic field is given as;
where;
B is the magnetic field strength
Energy density of electric field
where;
E is electric field strength
Take the ratio of the two fields energy density
But, Electric field potential, V = E x L = IR (I is current and R is resistance)
Now replace E x L with IR
Also, B = μ₀I / 2πr, substitute this value in the above equation
cancel out the current "I" and factor out μ₀
Finally, the equation becomes;
Therefore, the correct option is (d) μ₀/ϵ₀ (L /R 2πr)²
A. The angle at which the arrow must be released to hit the bull's-eye is 20.7 °
B. The arrow will go over the branch.
<h3>A. How to determine the angle</h3>
- Range (R) = 74 m
- Initial velocity (u) = 33 m/s
- Acceleration due to gravity (g) = 9.8 m/s²
- Angle (θ) = ?
R = u²Sine(2θ) / g
74 = 33² × Sine (2θ) / 9.8
Cross multiply
74 × 9.8 = 33² × Sine (2θ)
725.2 = 1098 × Sine (2θ)
Divide both sides by 1098
Sine (2θ) = 725.2 / 1098
Sine (2θ) = 0.6605
Take the inverse of sine
2θ = Sine⁻¹ 0.6605
2θ = 41.3
Divide both sides by 2
θ = 41.3 / 2
θ = 20.7 °
<h3>B. How to determine if the arrow will go over or under the branch</h3>
To determine if the arrow will go over or under the branch situated mid way, we shall determine the maximum height attained by the arrow. This can be obtained as follow:
- Initial velocity (u) = 33 m/s
- Acceleration due to gravity (g) = 9.8 m/s²
- Angle (θ) = 20.7 °
- Maximum height (H) = ?
H = u²Sine²θ / 2g
H = [33² × (Sine 20.7)²] / (2 ×9.8)
H = 6.94 m
Thus, the maximum height attained by the arrow is 6.94 m which is greater than the height of the branch (i.e 3.50 m).
Therefore, we can conclude that the arrow will go over the branch
Learn more about projectile motion:
brainly.com/question/20326485
#SPJ1
Given that,
Force F= 30,000 N
mass m= 4000 kg
acceleration = ?
since,
F= ma
a= F/m
a= 30,000/4000
a= 7.5 m/s²
acceleration of the given aeroplane will be 7.5 m/s².
Answer:
Explanation:
The proton is under a linear motion with constant acceleration. So, we use the kinemtic equations to calculate its final speed. We know its acceleration, its initial speed and its traveled distance. Thus, we use the following equation:
As long as the box was 1m off the floor. it's potential energy was m•g•h = 30•9.8 = 294 joules. When the person stopped restraining the box and allowed it to move, gravity did 294 joules of work on the box, and would have done more if the floor hadn't been in the way. The person did no work on the box.