<span>At an instant when the displacement is equal to a/2,
Potential energy U = 1/2ka(square) where a is displacement.
when a= a/2
U = 1/4ka(square)
U = E/4
Potential Energy = 1/4 Total energy</span>
Answer:
55.66 m
Explanation:
While falling by 50 m , initial velocity u = 0
final velocity = v , height h = 50 , acceleration g = 9.8
v² = u² + 2gh
= 0 + 2 x 9.8 x 50
v = 31.3 m /s
After that deceleration comes into effect
In this case final velocity v = 17 m/s
initial velocity u = 31.3 m/s
acceleration a = - 61 m/s²
distance traveled h = ?
v² = u² + 2gh
(17)² = (31.3)² - 2x 61xh
h = 690.69 / 2 x 61
= 5.66 m
Total height during which he was in air
= 50 + 5.66
= 55.66 m
Answer:
Explanation:
Given that,
Current in loops are
i1 = 12A
i2 = 20A
The loops are 3.4cm apart
The magnetic field at the center is found to be zero, so when want to find the radius of bigger loop
Magnetic Field is given as
B= μoi/2πr
Where,
μo is a constant = 4π×10^-7 Tm/A
r is the distance between the two wires
i is the current in the wires
B is the magnetic field
NOTE
Field due to large loop should be equal to the smaller loop.
B1 = B2
μo•i1 / 2π•r1 = μo•i2 / 2π•r2
Then, μo, 2π cancels out, so we have
i1 / r1 = i2 / r2
Make r2 subject of formula
i1•r2 = i2•r1
r2 = i2•r1 / i2
r2 = 20×3.4/12
r2 = 5.67cm
The radius of the bigger loop is 5.67cm.
Answer:
5.3 cm
Explanation:
This question is an illustration of real and apparent distance.
From the question, we have the following given parameters
Real Distance, R = 8.0cm
Refractive Index, μ = 1.5
Required
Determine the apparent distance (A)
The relationship between R, A and μ is:
μ = R/A
i.e.
Refractive Index = Real Distance ÷ Apparent Distance
Substitute values in the above formula
1.5 = 8/A
Multiply both sides by A
1.5 * A = A * 8/A
1.5A = 8
Divide both side by 1.5
1.5A/1.5 = 8/1.5
A = 8/1.5
A = 5.3cm
Hence, the letters would appear at a distance of 5.3cm
I think it is A. but then you can also produce your own energy
