B; to produce a hypothesis.
Answer:
converts high-voltage, low-current power into low-voltage, high-current power.
Explanation:
Answer:
Explanation:
We shall use Ampere's circuital law to find magnetic field at required point.
The point is outside the circumference of two given wires so whole current will be accounted for .
Ampere's circuital law
B = ∫ Bdl = μ₀ I
line integral will be over circular path of radius r = 41 cm .
Total current I = 5A -3A = 2A .
∫ Bdl = μ₀ I
2π r B = μ₀ I
2π x .41 B = 4π x 10⁻⁷ x 2
B = 2 x 10⁻⁷ x 2 / .41
= 9.75 x 10⁻⁷ T . It will be along - ve Y - direction.
The work done by the gravitational field of the earth on the rock is 9.998 x 10⁸ J.
<u>The given parameter include:</u>
the mass of the object, m₁ = 16 kg
Note: the mass of the earth, m₂ = 5.972 x 10²⁴ kg
- The work done by the gravitational field of the earth is given as;
Work done = gravitational force (F) x radius of the earth (R)

Therefore, the work done by the gravitational field of the earth on the rock is 9.998 x 10⁸ J.
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Refer to the diagram shown below.
Because of symmetry, equal forces, F, exist between the sphere of mass m and each of the other two spheres.
The acceleration of the sphere with mass m will be vertical as shown.
The gravitational constant is G = 6.67408 x 10⁻¹¹ m³/(kg-s²)
Calculate F.
F = [ (6.67408 x 10⁻¹¹ m³/(kg-s²))*(m kg)*(2.8 kg)]/(1.2 m)²
= 1.2977 x 10⁻¹⁰ m N
The resultant force acting on mass m is
2Fcos(30°) = 2*(1.2977 x 10⁻¹⁰m N)*cos(30°) = 2.2477 x 10⁻¹⁰m N
If the initial acceleration of mass m is a m/s², then
(m kg)(a m/s²) = (2.2477 x 10⁻¹⁰m N)
a = 2.2477 x 10⁻¹⁰ m/s²
Answer:
The magnitude of the acceleration on mass m is 2.25 x 10⁻¹⁰ m/s².
The direction of the acceleration is on a line that joins mass m to the midpoint of the line joining the known masses.