Answer:
Explanation:
Magnetic field due to circular wire at the center = μ₀ I / 2 r
I is current and r is radius . μ₀ = 4π x 10⁻⁷.
field B₁ due to inner loop
B₁ = 4π x 10⁻⁷ x 12 / 2 x .20
= 376.8 x 10⁻⁷
Field due to outer loop
B₂ = 4π x 10⁻⁷ x I / 2 x .30
For equilibrium
B₁ = B₂
376.8 x 10⁻⁷ = 4π x 10⁻⁷ x I / 2 x .30
I = 18 A.
The direction should be opposite to that in the inner wire . It should be anti-clockwise.
Answer:
The maximum range is 300% of the range of the projectile is projected at an angle of 9.74°.
None of the options are correct.
Explanation:
Normally, ignoring air resistance, for projectile motion, the range (horizontal distance travelled) of the motion is given as
R = (u² sin 2θ)/g
where
u = initial velocity of the projectile
θ = angle above the horizontal at which the projectile was launched
g = acceleration due to gravity = 9.8 m/s²
The range is maximum when θ = 45°
R when θ = 45° is
R = (u² sin 90°)/g = (u²/g)
We are then told that this maximum range is 300% of the value obtainable for the range at a particular angle
Maximum range = 3R
(u²/g) = 3(u² sin 2θ)/g
Sin 2θ = (1/3)
2θ = sin⁻¹ (1/3) = 19.47°
θ = (19.47°/2) = 9.74°
Hope this Helps!!!
Answer:
mercury
arsenic
ricin
ddt
Explanation:
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Answer:
a = 3.27 m/s²
F = 32.7 N
Explanation:
Draw a free body diagram. There are three forces:
Weight force mg pulling straight down.
Normal force N pushing perpendicular to the slope.
Friction force F pushing parallel up the slope.
Sum of forces in the parallel direction:
∑F = ma
mg sin θ − F = ma
Sum of torques about the cylinder's axis:
∑τ = Iα
Fr = ½ mr²α
F = ½ mrα
Since the cylinder rolls without slipping, a = αr. Substituting:
F = ½ ma
Two equations, two unknowns (a and F). Substituting the second equation into the first:
mg sin θ − ½ ma = ma
Multiply both sides by 2/m:
2g sin θ − a = 2a
Solve for a:
2g sin θ = 3a
a = ⅔ g sin θ
a = ⅔ (9.8 m/s²) (sin 30°)
a = 3.27 m/s²
Solving for F:
F = ½ ma
F = ½ (20 kg) (3.27 m/s²)
F = 32.7 N
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