Answer:
a
b
Explanation:
From the question we are told that
The initial position of the particle is 
The initial velocity of the particle is 
The acceleration is 
The time duration is 
Generally from kinematic equation
=> 
=>
Generally from kinematic equation
Here s is the distance covered by the particle, so
=> 
Generally the final position of the particle is
=> 
=>
It will be launched in the direction it was going in when the centripetal force is removed
Answer:
The force exerted by the ocean tide is directed to the right (east).
The force exerted by the wind is directed to the northwest (45° N of W).
We should separate the x- and y- components of the wind force, and evaluate each component separately.
We denote the direction to the right as the positive direction, so the x-component of the wind force is in the negative direction.
The resultant force is as follows:
 + [4242.64](\^{y})\\F_R = -2242.64\^{x} + 4242.64\^{y}](https://tex.z-dn.net/?f=F_R%20%3D%20%5B2000%20-%204242.64%5D%28%5C%5E%7Bx%7D%29%20%2B%20%5B4242.64%5D%28%5C%5E%7By%7D%29%5C%5CF_R%20%3D%20-2242.64%5C%5E%7Bx%7D%20%2B%204242.64%5C%5E%7By%7D)
The resultant acceleration can be found by Newton's Second Law:
The magnitude of the resultant acceleration is
Answer:
Explanation:
Rate of flow of liquid through a tube can be expressed by the following expression
V = π P r⁴ / 8ηl
P is pressure difference between end of tube = 618 Pa
r , radius of tube = .5 x 10⁻²
η is viscosity of liquid flowing = .63
l is length of tube = .10 m
V = 3.14 x 618 x ( .5 x 10⁻² )⁴ / (8 x .63 x .10 )
= 240.64 x 10⁻⁸ m³ /s
mass = 240.64 x 1260 x 10⁻⁸ kg / s
= 3.03 x 10⁻³ kg /s
= 3.03 gram /s .
Answer:
5.23 C
Explanation:
The current in the wire is given by I = ε/R where ε = induced emf in the wire and R = resistance of wire.
Now, ε = -ΔΦ/Δt where ΔΦ = change in magnetic flux = AΔB and A = area of loop and ΔB = change in magnetic field intensity = B₂ - B₁
B₁ = 0.670 T and B₂ = 0 T
ΔB = B₂ - B₁ = 0 - 0.670 T = - 0.670 T
A = πD²/4 where D = diameter of circular loop = 13.2 cm = 0.132 m
A = π(0.132 m)²/4 = 0.01368 m² = `1.368 × 10⁻² m²
ε = -ΔΦ/Δt = -AΔB/Δt = -1.368 × 10⁻² m² × (-0.670 T)/Δt= 0.9166 × 10⁻² Tm²/Δt
Now, the resistance R of the circular wire R = ρl/A' where ρ = resistivity of copper wire = 1.68 x 10⁻⁸ Ω.m, l = length of wire = πD and A' = cross-sectional area of wire = πd²/4 where d = diameter of wire = 2.25 mm = 2.25 × 10⁻³ m
R = ρl/A' = 1.68 x 10⁻⁸ Ω.m × π × 0.132 m÷π(2.25 × 10⁻³ m)²/4 = 0.88704/5.0625 = 0.1752 × 10⁻² Ω = 1.752 × 10⁻³ Ω
So, I = ε/R = 0.9166 × 10⁻² Tm²/Δt1.752 × 10⁻³ Ω
IΔt = 0.9166 × 10⁻² Tm²/1.752 × 10⁻³ Ω = 0.5232 × 10 C
Since ΔQ = It = 5.232 C ≅ 5.23 C
So the charge is 5.23 C
