Answer:
Hence the pressure is 
Explanation:
Given data
Q=1500 J system gains heat
ΔV=- 0.010 m^3 there is a decrease in volume
ΔU= 4500 J internal energy decrease
We know work done is
W= Q- ΔU
=1500-4500= -3000 J
The change in the volume at constant pressure is
ΔV= W/P
there fore P = W/ΔV= -3000/-0.01= 3×10^5
Hence the pressure is
Answer: 
Explanation:
Given
Initial position of object is (4.4 i+5 j)
Final position of object is (11.6 i -2 j)
Force acting (4i-9j)
Work done is given by
Initial kinetic energy
Change in kinetic energy is equal to work done by object
Answer:
Explanation:
Displacement vector along x axes = 4.5 - 2.5 = 2 m
Displacement vector along y axes = 3 - 2 = 1 m
Displacement vector along z axis = 3.5- 4 = - 0.5 m
Displacement vector = 2 i + j - 0.5 k m
Answer:
The velocity of charge is 1.1×10⁴
Option(D)
Explanation:
F = q(v×B)
3.5×10-²= (8.4×10-⁴)*(v)*(6.7×10-³)*sin35
v = (3.5)/(8.4×6.7×0.57) × 10⁵
v = 3.5/32.08 × 10⁵
v = 0.109 × 10⁵
v ≈ 1.1 × 10⁴ m/s
Answer: F = mg(1 + 4m / (½M + m))
Explanation:
"At this point seems" unclear. If the particle is at the top of the disc and angular velocity is negligible, then the force would equal the weight of the particle. F = mg
The more interesting question would be what force is needed to keep the particle attached when significant angular rotation has been achieved. The maximum point would be diametrically opposed to the starting point.
I will analyze it there
The potential energy will convert to kinetic energy
mgh = ½Iω²
mg(2R) = ½(½MR² + mR²)ω²
4mgR = R²(½M + m)ω²
ω² = 4mg / (R(½M + m))
With m at the lowest position, the force of attachment must support the weight of m and provide for the needed centripetal acceleration
F = m(g + ω²R)
F = m(g + 4mg / (R(½M + m))R)
F = mg(1 + 4m / (½M + m))
