Here it is given that initial speed of the package will be same as speed of the helicopter
displacement of the package as it is dropped on ground
acceleration is due to gravity
now by kinematics
by solving above equation we have
so it will take 5.2 s to reach the ground
Work done by the force = Force x displacement. Power = work done/time = F.s/t = F.u.t/t = F.u = 95 x 20 = 1900J. {S=ut because acceleration is zero since car is moving at constant velocity}.
Answer:
14.2 m
Explanation:
Using conservation of energy:
PE at top = KE at bottom
mgh = ½ mv²
h = v² / (2g)
h = (16.7 m/s)² / (2 × 9.8 m/s²)
h = 14.2 m
Using kinematics:
Given:
v₀ = 16.7 m/s
v = 0 m/s
a = -9.8 m/s²
Find: Δy
v² = v₀² + 2aΔy
(0 m/s)² = (16.7 m/s)² + 2 (-9.8 m/s²) Δy
Δy = 14.2 m
Answer:
Electric potential = 0.00054 V
Explanation:
We are given;
Charge; q = 3 pC = 3 × 10^(-12) C
Radius; r = 2 cm = 0.02 m
Formula for the electric potential of this surface will be;
V = kqr
Where;
K is a constant = 9 × 10^(9) N⋅m²/C².
Thus;
V = 9 × 10^(9) × 3 × 10^(-12) × 0.02
V = 0.00054 V
Answer:
Explanation:
This is a circular motion questions
Where the oscillation is 27.3days
Given radius (r)=3.84×10^8m
Circular motion formulas
V=wr
a=v^2/r
w=θ/t
Now, the moon makes one complete oscillation for 27.3days
Then, one complete oscillation is 2πrad
Therefore, θ=2πrad
Then 27.3 days to secs
1day=24hrs
1hrs=3600sec
Therefore, 1day=24×3600secs
Now, 27.3days= 27.3×24×3600=2358720secs
t=2358720secs
Now,
w=θ/t
w=2π/2358720 rad/secs
Now,
V=wr
V=2π/2358720 ×3.84×10^8
V=1022.9m/s
Then,
a=v^2/r
a=1022.9^2/×3.84×10^8
a=0.0027m/s^2
