The mass of the man is 63 kilograms
There is it written down properly. Its hard to type this stuff in here
There is it written down properly. Its hard to type this stuff in here
HELP!!!! WILL MARK BRAINLIEST!!!! IF YOU LEAVE AN ANSWER EXPLAIN! THANKS
1 answer:
You might be interested in
Answer:
a= 23.65 ft/s²
Explanation:
given
r= 14.34m
ω=3.65rad/s
Ф=Ф₀ + ωt
t = Ф - Ф₀/ω
= (98-0)×/3.65
98°= 1.71042 rad
1.7104/3.65
t= 0.47 s
r₁(not given)
assuming r₁ =20 in
r₁ = r₀ + ut(uniform motion)
u = r₁ - r₀/t
r₀ = 14.34 in= 1.195 ft
r₁ = 20 in = 1.67 ft
= (1.667 - 1.195)/0.47
0.472/0.47
u= 1.00ft/s
acceleration at collar p
a=rω²
= 1.67 × 3.65²
a = 22.25ft/s²
acceleration of collar p related to the rod = 0
coriolis acceleration = 2ωu
= 2× 3.65×1 = 7.3 ft/s²
acceleration of collar p
= 22.5j + 0 + 7.3i
√(22.5² + 7.3²)
the magnitude of the acceleration of the collar P just as it reaches B in ft/s²
a= 23.65 ft/s²
The electric field of a very large (essentially infinitely large) plane of charge is given by:
E = σ/(2ε₀)
E is the electric field, σ is the surface charge density, and ε₀ is the electric constant.
To determine σ:
σ = Q/A
Where Q is the total charge of the sheet and A is the sheet's area. The sheet is a square with a side length d, so A = d²:
σ = Q/d²
Make this substitution in the equation for E:
E = Q/(2ε₀d²)
We see that E is inversely proportional to the square of d:
E ∝ 1/d²
The electric field at P has some magnitude E. Now we double the side length of the sheet while keeping the same amount of charge Q distributed over the sheet. By the relationship of E with d, the electric field at P must now have a quarter of its original magnitude: