Answer:
We can use 2 g H = v2^2 - v1^2 or
v2^2 = 2 g H + v1^2
Since 88 ft/sec = 60mph we have 30 mph = 44 ft/sec
The object will return with the same speed that it had initially so the object
starts out with a downward speed of 44 ft/sec
Then v2^2 = 2 * 32 ft/sec^2 * 160 ft + 44 (ft/sec)^2
v2^2 = (2 * 32 * 160 + 44^2) ft^2 / sec^2 = 12180 ft^2/sec^2
v2 = 110 ft/sec
Answer:
it will remain same because mass of the body is constant everywhere
More mass and less difference
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
Answer:
Send a pic of it so i can mark it.
Explanation: