Answer:
The handrails must be approximately 10.63 meters long
Explanation:
The given parameters are;
The height of the bleachers, h = 8 m
The depth of the bleachers, d = 7 m
The length of the hand rails to go along the bleachers from bottom to top is given by Pythagoras' Theorem as follows;
The length of the hand rail = √(d² + h²)
∴ The length of the hand rail = √(7² + 8²) = √113 ≈ 10.63
In order for the handrails to go along the bleachers from top to bottom, they must be approximately 10.63 meters long.
Answer:
magnitude of the frictional torque is 0.11 Nm
Explanation:
Moment of inertia I = 0.33 kg⋅m2
Initial angular velocity w° = 0.69 rev/s = 2 x 3.142 x 0.69 = 4.34 rad/s
Final angular velocity w = 0 (since it stops)
Time t = 13 secs
Using w = w° + §t
Where § is angular acceleration
O = 4.34 + 13§
§ = -4.34/13 = -0.33 rad/s2
The negative sign implies it's a negative acceleration.
Frictional torque that brought it to rest must be equal to the original torque.
Torqu = I x §
T = 0.33 x 0.33 = 0.11 Nm
Answer:10.4 times of initial velocity
Explanation:
Given
Diameter reduced by 69 %
it approaches with velocity 
suppose its velocity is v during blocked passage
suppose d is the initial diameter and 
diameter is
As flow is constant
Answer:
A) 1.67 x 10 ⁻⁶ m/s
B)5.59 x 
%
Explanation:
A)
Given:
d = 5.0 km,
mₐ = 2.5 x 
kg
u₁ = 4.0 x 10⁴ m/s
= 5.98 x 10 ²⁴ kg
Solve using kinetic conserved energy
mₐ x u₁ + x u₂ = uₓ x (mₐ + )
(2.5 x ) (4.0 x 10⁴ )+ (5.98 x 10 ²⁴ )(0) = uₓ x (2.5 x + 5.98 x 10 ²⁴ )
uₓ = ( 2.5 x x 4.0 x 10⁴ ) / (2.5 x + 5.98 x 10 ²⁴ )
uₓ = 1.67 x 10 ⁻⁶ m/s
B) Assuming earth radius as a R = 1.5 x 10 ¹¹ m
t = 365 days x 24 hr / 1 day x 60 minute / 1 hr x 60s / 1 minute = 31536000 s
t = 31536000 s
D = 2 π R = 2 π( 1.5 x 10 ¹¹ )
D = 9.4247 x 10 ¹¹ m
u₂ = D / t = 9.4247 x 10 ¹¹ / 31536000
u₂ = 29885.775 m/s
% = ( 1.67 x 10 ⁻⁶ m/s ) / (29885.775 m/s) x 100
% = 5.59 x %
