Answer:
Maximum height reached by the rocket, h = 202.62 meters
Explanation:
It is given that,
Initial speed of the model rocket, u = 56.5 m/s
Constant upward acceleration,
Distance traveled by the engine until it stops, d = 198.8 m
Let v is the speed of the rocket when the engine stops. It can be calculated using the third equation of motion as :
v = 63.02 m/s
At the maximum height, v = 0 and the engine now decelerate under the action of gravity, a = -g. Let h is the maximum height reached by the rocket.
Again using third equation of motion as :
h = 202.62 meters
So, the maximum height reached by the rocket is 202.62 meters. Hence, this is the required solution.
Answer:
(I). The effective cross sectional area of the capillaries is 0.188 m².
(II). The approximate number of capillaries is
Explanation:
Given that,
Radius of aorta = 10 mm
Speed = 300 mm/s
Radius of capillary
Speed of blood
(I). We need to calculate the effective cross sectional area of the capillaries
Using continuity equation
Where. v₁ = speed of blood in capillarity
A₂ = area of cross section of aorta
v₂ =speed of blood in aorta
Put the value into the formula
(II). We need to calculate the approximate number of capillaries
Using formula of area of cross section
Put the value into the formula
Hence, (I). The effective cross sectional area of the capillaries is 0.188 m².
(II). The approximate number of capillaries is
The answer to your question I think is D
Answer:
Explanation:
Bridge length A B = 36.5 m
weight W = 2.56 x 10⁵ N acting at middle point that is at 18.25 m from either end.
weight of truck W₁ = 5.25 x 10⁴ N standing at 10.2 m from A .
Let force at A be F₁ and at B be F₂ .
Balancing total upward and downward force
F₁ + F₂ = ( 2.56 + .525 ) x 10⁵ = 3.085 x 10⁵ N
Taking torque of all the forces , for rotational balancing
2.56 x 10⁵ x 18.25 + .525 x 10⁵ x 10.2 = F₂ x 36.5
(46.72 + 5.355 )x 10⁵ = F₂ x 36.5
F₂ = 1.42 x 10⁵ N
F₁ = 1.665 x 10⁵ N