Answer:
m = 2.2 x 10⁻⁴ kg = 0.22 g
Explanation:
The surface tension of water is 0.072 N/m. So in order for the bug to avoid sinking, its weight per unit length of contact must be no more than the surface tension of water. Therefore,

where,
m = mass of bug = ?
g = acceleration due to gravity = 9.81 m/s²
L = Contact length = (contact length of each leg)(No. of Legs) = (5 mm)(6)
L = 30 mm = 0.03 m
Therefore,

<u>m = 2.2 x 10⁻⁴ kg = 0.22 g</u>
Answer:
V = 9.33 m/s
Explanation:
Given that,
Mass of the batsman, 
Mass of the boat,
Initial speed of the boat, v = 11 m/s
Let V is the velocity of the boat after Batman lands in it. The net momentum of the system remains constant. Using the conservation of linear momentum to find it as :


V = 9.33 m/s
So, the velocity of the boat after Batman lands in it 9.33 m/s. Hence, this is the required solution.
Answer:
Alpha = ω^2 R where R is radius of blade
g = w^2 r where r is distance from center
ω^2 R = 11.5 ω^2 r
R / r = 11.5 / 9.8 = 1.17
Or r = .852 R
Since the angular acceleration depends on both R and ω it seems that one can only get r as it depends on R
Answer:
more point plzzzzz... eeeeeeeee
<h3>Answer;</h3>
<em><u>Sand Spit or Spit </u></em>
<h3><u>Explanation;</u></h3>
- <em><u>Long shore drift is the process that occurs when a sheet of water moves on and off the beach, in other words the swash and back swash</u></em>, thus capturing and transporting sediment on the beach back out to the sea.
- <em><u>Sandbar</u></em> is normally formed when the sandspit stretches across a bay and connects the two sides. <em><u>Headland</u></em> is a high piece of land that extends out onto the sea. <em><u>Sea stacks </u></em>on the other hand results from the collapsing of the roof of the arch.