Answer: here you go I was looking for this answer everywhere,I have it now so it’s 6.30 x 10^-7 s
Explanation:
I hope this helps☺️
Answer:
4.75 m/s
Explanation:
The computation of the velocity of the existing water is shown below:
Data provided in the question
Tall = 2 m
Inside diameter tank = 2m
Hole opened = 10 cm
Bottom of the tank = 0.75 m
Based on the above information, first we have to determine the height which is
= 2 - 0.75 - 0.10
= 2 - 0.85
= 1.15 m
We assume the following things
1. Compressible flow
2. Stream line followed
Now applied the Bernoulli equation to section 1 and 2
So we get
where,
P_1 = P_2 = hydrostatic
z_1 = 0
z_2 = h
Now
= 4.7476 m/sec
= 4.75 m/s
The time component is needed. The acceleration is the change of velocity divided by the time in when this change of velocity happens.
Answer:
Tundra Biome
Explanation:
Permafrost is a type of soil that is frozen all year round. It consists of rocks, soils and ice. The ice or frost holds the earth materials together.
The tundra biome lies below the arctic circle close to the north pole. Most of the earth here is predominantly frozen all year round. A layer of glacier covers the surface and a deep lying layer of permafrost follows suit.
Some mountain tops capped with ice shows this tundra features.
Most tundras are termed cold deserts as they have little to no precipitation all year round. There is absence of vegetation cover as a result of low growing season of the plants.
Answer:
Speed =0.283m/ s
Direction = 47.86°
Explanation:
Since it is a two dimensional momentum question with pucks having the same mass, we derive the momentum in xy plane
MU1 =MU2cos38 + MV2cos y ...x plane
0 = MU2sin38 - MV2sin y .....y plane
Where M= mass of puck, U1 = initial velocity of puck 1=0.46, U2 = final velocity of puck 1 =0.34, V2 = final velocity of puck 2, y= angular direction of puck2
Substitute into equation above
.46 = .34cos38 + V2cos y ...equ1
.34sin38 = V2sin y...equ2
.19=V2cos Y...x
.21=V2sin Y ...y
From x
V2 =0.19/cost
Sub V2 into y
0.21 = 0.19(Sin y/cos y)
1.1052 = tan y
y = 47.86°
Sub Y in to x plane equ
.19 = V2 cos 47.86°
V2=0.283m/s
