Answer:
The propulsion force at the moment of departure, is 49 N
Explanation:
Given;
diameter of tubes = 5 cm = 0.05 m
volumetric flow rate, V = 50 L/s = 0.005 m³/s
density of water, ρ = 1000 Kg /m³
hydraulic / propulsion force, F = ρVg
where;
ρ is the density of the fluid (water)
V is the volumetric flow rate of water
g is acceleration due to gravity
propulsion force, F = ρVg
propulsion force, F = 1000 x 0.005 x 9.8
propulsion force, F = 49 N
Therefore, the propulsion force at the moment of departure, is 49 N
Answer:
1) t = 3.45 s, 2) x = 138 m, 3) v_{y} = -33.81 m /s, 4) v = 52.37 m / s
,
5) θ = -40.2º
Explanation:
This is a projectile exercise, as they indicate that the projectile rolls down the cliff, it goes with a horizontal speed when leaving the cliff, therefore the speed is v₀ₓ = 40 m / s.
1) Let's calculate the time that Taardaen reaches the bottom, we place the reference system at the bottom of the cliff
y = y₀ +
t - ½ g t²
When leaving the cliff the speed is horizontal v_{oy}= 0 and at the bottom of the cliff y = 0
0 = y₀ - ½ g t2
t = √ 2y₀ / g
t = √ (2 60 / 9.8)
t = 3.45 s
2) The horizontal distance traveled
x = v₀ₓ t
x = 40 3.45
x = 138 m
3) The vertical velocity at the point of impact
v_{y} = I go - g t
v_{y} = 0 - 9.8 3.45
v_{y} = -33.81 m /s
the negative sign indicates that the speed is down
4) the resulting velocity at this point
v = √ (vₓ² + v_{y}²)
v = √ (40² + 33.8²)
v = 52.37 m / s
5) angle of impact
tan θ = v_{y} / vx
θ = tan⁻¹ v_{y} / vx
θ = tan⁻¹ (-33.81 / 40)
θ = -40.2º
6) sin (-40.2) = -0.6455
7) tan (-40.2) = -0.845
8) when the projectile falls down the cliff, the horizontal speed remains constant and the vertical speed increases, therefore the resulting speed has a direction given by the angle that is measured clockwise from the x axis
Electromagnetic waves<span> transfer energy without going through a medium. ... Sometimes, a </span>transverse wave<span> and a </span>longitudinal wave can combine to form<span>another </span>kind<span> of </span>wave<span> called a surface </span>wave<span>. </span>Transverse Waves<span>. </span>Waves<span> in which the particles vibrate in an up-and-down motion
</span>
The solution for this problem is:
500 revolution per
minute = 8.33rev /s = 2π*8.33 rad /s = 52.36 rad /s
Angular velocity ω = 2π N
Angular acceleration α= (ω2 - ω1) /t
ω2 = 0
α = - ω1/t = -2π N /t
N = 500 rpm = 8.33 r p s.
α = -2π 8.33 /2.6 =- 20 rad/s^2
Answer:
It regulates the movement of proteins and RNAs into and out of the nucleus
Explanation:
The nuclear pore complex are protein channels connecting the outer membrane of the nucleus to the inner membrane of the nucleus. They securely regulates the almost all of the transport of protein and RNAs into and out of the nucleus.