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dangina [55]
3 years ago
15

A helicopter is moving horizontally to the right at a

Physics
1 answer:
In-s [12.5K]3 years ago
8 0

Answer:

Explanation:

Given

Weight of helicopter W=53800 N

inclination of blades with vertical \theta =21^{\circ}

Suppose F_l is the lifting Force

Now using F-B-D

F_{net}=0 as helicopter is moving with constant velocity

balancing forces in Vertical Force

F_l\cos \theta -W=0

F_l\cos \theta =W

F_l=\frac{W}{\cos \theta }

F_l=\frac{53800}{\cos 21}

F_l=57,627.6\ N

Now sin component of lift is air resistance

F_l\sin \theta -R=0

where R=air resistance

R=F_l\sin \theta

R=57,627.6\times \sin (21)

R=20,651.88\ N

 

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Solve this physics for me <br>please with steps<br>​
Mars2501 [29]

Answer:

The answers are located in each of the explanations showed below

Explanation:

a)

(i) Surface Tension: The tensile force that causes this tension acts parallel to the surface and is due to the forces of attraction between the molecules of the liquid. The magnitude of this force per unit of length is called surface tension.

σ = F/l [N/m]

where:

F = force [N]

l = length [m]

σ = Surface Tension [N/m]

(ii) Frequency is the number of repetitions per unit of time of any periodic event.

f = 1/T [1/s] or [s^-1] or [Hz]

where:

T = period [s] or [seconds]

f = frecuency [Hz] or [hertz]

(iii) Each of the units will be shown for each variable

v = velocity [m/s]

a = accelertion [m/s^2]

s = displacement [m]

[\frac{m}{s} ]^{2} =[\frac{m}{s} ]^{2} + 2* [\frac{m}{s^{2} } ]*[m]\\

[\frac{m^2}{s^2} ] =[\frac{m^2}{s^2} ] +  [\frac{m^{2} }{s^{2} } ]

[\frac{m^2}{s^2} ]

b) To find the velocity we must derivate the function X with respect to t because this derivate will give us the equation for the velocity, it means:

v=\frac{dx}{dt} \\v = 0.75*2*t+5*t

(i) X = 0.75*t^{2} +5*t+1\\X = 0.75*(4)^{2} +5*(4)+1\\X = 33 [m]

ii) replacing in the derivated equation.

v=1.5*(4)+5\\v=11[m/s]

iii) the average velocity is defined by the expresion v = x/t

v = \frac{x-x_{0} }{t-t_{0} } \\

x_{0}=0.75(2)^{2}+5(2)+1 \\ x_{0}=14[m]\\x=0.75(7)^{2}+5(7)+1\\x=72.75[m]\\t = 7 [s]t0= 2[s]Now replacing:[tex]v_{prom} = \frac{72.75-14}{7-2} \\v_{prom} = 11.75 [m/s]

2

a) Pascal's principle or Pascal's law, where the pressure exerted on an incompressible fluid and in balance within a container of indeformable walls is transmitted with equal intensity in all directions and at all points of the fluid.

Therefore:

P1 = pressure at point 1.

P2 = pressure at point 2.

P1 = F1/A1

P2= F2/A2

\frac{F_{1} }{A_{1} }=\frac{F_{2}}{A_{2} }  \\F_{1}=A_{1}*(\frac{F_{2}}{A_{2} })

b) One of the applications of the surface tension is the <u>capillarity</u> this is a property of liquids that depends on their surface tension (which, in turn, depends on the cohesion or intermolecular force of the liquid), which gives them the ability to climb or descend through a capillary tube.

Other examples of surface tension:

The mosquitoes that can sit on the water.

A clip on the water.

Some leaves that remain floating on the surface.

Some soaps and detergents on the water.

5 0
3 years ago
A change in which of the following effects the weight of an object?
Masja [62]
Acceleration due to gravity 
4 0
3 years ago
A 1,200 kg car travels at 20 m/s. what is it’s momentum ?
WINSTONCH [101]

Answer:c  vx v xv xv x

Explanation:

mlmcmcm

6 0
3 years ago
At a picnic, there is a contest in which hoses are used to shoot water at a beach ball from three different directions. As a res
hram777 [196]

Answer:

F₃ = 122.88 N

θ₃ = 20.63°

Explanation:

First we find the components of F₁:

For x-component:

F₁ₓ = F₁ Cos θ₁

F₁ₓ = (50 N) Cos 60°

F₁ₓ = 25 N

For y-component:

F₁y = F₁ Sin θ₁

F₁y = (50 N) Sin 60°

F₁y = 43.3 N

Now, for F₂. As, F₂ acts along x-axis. Therefore, its y-component will be zero and its x-xomponent will be equal to the magnitude of force itself:

F₂ₓ = F₂ = 90 N

F₂y = 0 N

Now, for the resultant force on ball to be zero, the sum of x-components of the forces and the sum of the y-component of the forces must also be equal to zero:

F₁ₓ + F₂ₓ + F₃ₓ = 0 N

25 N + 90 N + F₃ₓ = 0 N

F₃ₓ = - 115 N

for y-components:

F₁y + F₂y + F₃y = 0 N

43.3 N + 0 N + F₃y = 0 N

F₃y = - 43.3 N

Now, the magnitude of F₃ can be found as:

F₃ = √F₃ₓ² + F₃y²

F₃ = √[(- 115 N)² + (- 43.3 N)²]

<u>F₃ = 122.88 N</u>

and the direction is given as:

θ₃ = tan⁻¹(F₃y/F₃ₓ) = tan⁻¹(-43.3 N/-115 N)

<u>θ₃ = 20.63°</u>

7 0
3 years ago
What description best describes both mirrors and lenses?
Kipish [7]
The only reasonable choice from this list is choice-A.
8 0
3 years ago
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