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Montano1993 [528]
1 year ago
12

a plane flies 400 km east from city a to city b in 45.0 min and then 980 km south from city b to city c in 1.50 h. (assume the x

axis points east and the y axis points north.) (a) in magnitude-angle notation, what is the plane's displacement for the total trip?
Physics
1 answer:
maksim [4K]1 year ago
4 0

The plane's displacement for the total trip is, -63.2°.

Velocity, v = x/t

Vx = 459 km/2.483 hr = 184.8 km/h

Vy = -909 km/2.483 hr = - 366 km/hr

Magnitude, v =sqrt(Vx^2 + Vy^2) = 410 km/hr

Direction, theta = arctan (-366/184.8) = - 63.2°

<h3>What is in plane displacement?</h3>

Double symmetrical illumination of the object, successively for each illumination direction, is used to measure in-plane displacement as the loading is gradually increased. The impact response of structures is significantly impacted by the in-plane displacements at the supports.

This is so that axial membrane forces along the beam's center line can be introduced by the development of geometrical modifications. Given that the double-hull specimen should have been strengthened by sturdy brackets at the boundaries of the outer and inner plates, this anomaly is likely the result of insufficient experimental restrictions (top and bottom plates).

To learn more about plane displacement, visit:

brainly.com/question/13066887

#SPJ4

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A plane flies toward a stationary siren at 1/4 the speed of sound. Then the plane stands still on the ground and the siren is dr
melomori [17]

Answer:

See the attachments below.

Explanation:

4 0
3 years ago
When a planet is at its slowest orbital speed, its radius vector sweeps an area, A, in 45 days. What area will the radius vector
MArishka [77]

The area the radius vector will sweep is 0.889A

According to Kepler's second law, the radius vector <em>sweeps</em> out equal areas in equal times.

Let A = area and t = time period,

According to Kepler's law, A/t = constant

So, A₁/t₁ = A₂/t₂ where A₁ = area the radius vector sweeps at <em>slowest</em> orbital speed = A, t₁ = time period at <em>slowest </em>orbital speed = 45 days, A₂ = area the radius vector sweeps at<em> fastest</em> orbital speed, t₂ = time period at<em> fastest </em>orbital speed = 40 days.

Making A₂ subject of the formula, we have

A₂ = A₁t₂/t₁

Substituting the values of the variables into the equation, we have

A₂ = A₁t₂/t₁

A₂ = A × 40 days/45 days

A₂ = A × 40/45

A₂ = A × 8/9

A₂ = A × 0.889

A₂ = 0.889A

So, the area the radius vector will sweep is 0.889A

learn more about Kepler's second law here:

brainly.com/question/4639131

8 0
2 years ago
A 60 kg adult and a 30kg child are passengers on a rotor ride at an amusement park as shown in the diagram above. When the rotat
Ulleksa [173]

Answer:

 fr ’= ½ F

Explanation:

For this exercise we use the translational equilibrium equation, on the axis parallel to the wall

              fr - W = 0

              fr = W

for the adult man they indicate that the friction force is equal to F

              F = M g

we write the equilibrium equation for the child

             fr ’= w’

             fr ’= m g

in the statement they tell us that the mass of the adult is 2 times the mass of the child

             M = 2m

we substitute

            fr ’= M / 2 g

            fr ’= ½ Mg

we substitute

            fr ’= ½ F

therefore the force of friction in the child is half of the friction in the adult

8 0
3 years ago
A chain 72 meters long whose mass is 29 kilograms is hanging over the edge of a tall building and does not touch the ground. How
dangina [55]

Answer:

Work done required is 3567.2 J

Explanation:

Given :

Length of chain, l = 72 m

Mass of chain, M = 29 kg

Linear mass density of chain, μ = \frac{Mass\ of\ chain }{Length\ of\ chain} = \frac{29}{72}  = 0.40 kg/m

Let x be the length of the chain which lift to the top of the building.

Work done required to lift the chain is equal to the potential energy of the chain.

W = ∫μg (72 - x ) dx

Here g is acceleration due to gravity.

The limit of integration is from 0 to 14.

W = μg ( 72x - x²/2)

Substitute 0.40 kg/m for μ, 9.8 m/s² for g and 14 m for x in the above equation.

W = 0.40\times9.8\times(72\times14\ - \frac{14^{2} }{2})

W = 3567.2 J

5 0
3 years ago
sing a rope that will snap if the tension in it exceeds 361 N, you need to lower a bundle of old roofing material weighing 455 N
yaroslaw [1]

Answer:

a)-2m/s^2

b)27.2m/s

Explanation:

Hello! The first step to solve this problem is to find the mass of the block remembering that the definition of weight force is mass by gravity (g=9.8m / s ^ 2)

W=455N=weight

W=mg

W=455N=weight

m=\frac{W}{g} =\frac{455}{9.81}=46.38kg

The second step is to draw the free body diagram of the body (see attached image) and use Newton's second law that states that the sum of the forces is equal to mass by acceleration

a=\frac{T-W}{m} =\frac{361-455}{46.38kg} =-2m/s^2

for point b we use the equations of motion with constant acceleration to find the velocity

Vf=\sqrt{X(2)(a)+Vo^2}

Where

Vf = final speed

Vo = Initial speed =0

A = acceleration =2m/s

X = displacement =6.8m

Solving

Vf=\sqrt{(6.8)(2)(2)+0^2}=27.2m/s

4 0
3 years ago
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