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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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how much water is needed to produce 1kwh of electricity at a power plant that is 30% efficient if the temperature increase 10 C
Dimas [21]

The amount of water needed is 287 kg

Explanation:

The amount of energy that we need to produce with the power plant is

E=1 kWh = (1000W)(1h)=(1000W)(3600s)=3.6\cdot 10^6 J

We also know that the power plant is only 30% efficient, so the energy produced in input must be:

E_{in}=\frac{E}{0.30}=\frac{3.6\cdot 10^6}{0.3}=1.2\cdot 10^7 J

The amount of water that is needed to produce this energy can be found using the equation

E_{in}=mC\Delta T

where:

m is the amount of water

C=4186 J/kg^{\circ}C is the specific heat capacity of water

\Delta T=10^{\circ}C is the increase in temperature

And solving for m, we find:

m=\frac{E_{in}}{C\Delta T}=\frac{1.2\cdot 10^7}{(4186)(10)}=287 kg

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2 years ago
What is the speed of a commercial jet which travels form New York to Los Angeles (4800) in 6 hours
Colt1911 [192]

Answer:

Speed = 800km}/hr

Explanation:

Given

Distance = 4800km

Time = 6hr

Required

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The speed is calculated as:

Speed = \frac{Distance}{Time}

Substitute 4800 km for Distance and 6hr for Time

Speed = \frac{4800km}{6hr}

Speed = 800km}/hr

<em>Hence, the speed of the commercial jet is 800km/hr</em>

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2 years ago
In hydrogen, the transition from level 2 to level 1 has a rest wavelength of 121.6 nm.1).Find the speed for a star in which this
soldier1979 [14.2K]

Answer:

1). v = - 2960526m/s

2). Toward us

3). v = - 493421m/s

4). Toward us

5). v = 1480263m/s

6).  Away from us

7). v = 3207236m/s

8). Away from us

Explanation:

Spectral lines will be shifted to the blue part of the spectrum if the source of the observed light is moving toward the observer, or to the red part of the spectrum when it is moving away from the observer (that is known as the Doppler effect).

The wavelength at rest is 121.6 nm (\lambda_{0} = 121.6nm)

Redshift: \lambda_{measured} > \lambda_{0}

Blueshift: \lambda_{measured} < \lambda_{0}

Then, for this particular case it is gotten:

Star 1: \lambda_{measured} = 120.4nm

Star 2: \lambda_{measured} = 121.4nm

Star 3: \lambda_{measured} = 122.2nm

Star 4: \lambda_{measured} = 122.9nm

Star 1:

Blueshift: 120.4nm < 121.6nm

Toward us

Star 2:

Blueshift: 121.4nm < 121.6nm

Toward us

Star 3:

Redshift: 122.2nm > 121.6nm

Away from us

Star 4:

Redshift: 122.9nm > 121.6nm

Away from us

Due to that shift the velocity of the star can be determine by means of Doppler velocity.

v = c\frac{\Delta \lambda}{\lambda_{0}}  (1)

Where \Delta \lambda is the wavelength shift, \lambda_{0} is the wavelength at rest, v is the velocity of the source and c is the speed of light.

v = c(\frac{\lambda_{measured}- \lambda_{0}}{\lambda_{0}}) (2)

<em>Case for star 1 \lambda_{measured} = 120.4 nm:</em>

<em></em>

v = (3x10^{8}m/s)(\frac{120.4nm-121.6nm}{121.6nm})

v = - 2960526m/s

Notice that the negative velocity means that is approaching to the observer.

<em>Case for star 2 \lambda_{measured} = 121.4 nm:</em>

v = (3x10^{8}m/s)(\frac{121.4nm-121.6nm}{121.6nm})

v = - 493421m/s

<em>Case for star 3 \lambda_{measured} = 122.2 nm:</em>

v = (3x10^{8}m/s)(\frac{122.2nm-121.6nm}{121.6nm})

v = 1480263m/s

<em>Case for star 4 \lambda_{measured} = 122.9 nm:</em>

v = (3x10^{8}m/s)(\frac{122.9nm-121.6nm}{121.6nm})

v = 3207236m/s

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Answer:

before this type of attack, high acceleration is the most important thing.

Explanation:

As the zebra is ambushed by the crocodile the most important thing is a quick reaction, in this attack the most likely is that the crocodile is in the water so it cannot run after the zebra.

Consequently, before this type of attack, high acceleration is the most important thing.

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