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Montano1993 [528]

3 years ago

14

An airplane is heading south at a speed of 600km/h. If a wind begins blowing from the southwest at a speed of 100km/h, calculate

: a) The velocity (magnitude and direction) of the plane relative to the ground. b) How far away from its intended position will it be after 10min if the pilot takes no corrective action? c) In what direction should the pilot aim in order to fly due south?

1 answer:

Karo-lina-s [1.5K]3 years ago

8 0

Answer:

Here's what I got:

Let's assume that N and E are + directions while S and W are - directions.

Wind is blowing from SW; thus, it is blowing towards NE (or at 45 deg N of E).

Dividing the wind's speed into components:y-component: +70.71 km/h; x-component: +70.71 km/h

Dividing the airplane's speed into components:y-component: -600 km/h; x-component: 0 km/h

Adding the components to get the resulting components:y-component: -529.29 km/h; x-component: +70.71

Using the Pythagorean Theorem to find the resulting speed:v^2 = y^2 + x^2 so v = 533.99 km/h

To find the angle of direction, use arctan (y/x):arctan (529.29/70.71) = 82.39 deg

ANSWER: velocity = 533.99 km/h at 82.39 deg S of E

Explanation:

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Evgen [1.6K]

Answer:

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Explanation:

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7 0

3 years ago

A 23.0 kg iron weightlifting plate has a volume of 2920 cm3 . what is the density of the iron plate in g/cm3?

yanalaym [24]

The first thing you should know for this case is that density is defined as the quotient between mass and volume:
D = M / V
In addition, you should keep in mind the following conversion:
1Kg = 1000g
Substituting the values we have:
D = (23.0 * 1000) / (2920) = 7.88 g / cm ^ 3
answer
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8 0

3 years ago

An object is projected from the ground with an upward speed of ų m/s has a speed of 23m/s when it is at a height of 5m above the

vovikov84 [41]

Answer:

25.08m/s

Explanation:

mgh1 + 0.5mv1² = mgh2 + 0.5mv2²

h1 = 0m

v1 = u

h2 = 5m

v2 = 23m/s

putting the values into the formula above;

m(10)(0) + 0.5m(u²) = m(10)(5) + 0.5m(23²)

0 + 0.5mu² = 50m + 264.5m

0.5mu² = 314.5m

dividing through by m

0.5u² = 314.5

u² = 629

u = <u>2</u><u>5</u><u>.</u><u>0</u><u>8</u><u>m</u><u>/</u><u>s</u>

<u>Theref</u><u>ore</u><u>,</u><u> </u><u>the</u><u> </u><u>init</u><u>ial</u><u> </u><u>speed</u><u> </u><u>"</u><u>u</u><u>"</u><u> </u><u>=</u><u> </u><u>2</u><u>5</u><u>m</u><u>/</u><u>s</u>

6 0

3 years ago

An electron passes through a point 2.83 cm 2.83 cm from a long straight wire as it moves at 35.5 % 35.5% of the speed of light p

igor_vitrenko [27]

Answer:

The magnitude of electron acceleration is $2.34 \times 10^{15}$ $\frac{m}{s^{2} }$

Explanation:

Given:

Distance from the wire to the field point $r = 2.83 \times 10^{-2}$ m

Speed of electron $v = 35.5 \%c$

Current $I = 17.7$ A

For finding the acceleration,

First find the magnetic field due to wire,

$B = \frac{\mu _{o}I }{2\pi r }$

Where $\mu_{o} = 4\pi \times 10^{-7}$

$B = \frac{4\pi \times 10^{-7} \times 17.7 }{2\pi (2.83 \times 10^{-2} ) }$

$B = 12.50 \times 10^{-5}$ T

The magnetic force exerted on the electron passing through straight wire,

$F = qvB$

$F = 1.6 \times 10^{-19} \times 0.355 \times 3 \times 10^{8} \times 12.50 \times 10^{-5}$

$F = 21.3 \times 10^{-16}$ N

From the newton's second law

$F = ma$

Where $m =$ mass of electron $= 9.1 \times 10^{-31}$ kg

So acceleration is given by,

$a = \frac{F}{m}$

$a = \frac{21.3 \times 10^{-16} }{9.1 \times 10^{-31} }$

$a = 2.34 \times 10^{15}$ $\frac{m}{s^{2} }$

Therefore, the magnitude of electron acceleration is $2.34 \times 10^{15}$ $\frac{m}{s^{2} }$

7 0

3 years ago

A voltage of 220V (rms) line to netralvoltage is applied across a three phase Y connected resistive load. Each phase of the Y co

mrs_skeptik [129]

Answer:

Current through each phase Vp = 2.2A

Total three phase power Pt= 1.45kW

Power factor of the load pf = 1

Explanation:

i) Find current through each phase

Vp =220V (rms)

Z =100 Ω

I = Vp/Z

= 220/100

= 2.2A

ii) Find the total three phase power

for a resistive load, Power, P = VI

Power for each phase is given as:

P = 220 * 2.2

= 484 W

Total power TP =3* P

=484*3

= 1452W

=1.45kW

iii) Find the power factor of the load

Phase angle for a resistive load is 0.

α= 0

Hence, power factor of load = cos α

pf = cos 0

pf = 1

5 0

3 years ago