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3 years ago

I need this asap pls

Physics

1 answer:

user100 [1]3 years ago

3 0

Answer:

2 i think

Explanation:

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A migrating robin flies due north with a speed of 12 m/s relative to the air. The air moves due east with a speed of 6.3 m/s rel

Sonja [21]

consider east-west direction along X-axis and north-south direction along Y-axis

$V_{ra}$ = velocity of migrating robin relative to air = 12 j m/s

(where "j" is unit vector in Y-direction)

$V_{ag}$ = velocity of air relative to ground = 6.3 i m/s

(where "i" is unit vector in X-direction)

$V_{rg}$ = velocity of migrating robin relative to ground = ?

using the equation

$V_{rg}$ = $V_{ra}$ + $V_{ag}$

$V_{rg}$ = 12 j + 6.3 i

$V_{rg}$ = 6.3 i + 12 j

magnitude : sqrt((6.3)² + (12)²) = 13.6 m/s

direction : tan⁻¹(12/6.3) = 62.3 deg north of east

4 0

3 years ago

Gggcdfubrdegubtcwftvf y day rx u e HHS’s forget h

aev [14]

Answer:

um . . . yes ?

8 0

2 years ago

Read 2 more answers

Step 1, when solving a two dimensional, multi-charge problem, is to define the vectors. Please identify the next five steps, in

Masja [62]

Step 2: calculate A and B magnitudes
Step 3: calculate x, y components
Step 4: sum vector components
Step 5: calculate magnitude of R
Step 6: calculate direction of R

4 0

3 years ago

Read 2 more answers

If the radius of the equipotential surface of the point charge is 14.3 m at a potential of 2.20 kV, what will be the magnitude o

PSYCHO15rus [73]

Answer:

The magnitude of the point charge is 3.496 x 10⁻⁶ C

Explanation:

Given;

radius of the surface, r = 14.3 m

magnitude of the potential, V = 2.2 kV = 2,200 V

The magnitude of the point charge is calculated as follows;

$V = (\frac{1}{4\pi \epsilon _0} )(\frac{Q}{r} )\\\\V = \frac{KQ}{r} \\\\Q = \frac{Vr}{K} \\\\Q = \frac{2,200 \times 14.3}{9\times 10^9} \\\\Q = 3.496 \times 10^{-6} \ C\\\\Q = 3.496 \ \mu C$

Therefore, the magnitude of the point charge is 3.496 μC

6 0

2 years ago

The tires of a car make 60 revolutions as the car reduces its speed uniformly from 95.0 km/h to 60.0 km/h. The tires have a diam

yawa3891 [41]

Answer:

-2.869 rad/s2

Explanation:

Data given:

speed, vi at 95.0 km/h = 95 X (1 hour /3600 seconds) X (1000m / 1km)

Note that, for every 1 hour, there will be 60sec X 60sec = 3600 seconds

And for every 1km, there will be 1000m.

So, speed of 95.0 km/h = 26.389 m/s

speed, vi = r ω (radius X angular velocity)

angular velocity, ωi = v/r

ωi = 26.389 m/s ÷ half of 0.88 m diameter

= 59.975 rad/s

decelerating to speed, vf at 60.0 km/h = 60 X X (1 hour /3600 seconds) X (1000m / 1km)

= 16.667m/s

The angular velocity for this speed = 16.667m/s ÷ half of 0.88 m diameter

= 37.879rad/s

How far the car goes is equivalent to the angular acceleration which equals to (ωf^2 - ωi^2) ÷ 2θ

= (37.879rad/s)^2 - (59.975 rad/s)^2 ÷ 2 (60 rev X 2π rad/rev)

= -2.869 rad/s2

4 0

3 years ago

I need this asap pls ​

I need this asap pls