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3 years ago

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A wildlife researcher is tracking a flock of geese. The geese fly 4.0 km due west, then turn toward the north by 40º and fly ano

ther 4.0 km. What is the magnitude of their displacement? Calculate this and enter it, in km, to two significant figures.

1 answer:

Kobotan [32]3 years ago

7 0

Wildlife researcher starts from a and then reaches b, he turns towards north 40 degree to move towards c.

Total displacement is ac

Total horizontal displacement = 4+4 cos40 =7.06 km

Total vertical displacement = 4 sin40 =2.57 km

Total displacement $= \sqrt{7.06^2+2.57^2}$ = 7.51 km

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Answer:

Explanation:

Area of crossection, A = 7.80 cm²

Initial magnetic field, B = 0.5 T

Final magnetic field, B' = 3.3 T

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The induced emf is given by

$e=\frac{d\phi}{dt}=A\frac{B' - B}{t}$

where, Ф is the rate of change of magnetic flux.

e = 7.80 x 10^-4 x (3.3 - 0.5) / 1

e = 2.184 mV

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3 years ago

05. The time required to complete one lap around a perfectly circular track having a radius of 1,835 meters is 86

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Answer:

v = 134.06 m/s

Explanation:

Given that,

Radius of a circular track is 1,835 m

Time required to complete one lap around a perfectly circular track is 86 seconds

We need to find the car's velocity. Velocity is equal to,

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On circular path,

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So, car's velocity is 134.06 m/s.

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4 years ago

A proton is launched from an infinite plane of charge with surface charge density -1.10×10-6 C/m2. If the proton has an initial

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The distance covered by the proton is 48.4 m

Explanation:

The electric field produced by an electrically charged infinite plane is given by

$E=\frac{\sigma}{2\epsilon_0}$

where in this case,

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$\epsilon_0 = 8.85\cdot 10^{-12} F/m$ is the vacuum permittivity

Substituting,

$E=\frac{-1.10\cdot 10^{-6}}{2(8.85\cdot 10^{-12})}=-5.65\cdot 10^4N/C$

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The electric force on the proton due to this field is

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where

$q=1.6\cdot 10^{-19}C$ is the proton charge

Substituting,

$F=(1.6\cdot 10^{-19})(-5.65\cdot 10^4)=-9.0\cdot 10^{-15} N$

where the direction is toward the plane.

Now we can calculate the proton's acceleration using Newton's second law:

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

where

$m=1.67\cdot 10^{-27}kg$ is the proton mass

Substituting,

$a=\frac{-9.0\cdot 10^{-15}}{1.67\cdot 10^{-27}}=-5.4\cdot 10^{-12} m/s^2$

Now we can finally apply the following suvat equation for accelerated motion to find the distance travelled by the proton:

$v^2-u^2=2as$

where

v = 0 is the final velocity

$u=2.40\cdot 10^7 m/s$ is the initial velocity

a is the acceleration

s is the distance covered

And solving for s,

$s=\frac{v^2-u^2}{2a}=\frac{0-(2.4\cdot 10^7)^2}{2(-5.4\cdot 10^{12})}=53.3 m$

Therefore, the closest answer is 48.4 m.

Learn more about accelerated motion:

brainly.com/question/9527152

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3 years ago

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Answer:

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Explanation:

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Let Vector $\overrightarrow{OA}$ is the tidal current velocity as shown in the diagram.

In order to travel straight across the harbor, the vector addition of both the velocities (i.e the resultant velocity, $\vec {R}$ must be in the north direction.

Let $\overrightarrow{AB}$ is the speed of the kayaker having angle \theta measured north of east as shown in the figure.

For the resultant velocity in the north direction, the tail of the vector $\overrightarrow {OA}$ and head of the vector $\overrightarrow{AB}$ must lie on the north-south line.

Now, for this condition, from the triangle OAB

$|\overrightarrow{AB}|\sin \theta=|\overrightarrow{OA}|$

$\Rightarrow \sin\theta=\frac{|\overrightarrow{OA}|}{|\overrightarrow{AB}|}=\frac 2 3$

$\Rightarrow \theta=\sin^{-1}\frac23$

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4 years ago

If an ocean wave passes a stationary point every 5 s and has a velocity of 10 m/s, what is the wavelength of the wave

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Answer:

As Per Provided Information

Velocity of wave v is 10m/s

These ocean wave passes a stationary point every 5 s ( It's time period)

First we calculate the frequency of ocean wave .

<u>Using</u><u> Formulae</u>

$\blue{\boxed{\bf \: \nu = \cfrac{1}{v}}}$

here

v is the velocity of wave .

$\sf\longrightarrow \nu \: = \cfrac{1}{10} \\ \\ \\ \sf\longrightarrow \nu \: = 0.1Hz$

Now calculating the wavelength of the wave .

<u>Using </u><u>Formulae </u>

$\boxed{ \bf \lambda = \cfrac{v}{ \nu}}$

Substituting the value and we obtain

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<u>Therefore</u><u>,</u>

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3 years ago