1. Calculate the height of tree, 250 m away that produces

A fisherman sitting on the end of a pier notices that 6 wave crests pass him in 3 seconds. What is the frequency of the waves?.

dalvyx [7]

Answer:

2 Hz

Explanation:

f = frequency

n = wave crest

t = time

f = n/t → f = 6/3 = 2

4 0

3 years ago

A spaceship is 1600 m long when it is at rest. When it is traveling at a certain constant speed its length is measured by extern

Firdavs [7]

Answer:

speed of the spaceship is 0.495251 c

Time T = 1.4982 hr

Explanation:

given data

length L1 = 1600 m

length L2 = 1390 m

to find out

speed of the spaceship and time

solution

we know here

L2 = L1 × $\sqrt{1 - V^{2} }$

so we find V by put L1 and L2

1390 = 1600 × $\sqrt{1 - V^{2} }$

V = 0.495251 c

speed of the spaceship is 0.495251 c

and

we know T1 = 1.95 hr

so we find T2

T1 = T2 $\sqrt{1 - v^{2} }$

put here V and T1 so we get T2

1.95 = T2 $\sqrt{1 - 0.495251^{2} }$

Time T = 1.4982 hr

3 0

4 years ago

What would you expect to cause a drop in air pressure?

Murrr4er [49]

Answer:

D

Explanation:

6 0

3 years ago

Read 2 more answers

A copper wire has a square cross section 2.0 mm on a side. The wire is 5.0 m long and carries a current of 2.0 A. The density of

kondor19780726 [428]

Answer:

30.22 hours

Explanation:

Given data:

A= l² = (2 x $10^{-3}$ )² = 4 x $10^{-6}$ m²

Length 'L' = 5m

current ' $I$ ' = 2 A

density of free electrons 'n'= 8.5 x $10^{28}$ /m³

Current Density 'J' = $I$ / A

J= 2/4 x $10^{-6}$

J= 5 x $10^{5}$ A/m²

We can determine the time required for an electron to travel the length of the wire by

T= L/ Vd

Where,

L is length and Vd is drift velocity.

Vd can be defined by J/ n|q|

where,

n is the charge-carrier number density

|q| is is the charge carried by each charge carrier =>1.6 x $10^{-19}$ C

T= L/ Vd

Therefore,

T= L . n|q| / J

T= (4 x 8.5 x $10^{28}$ x |1.6 x $10^{-19}$ |)/5 x $10^{5}$

T= 108800 seconds =>1813.33 minutes

Converting minute into hours:

T= 30.22 hours

Thus, time that is required for an electron to travel the length of the wire is 30.22 hours

4 0

3 years ago

A flock of ducks is trying to migrate south for the winter, but they keep being blown off course by a wind blowing from the west

Minchanka [31]

The ducks' flight path as observed by someone standing on the ground is the sum of the wind velocity and the ducks' velocity relative to the wind:

ducks (relative to wind) + wind (relative to Earth) = ducks (relative to Earth)

or equivalently,

$\vec v_{D/W}+\vec v_{W/E}=\vec v_{D/E}$

(see the attached graphic)

We have

ducks (relative to wind) = 7.0 m/s in some direction θ relative to the positive horizontal direction, or

$\vec v_{D/W}=\left(7.0\dfrac{\rm m}{\rm s}\right)(\cos\theta\,\vec\imath+\sin\theta\,\vec\jmath)$

wind (relative to Earth) = 5.0 m/s due East, or

$\vec v_{W/E}=\left(5.0\dfrac{\rm m}{\rm s}\right)(\cos0^\circ\,\vec\imath+\sin0^\circ\,\vec\jmath)$

ducks (relative to earth) = some speed v due South, or

$\vec v_{D/E}=v(\cos270^\circ\,\vec\imath+\sin270^\circ\,\vec\jmath)$

Then by setting components equal, we have

$\left(7.0\dfrac{\rm m}{\rm s}\right)\cos\theta+5.0\dfrac{\rm m}{\rm s}=0$

$\left(7.0\dfrac{\rm m}{\rm s}\right)\sin\theta=-v$

We only care about the direction for this question, which we get from the first equation:

$\left(7.0\dfrac{\rm m}{\rm s}\right)\cos\theta=-5.0\dfrac{\rm m}{\rm s}$

$\cos\theta=-\dfrac57$

$\theta=\cos^{-1}\left(-\dfrac57\right)\text{ OR }\theta=360^\circ-\cos^{-1}\left(-\dfrac57\right)$

or approximately 136º or 224º.

Only one of these directions must be correct. Choosing between them is a matter of picking the one that satisfies both equations. We want

$\left(7.0\dfrac{\rm m}{\rm s}\right)\sin\theta=-v$

which means θ must be between 180º and 360º (since angles in this range have negative sine).

So the ducks must fly (relative to the air) in a direction 224º relative to the positive horizontal direction, or about 44º South of West.

8 0

3 years ago