A wave's frequency is 2Hz and its wavelength is 4 m. What is the wave's speed?

A ferry approaches shore, moving north with a speed of 6.2 m/s relative to the dock. A person on the ferry walks from one side o

jasenka [17]

Speed of Ferry is towards North with magnitude 6.2 m/s

Here if we assume that North direction is along Y axis and East is along X axis then we can say

$\vec v_f = 6.2 \hat j$

Now a person walk on ferry with speed 1.5 m/s towards east with respect to Ferry

so it is given as

$\vec v_{pf} = 1.5 \hat i$

also by the concept of relative motion we know that

$\vec v_{pf} = \vec v_p - \vec v_f$

now plug in all values in it

$1.5 \hat i = \vec v_p - 6.2 \hat j$

$\vec v_p = 1.5 \hat i + 6.2 \hat j$

now if we need to find the speed of the person then we need to find its magnitude

so it is given as

$v = \sqrt{1.5^2 + 6.2^2}$

$v = 6.37 m/s$

7 0

3 years ago

A river 800m wide flows at the rate of 5km/h . A swimmer who can swim at 10km/h in still water wants to cross the river straight

LuckyWell [14K]

Answer:

At an angle of $30^{\circ}$

Explanation:

Assume the river flows from East to West so for the swimmer to cross across it, assume he crosses it from West to East.

The resultant speed will be given by

$R= \sqrt {10^{2}-5^{2}=\sqrt {75}\approx 8.66 km/h\\Direction=sin^{-1}\frac {5}{10}\approx 30^{\circ}$

6 0

2 years ago

A circular ring with area 4.45 cm2 is carrying a current of 13.5 A. The ring, initially at rest, is immersed in a region of unif

Gwar [14]

Answer:

a) ( 0.0015139 i^ + 0.0020185 j^ + 0.00060556 k^ ) N.m

b) ΔU = -0.000747871 J

c) w = 47.97 rad / s

Explanation:

Given:-

- The area of the circular ring, A = 4.45 cm^2

- The current carried by circular ring, I = 13.5 Amps

- The magnetic field strength, vec ( B ) = (1.05×10−2T).(12i^+3j^−4k^)

- The magnetic moment initial orientation, vec ( μi ) = μ.(−0.8i^+0.6j^)

- The magnetic moment final orientation, vec ( μf ) = -μ k^

- The inertia of ring, T = 6.50×10^−7 kg⋅m2

Solution:-

- First we will determine the magnitude of magnetic moment ( μ ) from the following relation:

μ = N*I*A

Where,

N: The number of turns

I : Current in coil

A: the cross sectional area of coil

- Use the given values and determine the magnitude ( μ ) for a single coil i.e ( N = 1 ):

μ = 1*( 13.5 ) * ( 4.45 / 100^2 )

μ = 0.0060075 A-m^2

- From definition the torque on the ring is the determined from cross product of the magnetic moment vec ( μ ) and magnetic field strength vec ( B ). The torque on the ring in initial position:

vec ( τi ) = vec ( μi ) x vec ( B )

= 0.0060075*( -0.8 i^ + 0.6 j^ ) x 0.0105*( 12 i^ + 3 j^ -4 k^ )

= ( -0.004806 i^ + 0.0036045 j^ ) x ( 0.126 i^ + 0.0315 j^ -0.042 k^ )

- Perform cross product:

$\left[\begin{array}{ccc}i&j&k\\-0.004806&0.0036045&0\\0.126&0.0315&-0.042\end{array}\right] = \left[\begin{array}{ccc}-0.00015139\\-0.00020185\\-0.00060556\end{array}\right] \\\\$

- The initial torque ( τi ) is written as follows:

vec ( τi ) = ( 0.0015139 i^ + 0.0020185 j^ + 0.00060556 k^ )

- The magnetic potential energy ( U ) is the dot product of magnetic moment vec ( μ ) and magnetic field strength vec ( B ):

- The initial potential energy stored in the circular ring ( Ui ) is:

Ui = - vec ( μi ) . vec ( B )

Ui =- ( -0.004806 i^ + 0.0036045 j^ ) . ( 0.126 i^ + 0.0315 j^ -0.042 k^ )

Ui = -[( -0.004806*0.126 ) + ( 0.0036045*0.0315 ) + ( 0*-0.042 )]

Ui = - [(-0.000605556 + 0.00011)]

Ui = 0.000495556 J

- The final potential energy stored in the circular ring ( Uf ) is determined in the similar manner after the ring is rotated by 90 degrees with a new magnetic moment orientation ( μf ) :

Uf = - vec ( μf ) . vec ( B )

Uf = - ( -0.0060075 k^ ) . ( 0.126 i^ + 0.0315 j^ -0.042 k^ )

Uf = - [( 0*0.126 ) + ( 0*0.0315 ) + ( -0.0060075*-0.042 ) ]

Uf = -0.000252315 J

- The decrease in magnetic potential energy of the ring is arithmetically determined:

ΔU = Uf - Ui

ΔU = -0.000252315 - 0.000495556

ΔU = -0.000747871 J

Answer: There was a decrease of ΔU = -0.000747871 J of potential energy stored in the ring.

- We will consider the system to be isolated from any fictitious forces and gravitational effects are negligible on the current carrying ring.

- The conservation of magnetic potential ( U ) energy in the form of Kinetic energy ( Ek ) is valid for the given application:

Ui + Eki = Uf + Ekf

Where,

Eki : The initial kinetic energy ( initially at rest ) = 0

Ekf : The final kinetic energy at second position

- The loss in potential energy stored is due to the conversion of potential energy into rotational kinetic energy of current carrying ring.

-ΔU = Ekf

0.5*T*w^2 = -ΔU

w^2 = -ΔU*2 / T

Where,

w: The angular speed at second position

w = √(0.000747871*2 / 6.50×10^−7)

w = 47.97 rad / s

6 0

3 years ago

A seagull flies at a velocity of 9.00 m/s straight into the wind. (a) if it takes the bird 20.0 min to travel 6.00 km relative t

enot [183]

Here we will the speed of seagull which is v = 9 m/s

this is the speed of seagull when there is no effect of wind on it

now in part a)

if effect of wind is in opposite direction then it travels 6 km in 20 min

so the average speed is given by the ratio of total distance and total time

$v_{avg} = \frac{6000}{20*60}$

$v_{avg} = 5m/s$

now since effect of wind is in opposite direction then we can say

$V_{net} = v_{bird} - v_{wind}$

$5 = 9 - v_{wind}$

$v_{wind}= 4 m/s$

Part b)

now if bird travels in the same direction of wind then we will have

$v_{net}= v_{bird} + v_{wind}$

$v_{net} = 9 + 4 = 13 m/s$

now we can find the time to go back

$time = \frac{distance}{speed}$

$time = \frac{6000}{13}$

$time = 7.7 minutes$

Part c)

Total time of round trip when wind is present

$T = t_1 + t_2$

$T = 20 + 7.7 = 27.7 min$

now when there is no wind total time is given by

$T = \frac{6000}{9} + \frac{6000}{9}$

$T = 22.22 min$

So due to wind time will be more

4 0

3 years ago

A ball is falling at terminal velocity. Terminal velocity occurs when the ball is in equilibrium and the forces are balanced. Wh

Greeley [361]

Answer:

A free body diagram with 2 forces: the first pointing downward labeled F Subscript g Baseline 20 N and the second pointing upward labeled F Subscript air Baseline 20 N.

Explanation:

This is because at terminal velocity, the ball stops accelerating and the net force on the ball is zero. For the net force to be zero, equal and opposite forces must act on the ball, so that their resultant force is zero. That is F₁ + F₂ = 0 ⇒ F₁ = -F₂

Since F₁ = 20 N, then F₂ = -F₁ = -20 N

So, if F₁ points upwards since it is positive, then F₂ points downwards since it is negative.

So, a free body diagram with 2 forces: the first pointing downward labeled F Subscript g Baseline 20 N and the second pointing upward labeled F Subscript air Baseline 20 N best describes the ball falling at terminal velocity.

4 0

3 years ago

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