Grade 11 Physics Help

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

Rainbow [258]

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 .

Using Formulae

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

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 .

Using Formulae 

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

Substituting the value and we obtain

$\sf \longrightarrow \lambda \: = \cfrac{10}{0.1} \\ \\ \\ \sf \longrightarrow \lambda \: = \cancel\cfrac{10}{0.1} \\ \\ \\ \sf \longrightarrow \lambda \: =100m$

Therefore,

Wavelength of the wave is 100 metres.

8 0

3 years ago

Can someone help me?

shtirl [24]

D is the answer the force excerted on the ball is greater than gravity

4 0

4 years ago

When a driver presses the brake pedal, his car can stop with an acceleration of 5.4 m/s2. How far will the car travel while comi

notsponge [240]

$Given:\\a=5.4 \frac{m}{s^2} \\v_0= 25\frac{m}{s} \\\\Find:\\s=?\\\\Solution:\\\\s=v_0t- \frac{at^2}{2} \\\\a= \frac{\Delta v}{t} \\\\v_e=0\Rightarrow \Delta v=v_0\\\\t= \frac{v_0}{a} \\\\s= \frac{v_0^2}{a} - \frac{v_0^2}{2a} =\frac{v_0^2}{2a} \\\\s= \frac{ (25\frac{m}{s})2 }{2\cdot 5.4 \frac{m}{s^2} } \approx 57.9m\\\\C\rightarrow Correct\;answer$

3 0

3 years ago

A small cork with an excess charge of +6.0 μC (1 μC = 10 -6 C) is placed 0.12 m from another cork, which carries a charge of -4.

Alex

Answer:

a.16.125 N b. The force is an attractive force. c. 2.68 × 10¹³ electrons d. 3.75 × 10¹³ electrons

Explanation:

a. What is the magnitude of the electric force between the corks?

The electrostatic force of attraction between the two corks is given by

F = kq₁q₂/r² where k = 9 × 10⁹ Nm²/C², q₁ = +6.0 μC = +6.0 × 10⁻⁶ C, q₂ = -4.3 μC = -4.3 × 10⁻⁶ C and r = distance between the corks = 0.12 m

Substituting the values of the variables into the equation, we have

F = kq₁q₂/r²

F = 9 × 10⁹ Nm²/C² × +6.0 × 10⁻⁶ C × -4.3 × 10⁻⁶ C/(0.12 m)²

= -232.2 × 10⁻³ Nm²/(0.0144 m)²

= -16125 × 10⁻³ N

= -16.125 N

So, the magnitude of the force is 16.125 N

b. Is this force attractive or repulsive?

Since the direction of the force is negative, it is directed towards the positively charged cork, so the force is an attractive force.

c. How many excess electrons are on the negative cork?

Since Q = ne where Q = charge on negative cork = -4.3 μC = -4.3 × 10⁻⁶ C and n = number of excess electrons and e = electron charge = -1.602 × 10⁻¹⁹ C

So n = Q/e = -4.3 × 10⁻⁶ C/-1.602 × 10⁻¹⁹ C = 2.68 × 10¹³ electrons

d. How many electrons has the positive cork lost?

We need to first find the number of excess positive charge n'

Q' = n'q where Q = charge on positive cork = + 6.0 μC = + 6.0 × 10⁻⁶ C and n = number of excess protons and q = proton charge = +1.602 × 10⁻¹⁹ C

So n' = Q'/q = +6.0 × 10⁻⁶ C/+1.602 × 10⁻¹⁹ C = 3.75 × 10¹³ protons

To maintain a positive charge, the number of excess protons equals the number of electrons lost = 3.75 × 10¹³ electrons

4 0

3 years ago

An airplane is traveling 25° west of north at 300 m/s when a wind with velocity 100 m/s directed 35° east of north begins to blo

hodyreva [135]

Answer:

The resultant velocity is 360.5 m/s and direction 79° north of east.