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2 years ago

An ocean wave has a wavelength of 10 m and a frequency of 4.0 Hz. What is the velocity of the

Physics

1 answer:

aleksandr82 [10.1K]2 years ago

3 0

The velocity of the wave is 40 m/s

Explanation:

The relationship between the velocity of a wave and its frequency and wavelength is given by

$v=f \lambda$

where

v is the velocity

f is the frequency

$\lambda$ is the wavelength

For the ocean wave in this problem, we have:

$\lambda = 10 m$ (wavelength)

f = 4.0 Hz (frequency)

Therefore, its velocity is

$v=(4.0)(10)=40 m/s$

Learn more about waves:

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A helicopter flies 250 km on a straight path in a direction 60° south of east. The east component of the helicopter’s displaceme

GaryK [48]

Given that,

Distance in south-west direction = 250 km

Projected angle to east = 60°

East component = ?

since,

cos ∅ = base/hypotenuse

base= hyp * cos ∅

East component = 250 * cos 60°

East component = 125 km

8 0

2 years ago

Read 2 more answers

If a rock is thrown upward on the planet mars with a velocity of 14 m/s, its height (in meters) after t seconds is given by h =

crimeas [40]

Answer:

Velocity of rock after 2 seconds = 6.56 m/s

Explanation:

We have equation of motion , $s= ut+\frac{1}{2} at^2$ , s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

Here height of rock in meters, h = $14t-1.86t^2$

Comparing both the equations

We will get initial velocity = 14 m/s(already given) and $\frac{1}{2} a = -1.86$

So, Acceleration, a = -3.72 $m/s^2$

Now we have equation of motion, v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration and t is the time taken.

When time is 2 seconds we need to find final velocity.

v = 14 - 3.72 * 2 = 6.56 m/s.

So, Velocity of rock after 2 seconds = 6.56 m/s

6 0

2 years ago

Please help! will give brainlest!!!!!!!!!!!!

eimsori [14]

The force of friction is 19.1 N

Explanation:

According to Newton's second law, the net force acting on the bag is equal to the product between its mass and its acceleration:

$\sum F = ma$

where

$\sum F$ is the net force

m is the mass

a is the acceleration

The bag is moving at constant speed, so its acceleration is zero:

$a=0$

Therefore the net force is zero as well:

$\sum F = 0$

Here we are interested only in the forces acting along the horizontal direction, therefore the net force is given by:

$\sum F = F cos \theta - F_f = 0$

where

$F cos \theta$ is the horizontal component of the applied force, with

F = 22.5 N

$\theta=32.0^{\circ}$

$F_f$ is the force of friction

And solving for $F_f$ , we find

$F_f =Fcos \theta=(22.5)(cos 32.0^{\circ})=19.1 N$

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7 0

2 years ago

A child of mass 27 kg swings at the end of an elastic cord. At the bottom of the swing, the child's velocity is horizontal, and

snow_tiger [21]

Answer:

The magnitude of the rate of change of the child's momentum is 794.11 N.

Explanation:

Given that,

Mass of child = 27 kg

Speed of child in horizontal = 10 m/s

Length = 3.40 m

There is a rate of change of the perpendicular component of momentum.

Centripetal force acts always towards the center.

We need to calculate the magnitude of the rate of change of the child's momentum

Using formula of momentum

$\dfrac{dp}{dt}=F$

$\dfrac{dP}{dt}=\dfrac{mv^2}{r}$

Put the value into the formula

$\dfrac{dP}{dt}=\dfrac{27\times10^2}{3.40}$

$\dfrac{dP}{dt}=794.11\ N$

Hence, The magnitude of the rate of change of the child's momentum is 794.11 N.

7 0

2 years ago

gaseous h2 and br2 are added to an evacuated 1.15L container kept at 298K. The intial partial pressurre of H2(g) is 0.782 atm an

Nastasia [14]

The partial pressures of HBr when the system reaches equilibrium is 2.4 X 10⁻¹¹ atm

Explanation:

H₂ + Br₂ ⇒ 2HBr

PH₂ = 0.782atm

PBr₂ = 0.493atm

Kp = (PHBr)²/ (PH₂) (PBr₂) = 1.4 X 10⁻²¹

At equilibrium:

Let 2x = pressure of HBr

PH₂ = 0.782 -x

PBr₂ = 0.493 - x

Kp = (2x)^2 / (0.782-x)(0.493-x)

Now, because Kp is very small, x will be very small compared to 0.782 and 0.493.

Then,

Kp = 1.4X10⁻²¹ = (4x²) / (0.782)(0.493)

x = 1.2X10⁻¹¹

PHBr = 2x = 2.4 X 10⁻¹¹ atm

Therefore, the partial pressures of HBr when the system reaches equilibrium is 2.4 X 10⁻¹¹ atm

3 0

2 years ago