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FrozenT [24]
3 years ago
5

A river current has a velocity of 5 km/h relative to the shore. A boat moves in the same direction as the current at 4 km/h rela

tive to the river. How do you calculate the velocity of the boat relative to the shore?
subtract the boat’s velocity from the river’s velocity

add the river’s velocity to the boat’s velocity

divide the river’s velocity by the boat’s velocity
Physics
2 answers:
AleksAgata [21]3 years ago
7 0
<span>Add the river’s velocity to the boat’s velocity.</span>
Alona [7]3 years ago
7 0

Answer:

Add the river’s velocity to the boat’s velocity

Explanation:

It is given that,

A river current has a velocity of 5 km/h relative to the shore. A boat moves in the same direction as the current at 4 km/h relative to the river. We need to calculate the velocity of the boat relative to the shore.

As the two vectors are going in same direction. So, the velocity of the boat relative to the shore can be calculated by adding the river's velocity to the boat's velocity. So, the correct option is (c) "add the river’s velocity to the boat’s velocity".

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A source emits sound uniformly in all directions. There are no reflections of the sound. At a distance of 12 m from the source,
yaroslaw [1]

Answer:

1.58 W

Explanation:

Since the sound spreads uniformly in all directions, it must be in a form of a circle with radius of 12 m. So the area of the circle is

A = \pi r^2 = \pi 12^2 = 452.389 m^2

From the intensity of the sound we can calculate the power at 12 m

P = AI = 452.389 * 3.5\times10^{-3} = 1.58 W

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3 years ago
For a given wave IF frequency doubles the wavelength
motikmotik

Answer:

C is halved

Explanation:

The frequency and the wavelength of a wave are related by the equation:

v=f\lambda

where

v is the speed of the wave

f is the frequency

\lambda is the wavelength

From the equation above, we see that for a given wave, if the wave is travelling in the same medium (and so, its speed is not changing), then the frequency and the wavelength are inversely proportional to each other.

Therefore, if the frequency doubles, the wavelength will halve in order to keep the speed constant:

\lambda = \frac{v}{f}\\\lambda' = \frac{v}{f'}=\frac{v}{2f}=\frac{1}{2}(\frac{v}{f})=\frac{\lambda}{2}

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3 years ago
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What is the magnitude of the total acceleration of point A after 2 seconds? The bar starts from rest and has a constant angular
monitta

Answer:

a_total = 2 √ (α² + w⁴) ,   a_total = 2,236 m

Explanation:

The total acceleration of a body, if we use the Pythagorean theorem is

          a_total² = a_T²2 + a_{c}²

where

the centripetal acceleration is

  a_{c} = v² / r = w r²

tangential acceleration

   a_T = dv / dt

angular and linear acceleration are related

         a_T = α  r

we substitute in the first equation

       a_total = √ [(α r)² + (w r² )²]

       a_total = 2 √ (α² + w⁴)

Let's find the angular velocity for t = 2 s if we start from rest wo = 0

        w = w₀ + α t

        w = 0 + 1.0 2

        w = 2.0rad / s

       

we substitute

        a_total = r √(1² + 2²) = r √5

        a_total = r 2,236

In order to finish the calculation we need the radius to point A, suppose that this point is at a distance of r = 1 m

         a_total = 2,236 m

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