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kumpel [21]
3 years ago
10

A river flows due south at 5 mi/h. A swimmer attempting to cross the river heads due east swimming at 3 mi/h relative to the wat

er. Find the true velocity of the swimmer as a vector. (Assume that the i vector points east, and the j vector points north.)

Physics
1 answer:
Dafna11 [192]3 years ago
3 0

Answer:

<u>velocity of swimmer relative to ground = 3 i -5 j</u>

Explanation:

  • To cross a river the swimmer swims relative to river in perpendicular direction.

Velocity of river = -5 j (south)

Velocity of swimmer relative to river = 3 i(north)

So

<h2>Velocity of swimmer relative to ground = Velocity of swimmer relative to river + Velocity of river</h2>

Velocity of swimmer relative to ground = 3 i -5 j

So magnitude of total velocity is \sqrt{3^2+(-5)^2} =\sqrt{9+25} = \sqrt{34}

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3 years ago
g A box having mass 0,5kg is placed in front of a 20 cm compressed spring. When the spring released, box having mass m1, collide
vagabundo [1.1K]

Answer:Expression given below

Explanation:

Given mass of spring\left ( m_1\right )=0.5 kg

Compression in the spring\left ( x\right )=20 cm

Let the spring constant be K

Using Energy conservation

potential energy stored in spring =Kinetic energy of Block\left ( m_1\right )

\frac{1}{2}Kx^2=\frac{1}{2}m_1v^2

v=x\sqrt{\frac{k}{m_1}}

now conserving momentum

m_1v=\left ( m_1+m_2\right )v_0

v_0=\frac{m_1}{m_1+m_2}v

where v_0 is the final velocity

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3 years ago
what is the ratio of ey,final, the maximum value of the y-component of the electric field immediately following the last polariz
OleMash [197]

That ratio is 2 .

<h3>What is ratio?</h3>

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-1/4-1/8

When dividing fractions, we multiply by the reciprocal:

-1/4x-8/1

To multiply fractions, multiply straight across:

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8 0
1 year ago
Two radio antennas are 120 m apart on a north-south line, and they radiate in phase at a frequency of 3.4 MHz. All radio measure
chubhunter [2.5K]

Answer:

the smallest angle from the antennas is <em>47.3°</em>

Explanation:

We first need to write the expression for the relation between the wavelength (λ) and the frequency (f) of the wave, and then solve for the wavelength.

Therefore, the relation is:

λ = c /f

where

  • c is the speed of light constant
  • λ is the wavelength
  • f is the frequency

Thus,

λ = (3 × 10⁸ m/s) / (3.4 MHz)

  = (3 × 10⁸ m/s) / (3.4 MHz)(10⁶ Hz/1 MHz)

  = 88.235 m

Therefore, the smallest angle measured (from the north of east) from the antennas for the constructive interference of the two-radio wave can be calculated as

θ = sin⁻¹(λ / d)

where

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Thus,

θ = sin⁻¹(88.235 / 120)

<em>θ = 47.3 °</em>

<em></em>

Therefore, the smallest angle from the antennas, measured north of east, at which constructive interference of two radio waves occurs is <em>47.3 °</em>.

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3 years ago
Please help me look at the image below
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