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dalvyx [7]
3 years ago
15

You use 8x binoculars were used on a warbler (14cm long) in a tree 18cm away. What angle (in degrees) does the image of the warb

ler subtend on your retina?
Physics
1 answer:
mafiozo [28]3 years ago
7 0

Answer:

The angle it subtend on the retina is  \theta_z = 0.44586^o    

Explanation:

From the question we are told that

     The length of the warbler is  L = 14cm = \frac{14}{100} = 0.14m

      The distance from the binoculars is    d = 18cm = \frac{18}{100} = 0.18m

        The magnification of the binoculars is  M =8

Without the 8 X binoculars the  angle made with the angular size of the object  is mathematically represented as

          \theta = \frac{L}{d}

        \theta  = \frac{0.14}{0.18}

           = 0.007778 rad

Now magnification can be represented mathematically as

         M = \frac{\theta _z}{\theta}

Where \theta_z is the angle the image of the warbler subtend on your retina when the   binoculars i.e the  binoculars zoom.

So

      \theta_z = M * \theta

=>    \theta_z =8 * 0.007778

            = 0.0622222224

Generally the conversion to degrees can be mathematically evaluated as

             \theta_z = 0.062222224 * (\frac{360 }{2 \pi rad} )

              \theta_z = 0.44586^o  

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The mass of a regulation tennis ball is 57 g. The ball is in contact with the tennis racket for 30 ms. The ball was served at 73
yaroslaw [1]

Answer:

Correct answer:  F₂ = 104.5 N

Explanation:

Given:

m = 57 g = 57 · 10⁻³ kg

Δt = 30 ms = 30 · 10 ⁻³ seconds

V₁ = 73.14 m/s   service speed

V₂ = 55 m/s  returned speed

M = m · V  Momentum or Impulse

You forgot to indicate what time the ball contact when returning.

We will assume that the time is the same Δt = 30 ms = 30 10 ⁻³ seconds.

The formula for calculating force is according to Newton's second law is:

F = ΔM / Δt = m · ΔV / Δt

Force during service is:

F₁ = 57 · 10⁻³ · 73.14 / 30 · 10 ⁻³ = 138.97 N

F₁ = 138.97 N

Returned force:

F₂ =  57 · 10⁻³ · 55 / 30 · 10 ⁻³ = 104.5 N

F₂ =  104.5 N

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A sled is moving down a steep hill. The mass of the sled is 50 kg and the net force acting on it is 20 N. What must be done to f
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You need to first measure the angle of descent, i.e. the angle the hill makes with the ground. Then identify the forces acting on the sled, split them up into horizontal and vertical components, or into components that are parallel and perpendicular to the hill, and use Newton's second law to determine the components of the sled's acceleration vector.

There are at least 2 forces acting on the sled:

• its weight, pointing downward with magnitude <em>W</em> = <em>m g</em>

• the normal force, pointing perpendicular to the hill and away from the ground with mag. <em>N</em>

The question doesn't specify, but there might also be friction to consider, indicated in the attachment by the vector <em>F</em> pointing parallel to the slope of the hill and opposing the direction of the sled's motion with mag. <em>F</em>.

Splitting up the forces into parallel/perpendicular components is less work. By Newton's second law, the net force (denoted with ∑ or "sigma" here) in a particular direction is equal to the mass of the sled times its acceleration in that direction:

∑ (//) = <em>W</em> (//) = <em>m</em> <em>a</em> (//)

∑ (⟂) = <em>W</em> (⟂) + <em>N</em> = <em>m </em><em>a</em> (⟂)

where, for instance, <em>W</em> (//) denotes the component of the sled's weight in the direction parallel to the hill, while <em>a</em> (⟂) denotes the component of the sled's acceleration perpendicular to the hill. If there is friction, you need to add -<em>F</em> to the first equation.

If the hill makes an angle of <em>θ</em> with flat ground, then <em>W</em> makes the same angle with the hill so that

<em>W</em> (//) = -<em>m g </em>sin(<em>θ</em>)

<em>W</em> (⟂) = -<em>m g</em> cos(<em>θ</em>)

So we have

<em>-m g </em>sin(<em>θ</em>) = <em>m</em> <em>a</em> (//)   →   <em>a</em> (//) = -<em>g </em>sin(<em>θ</em>)

<em>-m g</em> cos(<em>θ</em>) + <em>N</em> = <em>m </em><em>a</em> (⟂)   →   <em>a</em> (⟂) = 0

where the last equality follows from the fact that the normal force exactly opposes the perpendicular component of the weight. This is because the sled is moving along the slope of the hill, and not into the air or into the ground.

Then the acceleration vector is

<em>a</em> = <em>a</em> (//)

with magnitude

||<em>a</em>|| = <em>a</em> = <em>g </em>sin(<em>θ</em>).

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Explanation:

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