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ivann1987 [24]
3 years ago
6

An impala is an African antelope capable of a remarkable vertical leap. In one recorded leap, a 45 kg impala went into a deep cr

ouch, pushed straight up for 0.21 s , and reached a height of 2.5 m above the ground
Physics
1 answer:
zzz [600]3 years ago
8 0

Two missing questions:

<em>"A) To achieve this vertical leap, with what force did the impala push down on the ground? </em>

<em> Express your answer to two significant figures and include the appropriate units. </em>

<em> B) What is the ratio of this force to the antelope's weight? </em>

<em>Express your answer using two significant figures."</em>

A) 1500 N

First of all, we have to calculate the velocity at which the impala jumps upward. This can be done by using the motion equation:

v^2-u^2 = 2ah

where

v = 0 is the velocity of the impala as it reaches the maximum height

u is the initial velocity of the impala

a = g = -9.8 m/s^2 is the acceleration of gravity

h = 2.5 m is the maximum height reached

Solving  for u,

u=\sqrt{-2ah}=\sqrt{-2(-9.8)(2.5)}=7 m/s

Now we know that the impulse on the impala is equal to its change in momentum, so we can write:

I=\Delta p \rightarrow F \Delta t = m \Delta v

where

F is the force exerted by the ground on the impala

\Delta t = 0.21 s is the duration of the push

m = 45 kg is the mass

\Delta v = 7 m/s is the gain in velocity of the impala

Solving for F,

F=\frac{m\Delta v}{\Delta t}=\frac{(45)(7)}{0.21}=1500 N

And according to Newton's third law, the force exerted by the impala on the ground is equal (and opposite) to the force exerted by the ground on the impala, so still 1500 N.

B) 3.4

The weight of the impala is given by

W=mg

where

m = 45 kg is its mass

g = 9.8 m/s^2 is the acceleration of gravity

Substituting,

W=(45)(9.8)=441 N

So the ratio of the force calculated at point A) to the weight of the impala is

r=\frac{F}{W}=\frac{1500}{441}=3.4

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The normal eye, myopic eye and old age
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Answer:

1)    f’₀ / f = 1.10, the relationship between the focal length (f'₀) and the distance to the retina (image) is given by the constructor's equation

2) the two diameters have the same order of magnitude and are very close to each other

Explanation:

You have some problems in the writing of your exercise, we will try to answer.

1) The equation to be used in geometric optics is the constructor equation

          \frac{1}{f} = \frac{1}{p} + \frac{1}{q}

where p and q are the distance to the object and the image, respectively, f is the focal length

* For the normal eye and with presbyopia

the object is at infinity (p = inf) and the image is on the retina (q = 15 mm = 1.5 cm)

        \frac{1}{f'_o} = 1/ inf + \frac{1}{1.5}

        f'₀ = 1.5 cm

this is the focal length for this type of eye

* Eye with myopia

the distance to the object is p = 15 cm the distance to the image that is on the retina is q = 1.5 cm

           1 / f = 1/15 + 1 / 1.5

           1 / f = 0.733

            f = 1.36 cm

this is the focal length for the myopic eye.

In general, the two focal lengths are related

         f’₀ / f = 1.5 / 1.36

         f’₀ / f = 1.10

The question of the relationship between the focal length (f'₀) and the distance to the retina (image) is given by the constructor's equation

2) For this second part we have a diffraction problem, the point diameter corresponds to the first zero of the diffraction pattern that is given by the expression for a linear slit

          a sin θ= m λ

the first zero occurs for m = 1, as the angles are very small

          tan θ = y / f = sin θ / cos θ

for some very small the cosine is 1

          sin θ = y / f

where f is the distance of the lens (eye)

           y / f = lam / a

in the case of the eye we have a circular slit, therefore the system must be solved in polar coordinates, giving a numerical factor

           y / f = 1.22 λ / D

           y = 1.22 λ f / D

where D is the diameter of the eye

          D = 2R₀

          D = 2 0.1

          D = 0.2 cm

           

the eye has its highest sensitivity for lam = 550 10⁻⁹ m (green light), let's use this wavelength for the calculation

         

* normal eye

the focal length of the normal eye can be accommodated to give a focus on the immobile retian, so let's use the constructor equation

      \frac{1}{f} = \frac{1}{p} + \frac{1}{q}

sustitute

       \frac{1}{f} = \frac{1}{25} + \frac{1}{1.5}

       \frac{1}{f}= 0.7066

        f = 1.415 cm

therefore the diffraction is

        y = 1.22  550 10⁻⁹  1.415  / 0.2

        y = 4.75 10⁻⁶ m

this is the radius, the diffraction diameter is

       d = 2y

       d_normal = 9.49 10⁻⁶ m

* myopic eye

In the statement they indicate that the distance to the object is p = 15 cm, the retina is at the same distance, it does not move, q = 1.5 cm

       \frac{1}{f} = \frac{1}{15} + \frac{1}{ 1.5}

        \frac{1}{f}= 0.733

         f = 1.36 cm

diffraction is

        y = 1.22 550 10-9 1.36 10-2 / 0.2 10--2

        y = 4.56 10-6 m

the diffraction diameter is

        d_myope = 2y

         d_myope = 9.16 10-6 m

         \frac{d_{normal}}{d_{myope}} = 9.49 /9.16

        \frac{d_{normal}}{d_{myope}} =  1.04

we can see that the two diameters have the same order of magnitude and are very close to each other

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A particle with a charge of − 5.10 nC is moving in a uniform magnetic field of B⃗ =−( 1.20 T )k^. The magnetic force on the part
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Answer:

Explanation:

Given that,

Charge q=-5.10nC

Magnetic field B= -1.2T k

And the magnetic force

F =−( 3.30×10−7N )i+( 7.60×10−7N )j

Let the velocity be V(xi + yj + zk)

Then, the force is given as

Note i×i=j×j×k×k=0

i×j=k. j×i=-k

j×k=i. k×j=-i

k×i=j. i×k=-j

F= q(v×B)

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q(xi + yj + zk) × -1.2k

−( 3.30×10−7N )i+( 7.60×10−7N )j=

q( -1.2x i×k - 1.2y j×k - 1.2z k×k)

−( 3.30×10−7N )i+( 7.60×10−7N )j=

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y= -3.3×10^-7/-1.2q

y= -3.3×10^-7/-1.2×-5.10×10^-9)

y=-53.92m/s

Let compare y-axisaxis

7.6×10−7N j = 1.2qx j

7.6×10−7N = 1.2qx

x= 7.6×10^-7/-1.2q

x= 7.6×10^-7/1.2×-5.10×10^-9)

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a. Then, the velocity of the x component is x= -124.18m/s

b. Also, the velocity component of the y axis is =-53.92m/s

c. We will compute

V•F

V=-124.18i -53.92j

F=−( 3.30×10−7 N )i+( 7.60×10−7 N )j

Note

i.j=j.i=0. Also i.i = j.j =1

V•F is

(-124.18i-53.92j)•−(3.30×10−7N)i+(7.60×10−7 N )j =

4.1×10^-5 - 4.1×10^-5=0

V•F=0

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Since the dot product is zero, from vectors , if the dot product of two vectors is zero, then the vectors are perpendicular to each other

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