According to Hookes Law, in order for a spring to apply a restorative force, what must be true about the displacement of the spring? a. The displacement must be proportional to the spring constant b. The displacement must be towards the spring, to compress it and cause it to force back c. The displacement must be measured away from a starting rest position in order to have a force from the spring d. The displacement must be away from the spring, to stretch it and cause it to pull back

Answer:

Explanation:

A ball thrown horizontally is a type of projectile which moves in a trajectory path. It moves with respect to its momentum and gravitational pull of the earth.

The motion of the thrown ball can be described in two components, horizontal and vertical. In horizontal component of motion of the thrown ball, gravitational force is zero. While in the case of vertical component, the gravitational pull causes the path followed.

The force of gravity pulls the ball towards the surface of the earth gradually, thus forming a trajectory path.

Answer:  Kb = 1.89 × 10⁻⁶

Explanation:

R-NH₃+ (aq) <----------> R-NH₂ (aq) + H^+ (aq)

Ecell = - ( 0.0592 / n ) log Kₐ

Where, Ecell = 0.731 - 0.241 = 0.490 V

Therefore, 0.490 = - (0.0592 / 1 ) log Kₐ

Therefore, Kₐ = 5.30 × 10⁻⁹

Thus, Kb = Kw / Ka = ( 1.0 × 10⁻¹⁴ / 5.30 × 10⁹ ) = 1.89 × 10⁻⁶

Complete Question

The complete question is shown on the first uploaded image  

Answer:

The theoretical angular magnification lies within the angular magnification range

Explanation:

From the question we are told that

   The  focal length of  B  is  [tex]f_{objective } = 43.0 \ cm[/tex]

    The focal length of  A  is   [tex]f_{eye} = 10.4 \ cm[/tex]

The  theoretical angular  magnification is mathematically represented as

           [tex]m = \frac{f_{objective }}{f_{eye}} = \frac{43.0}{10.4}[/tex]

            [tex]m = \frac{f_{objective }}{f_{eye}} = 4.175[/tex]

Form the question the measured angular magnification ranges from 4 -5

So from the value calculated and the value given we can deduce that the theoretical angular  magnification lies within the angular magnification range

Answer: 0.757m; 0.881m; 432.70Hz; 371.89Hz

Explanation:

Give the following :

Velocity of train (Vt) = 26m/s

Frequency of sound (Fs) = 420Hz

Speed of sound (Vs) = 344m/s

1) wavelength = (Vs - Vt) / Fs

Wavelength = (344 - 26) / 420 = 318/420 = 0.757m

11) Wavelength = (Vs + Vt) / Fs

Wavelength = (344 + 26) / 420 = 370/420 = 0.881m

111) According to the doppler effect :

Fl = [(V + Vl) / (V + Vs)] * fs

Fl = frequency of listener ; fs = frequency of sound source ; V = speed of sound ; Vl = Velocity of listener ; Vs = speed of sound source

Vs = - ve (train moving towards listener)

Fl = [(V + Vl) / (V + Vs)] * fs

Fl = [(344 + 0) / (344 - 26)] * 400

Fl = (344 / 318) * 400 = 432.70Hz

1V) Vs = + ve (train moving away listener)

Fl = [(V + Vl) / (V + Vs)] * fs

Fl = [(344 + 0) / (344 + 26)] * 400

Fl = (344 / 370) * 400 = 371.89Hz

Answer:

1) A) Lens converging, B) lens 1 converging

2) A and B object is at a greater distance than its focal

3)

Explanation:

For this exercise we will give some more important characteristics of the lenses.

we use the equation of the constructor

         1 / f = 1 / p + 1 / q

where f is the focal length and p and qlas the distance to the object and image, respectively.

We must also use how the lenses are combined

       1 / = 1 / f₁ + 1 / f₂

where feq is the equivalent focal length and it is assumed that the two lenses are in contact.

By applying the equation of the constructor you get some characteristics of the image

Converging lens

* Object farther than focal length real and inverted image

* Oject after focal virtual image and right

Diverging lens

* for all virtual image distance and right

with this information let's analyze the different configurations

1) let's do an analysis for each scenario

A) since lens 1 and lens 2 create real image they must be converging lenses

B) Lens 1 must be convergent and lens 2 can be of both types

C) Lens 1 can be convergent with the object that is at a shorter distance than at focal point (p <f₁). The same analysis diverges for lens 2, the object that is the image of the other lens is shorter than its focal distance p₂ <f₂

D) the object is at a shorter distance than the focal length p₁ <f₁, the object for this lens is at a greater distance than its focal point p₂> f₂

2) let's analyze for each scenario

A) for each lens the object is at a greater distance than its focal length p₁> f₁ and p₂> f₂

B) for lens 1 p₁> f₁

for lens 2

if it is convergent p₂ <f₂

and if it is divergent any distance is possible

C) lens 1 p₁ <f₁ and for lens 2 p₂ <f₂

D) lens 1 p₁ <f₁ and for lens 2 p₂> f₂

3) see attached

The distance with subscript 1 is measured with respect to line1 and the distance with subscript 2 is measured with respect to the lens with subscript 2