You don't learn any movement concepts until high school. A. True B. False

Answers

Answer 1

Answer:

B

Explanation:

Answer 2

Answer:

false

Explanation:


Related Questions

The flight path of a jet aircraft as it takes off is defined by the parmetric equations x=1.25 t2 and y=0.03 t3, where t is the time after take-off, measured in seconds, and x and y are given in meters. If the plane starts to level off at t = 40 s, determine at this instant:a. The horizontal distance it is from the airport b. Its altitudec. Its speed.

Answers

Answer:

A. We are given t= 40s, x=1.25t²

So The horizontal distance it is from the airport will be = 1.25(40)²= 2000meters

B. y= 0.03t³

So y= 0.03(40)³= 1920m altitude

C. To find speed

We say

dx/dt=Vx= 2.5t= 100

dy/dy = 0.09t= 144

So speed= √100²+144²= 175.31m/s

(Application) You are asked to design a multiple lens system for the scenarios below. The lenses youhave available to you are:1. converging (f =7.5 cm)2. converging (f =15.0 cm)3. diverging (f = -15.0 cm)Think about the following scenarios:A) Lens 1 makes a real image i1, of which lens 2 makes a real image i2.B) Lens 1 makes a real image i1, of which diverging lens 2 makes a virtual image i2.C) Converging lens 1 makes a virtual image i1, of which converging lens 2 makes a virtual image i2.D) Converging lens 1 makes a virtual image i1, of which lens 2 makes a real image i2.For each scenario, decide:1) What type of lens is lens 1? Lens 2? Why? 2) Where should the object be placed with respect to the focal point to make the indicated type of image?3) Make a schematic drawing for Scenario D only. If you want extra practice, try to makeschematic drawings of the other three scenarios.

Answers

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

8. How did the measured angular magnification of the telescope compare with the theoretical prediction?

Answers

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

when a marble rolls down a slope which forces acts on it ​

Answers

Answer:

What forces act on a marble rolling down a ramp?

Answer: Gravity acts vertically downward. A normal force acts from the ruler toward the marble/ball in a direction that is perpendicular to the plane of the ruler. Friction acts in the direction opposite to which the marble/ball is moving. ... Friction slows down the marble/ball.

Calculate the maximum height to which water could be squirted with the hose if it emerges from the nozzle at 16.2 m/s.

Answers

Answer:

The maximum height to which water could be squirted with the hose is 13.380 meters.

Explanation:

A line of current of a fluid can be explained sufficiently by Bernoulli's Theorem. In this case, the system can be simplified due to neglectance of changes in absolute pressure. Water is squirted with an initial speed and reaches its maximum height, where final speed is zero. That is to say:

[tex]\frac{v_{1}^{2}}{2\cdot g} +z_{1} = \frac{v_{2}^{2}}{2\cdot g} +z_{2}[/tex]

Where:

[tex]z_{1}[/tex], [tex]z_{2}[/tex] - Initial and final height of water, measured in meters.

[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.

[tex]v_{1}[/tex], [tex]v_{2}[/tex] - Initial and final speed of water, measured in meters per second.

If [tex]z_{1} = 0\,m[/tex], [tex]v_{1} = 16.2\,\frac{m}{s}[/tex], [tex]v_{2} = 0\,\frac{m}{s}[/tex] and [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], then:

[tex]z_{2} = \frac{v_{1}^{2}-v_{2}^{2}}{2\cdot g} +z_{1}[/tex]

[tex]z_{2} = \frac{\left(16.2\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}}{2\cdot \left(9.807\,\frac{m}{s^{2}} \right)} +0\,m[/tex]

[tex]z_{2} = 13.380\,m[/tex]

The maximum height to which water could be squirted with the hose is 13.380 meters.

Why is the forward biased voltage for a typical red LED so much greater than the forward biased voltage for a typical silicon diode?

Answers

Answer:

This is because as we know the magnitude of a forward voltage drop depends on the material used for constructing the diode and also the magnitude of the band gap, so since LED material are generally gallium arsenide and gallium phosphide which has the higher band gap than silicon and since the higher the band gap the higher the voltage drop so LED has the greater forward voltage drop then the silicon diode

What do we measure sound intensity in?

Answers

Answer:

we measure sound intensity in Decibels.

Answer:

Decibels

Explanation:

We measure  sound power or sound pressure in decibels.

They  were named in honour of Alexander Graham Bell,( the inventor of both the telephone and the audiometer).

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

Answers

Answer:

Its not A..

Explanation:

That was my choice and it was incorrect

The displacement must be towards the spring, to compress it and cause it to force back.

The displacement must be away from the spring, to stretch it and cause it to pull back.

What is the spring constant?

The spring constant is used to define the stiffness of the spring, the greater the value of the spring constant stiffer the spring and it is more difficult to stretch the spring.

The spring's restoring force attempts to return it to its equilibrium position; as a result, the displacement must be away from the spring in order to stretch it and make it a drawback, and it must be toward the spring in order to compress it and force it back.

Thus, the correct answers are option B and option D.

Learn more about spring constant,

brainly.com/question/14670501

#SPJ2

g Explain completely, but in the fewest words possible, why a ball thrown horizontally follows the path that it does

Answers

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.

Why was miasma theory replaced?

Answers

In the mid of the 19th century the miasma theory was replaced by the germ theory of diseases (Maia 2013) The Greek physician Hippocrates (c.460- 377 B.C.E) believed that bad air could be the cause of any pestilences, the fatal epidemic.


Hope that helps!

A railroad train is traveling at a speed of 26.0 m/s in still air. The frequency of the note emitted by the locomotive whistle is 420 Hz.
1. What is the wavelength of the sound waves in front of the locomotive? Use 344 m/s for the speed of sound in air.
2. What is the wavelength of the sound waves behind the locomotive? Use 344 m/s for the speed of sound in air.
3. What is the frequency of the sound heard by a stationary listener in front of the locomotive? Use 344 m/s for the speed of sound in air.
4. What is the frequency of the sound heard by a stationary listener behind the locomotive? Use 344 m/s for the speed of sound in air.

Answers

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

Part B
Do the atoms in salt follow a pattern?

Answers

Answer: Yes.

Explanation: Because of the crystalline structure.

The voltage for the following cell is +0.731 V. Find Kb for the organic base RNH2. Use E_SCE = 0.241V.
Pt(s)│H2 (1.00 atm)│RNH2 (0.100M), RNH3 + (0.0500M)║SCE

Answers

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

When you attempt to swim, you brush your hands in a wide motion to push the water around you back behind you. This in turn, causes the water around you to push your forward. Which of Newton's Laws would explain this action? a. Newton's First Law b. Newton's Second Law c. Newton's Third Law d. None of Newton's Laws

Answers

Answer:

C. Newton's 3rd law

Explanation:

Swimmers must stroke downward in the water to stay afloat and propel forward. This movement is equal and opposite to the force the water exerts against the swimmer to stop them from moving.

Answer:

c. Newton's Third Law

Explanation:

I got it correct on the test

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