What is the total displacement of a bee that flies 2 meters east, 5 meters north, and 3 meters east? explain how you did it

Answers

Answer 1

Answer:

The displacement covered by the bee that flies 2 meters east, 3 meters east, and 5 meters north is 7.05 meters.

What is Displacement?

Displacement of an object can be defined as the actual change in the position of an object. It is a vector quantity because it has both the direction and the magnitude.

The bee travels 2 meter east and 3 meters east. Therefore, the total distance covered in the east direction is (2+3) meters = 5 meters.

The bee travels 5 meters in the north direction. This makes a square of 5 meters dimension and in this condition, the displacement covered by the bee will be the diagonal of the square. Therefore, the total displacement of the bee can be calculated as:

Diagonal of square = $√(2)$ a

where, a = side of the square

Side of the square = 5 meters

Diagonal of the square = $√(2)$ × 5

$√(2)$ = 1.41

Therefore, diagonal = 1.41 × 5

Displacement of the bee = 7.05 meters

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Answer 2

Answer: It flies 2+3meters east and 5 meters north so it flies 5 east and 5 north. Then the way of it would be diagonal of 5x5 square it equals 5√2 which is 7.07.

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(d) You observe someone pulling a block of mass 41 kg across a low-friction surface. While they pull a distance of 4 m in the direction of motion, the speed of the block changes from 5 m/s to 7 m/s. Calculate the magnitude of the force exerted by the person on the block.

Answers

Answer:

The magnitude of the force is 124.23 N.

Explanation:

To known the magnitude of the force is necessary to find the acceleration, that can be done by means of the equations for a a Uniformly Accelerated Motion:

$v_(f) = v_(i) + at$ (1)

Where $v_(f)$ is the final velocity, $v_(i)$ is the initial velocity, a is the acceleration and t is the time.

$a = (v_(f)-v_(i))/(t)$ (2)

$d = v_(i)t + (1)/(2)at^(2)$ (3)

By replacing (2) in equation (3) it is gotten:

$d = v_(i)t + (1)/(2)((v_(f)-v_(i))/(t))t^(2)$

$d = v_(i)t + (1)/(2)(v_(f)-v_(i))t$

$d = v_(i)t + (1)/(2)v_(f)t-(1)/(2)v_(i)t$

Therefore, by subtracting the first and third term ( $v_(i)t-(1)/(2)v_(i)t$ ⇒ $(1)/(2)v_(i)t$ ) it is got:

$d = (1)/(2)v_(f)t + (1)/(2)v_(i)t$

Applying common factor for $(1)/(2)t$ :

$d = (1)/(2)(v_(f) + v_(i))t$ (4)

Equation (4) can be rewritten in terms of t:

$t = (2d)/((v_(f) + v_(i)))$

$t = ((2)(4m))/((7m/s + 5m/s))$

$t = (8m)/(12m/s)$

$t = 0.66s$

By knowing the time is possible to determine the acceleration by means of equation (2):

$a = (7m/s-5m/s)/(0.66s)$

$a = (2m/s)/(0.66s)$

$a = 3.03m/s^(2)$

The magnitude of the force exerted by the person on the block can be determine by means of Newton's second law:

$F = ma$ (5)

$F = (41Kg)(3.03m/s^(2))$

$F = 124.23 Kg.m/s^(2)$

$F = 124.23 N$

Hence, the magnitude of the force is 124.23 N.

A glass lens, n glass = 1.52 , has a 131 nm thick antireflective film coating one side, n film = 1.25. White light, moving through the air, is perpendicularly incident on the coated side of the lens. What is the largest wavelength of the reflected light that is totally removed by the coating?

Answers

Answer:

655 nm

Explanation:

When the intereference is destructive then the thickness, d of antireflective film coating one side is given by

2d=w/2n

Where w is wavelength and n is the reflective index of the film

Making w the subject of formula then

w=4nd

Substituting 1.25 for n and 131 nm for d then the wavelength will be

w=4*1.25*131=655 nm

Therefore, the wavelength is equivalent to 655 nm

The formula for calculating the wavelength in an antireflective film involves thickness (d) and refractive index (n). For n = 1.25 and d = 131 nm, the resulting wavelength is 655 nm.

When light waves encounter a thin film, some of the waves are reflected from the top surface of the film, and some pass through it. These waves can interfere with each other, leading to constructive or destructive interference. In the case of antireflective coatings, destructive interference is desired to minimize reflection.

The formula you mentioned is used to calculate the thickness (d) of an antireflective film that results in destructive interference for a specific wavelength (w) of light. The formula is:

2d = w / (2n)

Where:

d is the thickness of the film.

w is the wavelength of light.

n is the refractive index of the film.

To find the wavelength (w) when given the thickness (d) and refractive index (n), you can rearrange the formula:

w = 4 * n * d

Now, let's calculate the wavelength using the provided values:

n = 1.25 (refractive index)

d = 131 nm (thickness in nanometers)

Substitute these values into the formula:

w = 4 * 1.25 * 131 = 655 nm

Therefore, the calculated wavelength (w) is 655 nanometers (nm). This means that for a film with a refractive index of 1.25 and a thickness of 131 nm, destructive interference occurs at a wavelength of 655 nm.

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Which particle builds a static electric charge when it is transferred from one object to another? A.proton

B.atom

C.electron

D.molecule

Answers

c, because of its electrical charge

Tech A says that a battery stores electrical energy in chemical form. Tech B says that a battery creates direct current. Who is correct?

Answers

Answer:

Battery stores the electrical energy in the chemical form. When we connect it's positive and negative terminals and provide the path for the flow of electrons then it converts that chemical energy into the electrical energy and hence the direct current is created. So both of the technicians are right.

Explanation:

Which scientist first made a model of the atom called the plum pudding model?John Dalton
J.J. Thomson
Ernest Rutherford
Neils Bohr

Answers

The plum pudding model of theatom by the scientist Sir Joseph John Thompson or J.J. Thompson. It wastheorized after the discovery that electrons are present in the atom. In thismodel, the atom is composed mainly of electrons. It was describe as plum puddingbecause the electrons are dispersed within the atom. John Dalton discovered theatom. Ernest Rutherford discovered the half – life of an atom and Neils Bohrexplain the quantum mechanics. So the answer is J.J. Thompson.

How much amount of heat is required to raise the temperature of 20 grams of water from 10°C to 30°C? The specific heat of water is 4.18 J/g°C.