You move a box 5 meters and perform 900 joules of work. How much force did you apply to the box?

When a honeybee flies through the air, it develops a charge of +20 pC . Part A How many electrons did it lose in the process of

Yuri [45]

Answer:

1.3 × 10⁸ e⁻

Explanation:

When a honeybee flies through the air, it develops a charge of +20 pC = + 20 × 10⁻¹² C. This is a consequence of losing electrons (negative charges). The charge of 1 mole of electrons is 96468 C (Faraday's constant). The moles of electrons representing 20 pC are:

20 × 10⁻¹² C × (1 mol e⁻/ 96468 C) = 2.1 × 10⁻¹⁶ mol e⁻

1 mole of electrons has 6.02 × 10²³ electrons (Avogadro's number). The electrons is 2.1 × 10⁻¹⁶ moles of electrons are:

2.1 × 10⁻¹⁶ mol e⁻ × (6.02 × 10²³ e⁻/ 1 mol e⁻) = 1.3 × 10⁸ e⁻

7 0

3 years ago

Student Exploration: Nuclear Decay. Has anyone done a Gizmos lab on this?

LUCKY_DIMON [66]

Answer:NOOPE

Explanation:IM THE MYSTERY MAN WHOOOOSHHHHHHHHHHHHHHH ???????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????

6 0

2 years ago

A screen is placed 1.20m behind a single slit. The central maximum in the resulting diffraction pattern on the screen is 1.40cm

andrew11 [14]

Answer:

2.8 cm

Explanation:

$y_1$ = Separation between two first order diffraction minima = 1.4 cm

D = Distance of screen = 1.2 m

m = Order

Fringe width is given by

$\beta_1=\dfrac{y_1}{2}\\\Rightarrow \beta_1=\dfrac{1.4}{2}\\\Rightarrow \beta_1=0.7\ cm$

Fringe width is also given by

$\beta_1=\dfrac{m_1\lambda D}{d}\\\Rightarrow d=\dfrac{m_1\lambda D}{\beta_1}$

For second order

$\beta_2=\dfrac{m_2\lambda D}{d}\\\Rightarrow \beta_2=\dfrac{m_2\lambda D}{\dfrac{m_1\lambda D}{\beta_1}}\\\Rightarrow \beta_2=\dfrac{m_2}{m_1}\beta_1$

Distance between two second order minima is given by

$y_2=2\beta_2$

$\\\Rightarrow y_2=2\dfrac{m_2}{m_1}\beta_1\\\Rightarrow y_2=2\dfrac{2}{1}\times 0.7\\\Rightarrow y_2=2.8\ cm$

The distance between the two second order minima is 2.8 cm

8 0

2 years ago

An aircraft has a liftoff speed of 33 m/s. What is the minimum constant acceleration an

JulsSmile [24]

Answer: The minimum acceleration for the air plane is 2.269m/s2.

Explanation: To solve such problem the equation of motion are applicable.

The initial velocity is 0 since the airplane was initially standing. We are going to use this equation

V^2=U^2+2as

33^2=0+2a (240)

a= 2.269m/s2

5 0

2 years ago

A 5.10 kgkg watermelon is dropped from rest from the roof of a 18.5 mm-tall building and feels no appreciable air resistance.

VARVARA [1.3K]

Work done is by the change in the potential energy of the system. The work done by gravity is 924.63 J.

<h3>What is the Kinetic Energy?</h3>

Potential energy in physics is the energy that an item retains as a result of its position in relation to other objects, internal tensions, electric charge, or other elements.
The gravitational potential energy of an object, which is based on its mass and distance from another object's center of mass, the elastic potential energy of an extended spring, and the electric potential energy of an electric charge in an electric field are examples of common types of potential energy. The joule, denoted by the letter J, is the energy unit in the International System of Units (SI).

Solution:

mass = 5.10 kg

height = 18.5 mm

We know that work done by the gravity on the watermelon is the change in the potential energy of the watermelon, therefore,

Work done due to gravity = change in the potential energy of the system

W = $\Delta PE$

W = mg (h₀ - h₁)

W = 5.10 × 9.8 × 18.5

W = 924.63 J

know more about potential energy brainly.com/question/24284560

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7 0

1 year ago