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adoni [48]
3 years ago
12

Taylor places a nail on a bar magnet. The nail sticks to the magnet when lifted up off the table. She touches a paperclip to the

nail and it sticks to the nail. Explain what happened to the magnetic domains of the nail before and after touching it to the bar magnet
Physics
1 answer:
Bas_tet [7]3 years ago
3 0

Answer:

When touching the bar magnet ,the nail gets attached to the magnet from its metallic field is used to connect when taylor touched the nail to the bar magnet,the magnetic fields were ranged,and made a temporary magnet.

Explanation:

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What are the periodic variations in Earth's rotation and orbit around the sun that alter the way solar radiation is distributed
Dahasolnce [82]

Answer:

1. The precession of the equinoxes.

2. Changes in the tilt angle of Earth’s rotational axis relative to the plane of Earth’s orbit around the Sun.

3. Variations in the eccentricity

Explanation:

These variations listed above;  the precession of the equinoxes (refers, changes in the timing of the seasons of summer and winter), this occurs on  a roughly about 26,000-year interval; changes in the tilt angle of Earth’s rotational axis relative to the plane of Earth’s orbit around the Sun, this occurs roughly in a 41,000-year interval; and changes in the eccentricity (that is a departure from a perfect circle) of Earth’s orbit around the Sun, occurring on a roughly 100,000-year timescale. which influences the mean annual solar radiation at the top of Earth’s atmosphere.

5 0
3 years ago
A hobby rocket reaches a height of 72.3 m and lands 111 m from the launch point with no air resistance. What was the angle of la
Deffense [45]

Answer:

The angle of launch is 52.49 Degree.

Explanation:

The Range R and Height H of a thrown object is calculated using the formula,

R=V₀² sin(2φ)/g

H=V₀²sin²(φ)/g

From these equations it can be written,

V₀²=R g/ sin(2φ)

V₀²=H g/ sin²(φ)

These values are equal so it can be written by equating these equations,

R g/sin(2φ)=H g/sin²(φ)

tan(φ)= 2H/R

Given H=72.3 m and R=111 m, the angle of launch is,

tan(φ)= 2*72.3/111

φ= 52.49 Degree.

Check out other solutions,

brainly.com/question/1495042

#SPJ10

7 0
2 years ago
if you apply a Force of F1 to area A1 on one side of a hydraulic jack, and the second side of the jack has an area that is twice
Firlakuza [10]

Answer:

2F_{1}

Explanation:

F₁ = Force on one side of the jack

A₁ = Area of cross-section of one side of the jack

F₂ = Force on second side of the jack

A₂ = Area of cross-section of second side of the jack = 2 A₁

Using pascal's law

\frac{F_{1}}{A_{1}}= \frac{F_{_{2}}}{A_{_{2}}}

\frac{F_{1}}{A_{1}}= \frac{F_{_{2}}}{2A_{_{1}}}

F_{1}= \frac{F_{2}}{2}\\

F_{2}= 2F_{1}

7 0
4 years ago
Please need some help on this thank you so much
fomenos
Weight is based on density mass is not.
8 0
4 years ago
18) A charm quark has a charge of approximately
Allushta [10]

Based on scientific records, a charm quark has a charge that's approximately equal to: 2) 1.07 × 10⁻¹⁹ C.

<h3>What is a charge?</h3>

A charge simply refers to a fundamental, physical property of matter that governs how the particles of a substance are affected by an electromagnetic field, especially due to the presence of an electrostatic force (F).

Also, charge is typically measured in Coulombs and a charm quark (elementary particle) has a charge that's approximately equal to 1.07 × 10⁻¹⁹ Coulomb.

Read more on charges here: brainly.com/question/4313738

#SPJ1

3 0
2 years ago
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