What is an example of a sting nuclear force ?

Match each action to its result. The strength of the electromagnet increases. The electromagnet turns off. The poles of the elec

IceJOKER [234]

Answer:

1) The strength of the electromagnet increases → Place a magnetic core inside the coil of wire

2) The electromagnet turns off → Turn off the battery supply

3) The poles of the electromagnet reverse → Change the direction in which the current flows

Explanation:

when current passes through a coil it behaves a an electromagnet.

Magnetic field strength is given by

B = μ N I

N is no of turns and

I is the current through coil

μ is permeability of the medium or core in the coil.

1). Magnetic core increase permeability μ so it will strengthen magnetic field:

B = <u>μ</u> N I

2). When the battery turns off current becomes zeroi.e I=0

So B = μ N * 0

⇒ B = 0

so electromagnet turns off

3). Direction of magnetic field can be determine by right hand rule, i.e curl the fingers in the direction of current, thumb will point in the direction of north pole.

so changing current direction will change direction of magnetic field.

5 0

3 years ago

Estimate humanity’s current consumption of energy in one day (in Joule).

Alex73 [517]

Based on data from world energy consumption, the estimated humanity’s current consumption of energy in one day is above 208 million joules per day.

<h3>What is energy consumption?</h3>

Energy is the ability to do work. Therefore, energy consumption refers to the amount of energy used to do work.

Humans need energy for various purposes such as;

lighting
heating
operation of machines
feeding, etc.

Due to rapid growth on population as well as technology, the human energy consumption is on the increase.

In 2020, the world primary energy consumption was 71,4 GJ per person for a world population of about 7.7 billion people.

Therefore, the estimated humanity’s current consumption of energy in one day is above 208 million joules per day.

Learn more about energy at: brainly.com/question/6572692

6 0

2 years ago

A 25 kg child is riding on a swing. If the child travels 8.9 m/s at the bottom of their swing, how high into the air is the chil

Setler [38]

Answer:

h = 4.04 m

Explanation:

Given that,

Mass of a child, m = 25 kg

The speed of the child at the bottom of the swing is 8.9 m/s

We need to find the height in the air is the child is able to swing. Let the height is h. Using the conservation of energy such that,

$mgh=\dfrac{1}{2}mv^2\\\\h=\dfrac{v^2}{2g}$

Put all the values,

$h=\dfrac{(8.9)^2}{2\times 9.8}\\\\h=4.04\ m$

So, the child is able to go at a height of 4.04 m.

7 0

3 years ago

You illuminate the grating in a spectrometer at normal incidence θi=0° with a beam of light that has a wavelength of 6562.8 Å. T

monitta

Answer:

a) θ₁ = 23.14 ° , b) θ₂ = 51.81 °

Explanation:

An address network is described by the expression

d sin θ = m λ

Where is the distance between lines, λ is the wavelength and m is the order of the spectrum

The distance between one lines, we can find used a rule of proportions

d = 1/600

d = 1.67 10⁻³ mm

d = 1-67 10⁻³ m

Let's calculate the angle

sin θ = m λ / d

θ = sin⁻¹ (m λ / d)

First order

θ₁ = sin⁻¹ (1 6.5628 10⁻⁷ / 1.67 10⁻⁶)

θ₁ = sin⁻¹ (3.93 10⁻¹)

θ₁ = 23.14 °

Second order

θ₂ = sin⁻¹ (2 6.5628 10⁻⁷ / 1.67 10⁻⁶)

θ₂ = sin⁻¹ (0.786)

θ₂ = 51.81 °

3 0

3 years ago

A planet's moon travels in an approximately circular orbit of radius 7.0 ✕ 10⁷ m with a period of 6 h 38 min. Calculate the mass

natka813 [3]

Answer:

3.56×10²⁶ Kg.

Explanation:

Note: The gravitational force is acting as the centripetal force.

Fg = Fc........................... Equation 1

Where Fg = gravitational Force, Fc = centripetal force.

Recall,

Fg = GMm/r²......................... Equation 2

Fc = mv²/r............................. Equation 3

Where M = mass of the planet, m = mass of the moon, r = radius of the orbit and G = Universal gravitational constant.

Substituting equation 2 and 3 into equation 1

GMm/r² = mv²/r

Simplifying the equation above,

M = v²r/G .............................. Equation 4.

The period of the moon in the orbit

T = 2πr/v

Making v the subject of the equation,

v = 2πr/T............................. Equation 5

where r = 7.0×10⁷ m, T = 6 h 38 min = (6×3600 + 38×60) s = (21600+2280) s

T = 23880 s, π = 3.14

v = (2×3.14×7.0×10⁷ )/23880

v = 18409 m/s

Also Given: G = 6.67×10⁻¹¹ Nm²/kg²

Also substituting into equation 4

M = 18409²×7.0×10⁷ /(6.67×10⁻¹¹)

M = 3.56×10²⁶ Kg.

Thus the mass of the planet = 3.56×10²⁶ Kg.

5 0

3 years ago