In this video, we see that a _____ magnetic field can create an electric current. In this video, we see that a _____ magnetic field can create an electric current. Changing Perpendicular Strong Dipole

Explanation:

momentum = mass x velocity

initial momentum = 100 x 15 = 1500kgm/s

after momentum = 100 x 20 = 2000kgm/s

a =(v-u)/t

a = (20-15)/10

a = 5/10

a = 0.5m/s²

f = ma

f = 100 x 0.5

f = 50N

Question

The large-scale distribution of galaxies in the universe reveals

A) a smooth, continuous, and homogenous arrangement of clusters

B) large voids, with most of the galaxies lying in filaments and sheets a large supercluster at the center of the universe

C) a central void with walls of galaxies at the edge of the universe

Group of answer choices

Answer:

The correct answer is B)

Explanation:

The universe is arranged in a filamentary structure. Filamentary structures are very large. They are the largest kind of structures in the universe and comprise mostly of galaxies that are held together by gravity.

The structures found within Galaxy filaments have thread-like qualities spanning 52 to 78.7 megaparsecs h⁻¹ in lenght.

Other phenomena associated with the nature fo the universe is the existence of void spaces.

Cheers!

Answer:

20 m

Explanation:

Given:

v = 0 m/s

a = -10 m/s²

t = 4 s / 2 = 2 s

Find: Δy

Δy = vt − ½ at²

Δy = (0 m/s) (2 s) − ½ (-10 m/s²) (2 s)²

Δy = 20 m

Answer:

Explanation:

Given,

Force ( f ) = 30 N

Acceleration(a) = 7.6 m/s

Now, Let's find the mass of the ball

Using the Newton's second law of motion:

We get:

[tex]force \: = mass \: \times acceleration[/tex]

plug the value

[tex]30 \: = m \: \times 7.6[/tex]

Use the commutative property to reorder the terms

[tex] 30 = 7.6 \: m[/tex]

Swap the sides of the equation

[tex]7.6m = 30[/tex]

Divide both sides of the equation by 7.6

[tex] \frac{7.6 \: m}{7.6} = \frac{30}{7.6} [/tex]

Calculate

[tex]m = 3.94 \: kg[/tex]

Hope this helps..

Best regards!!

Answer:

[tex]\displaystyle \boxed{\mathrm{3.95 \: kg }}[/tex]

Explanation:

[tex]\mathrm{force \: (N) = mass \: (kg) \times acceleration \: (m/s^2)}[/tex]

[tex]\mathrm{force = 30N}[/tex]

[tex]\mathrm{acceleration = 7.6 \: m/s^2 }[/tex]

[tex]\mathrm{Find \: the \: mass.}[/tex]

[tex]\mathrm{30 = m \times 7.6}[/tex]

[tex]\displaystyle \mathrm{m =\frac{30}{7.6} }[/tex]

[tex]\displaystyle \mathrm{m = 3.947... }[/tex]

I would suggest an easier route. So for example, I would research light energy, where it comes from, and what causes it. I won't be able to do the work for you though.

Answer:

0.12 kW

Explanation:

Given that

The flow rate of air (V)=0.3 ft³/s

V=0.008 m³/s

Pressure, P=14.7 psia

P=1.013529 atm=101.325  kPa

Inlet temperature = 70° F=294.261 K

Exit temperature = 83° F=301.483 K

We know that , specific heat capacity of the air

Cp=1.005 kJ/kg.K

The mass flow rate of air is given as

[tex]\dot{m}=\dfrac{P\times V}{R\times T}\\\dot{m}=\dfrac{101.325\times 0.008}{0.287\times 294.261}\\\dot{m}= 0.0095\ kg/s[/tex]

By using energy conservation

[tex]Electric\ power =\dot{m}\times C_p\times (T_2-T_1)\\Electric\ power =0.0095\times 1.005\times (83-70)=0.12\ kW[/tex]

Therefore electric power dissipate by components will be 0.12 kW.

Answer:

A. 119kg

B.0.53m from head

Explanation:

A. Weight = 730+340.

=1170N

F= Wg

W = 1170/9.8

= 119kg

If x is distance from head to CG then 1.92–x is the other distance.

Moments must equal

470x = 330(1.89–x)

470x = 623.7 – 330x

1170x = 623.7

x = 0.53m from head

Answer:

the answer B

Explanation:

Answer:

Infrasound vs. Ultrasound: Infrasound is sound that is below the lower limit of human hearing, below 20 Hz, and ultrasound is above the upper limit of human hearing, above 20,000 Hz.  Individuals use infrasound - this recurrence run for checking seismic tremors and volcanoes, graphing rock and oil developments underneath the earth. Infrasound is described by a capacity to get around hindrances with little scattering.  

For instance, a few creatures, for example, whales, elephants and giraffes convey utilizing infrasound over significant distances. Torrential slides, volcanoes, seismic tremors, sea waves, water falls and meteors produce infrasonic waves. Symptomatic ultrasound, additionally called sonography or demonstrative clinical sonography, is an imaging technique that utilizes high-recurrence sound waves to create pictures of structures inside your body. The pictures can give important data to diagnosing and treating an assortment of ailments and conditions.

Explanation:

idk how many words this is but its a start for u to add on to and i hope this helps and its in my own words - pls mark me brainiest

Answer:

Ultrasound vs. Infrasound Research Exploration

Beyond the limit of human hearing, ultrasound is above 20,000 Hz. Under the limit of human hearing, infrasound is below 20 Hz. Individuals use infrasound - this recurrence run for checking seismic tremors and volcanoes as well as graphing rock and oil developments beneath the earth. Infrasound is described by a capacity to get around hindrances with little scattering.  

For instance, creatures like whales, elephants and giraffes convey utilizing infrasound over significant distances. Torrential slides, volcanoes, seismic tremors, sea waves, waterfalls and meteors produce infrasonic waves. Symptomatic ultrasound, additionally called sonography or demonstrative clinical sonography, is an imaging technique that utilizes high-recurrence sound waves to create pictures of structures inside your body. The pictures can give important data to diagnosing and treating an assortment of ailments and conditions.

(Not turned in yet, but this is what i have so far. Good luck 8th graders <33)

-Sav xx

Answer:

A car driving in a straight line 20 m/s

Explanation:

ayepecks silly

ANSWER:

0.4081%

Explanation:

Difference=24.6-24.5=0.1

Relative error = 0.1/24.5*100=0.4081%

Relative error is equal to the = difference between both the values/The true value *100

Explanation:

It is given that,

Object distance from the mirror, u = -30 cm

Focal length of the mirror, f = +15 cm

Size of the object, h = 7.5 cm

We need to find the image distance and the size of the image.

Mirror's formula, [tex]\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}[/tex]

v is image distance

[tex]\dfrac{1}{v}=\dfrac{1}{f}-\dfrac{1}{u}\\\\\dfrac{1}{v}=\dfrac{1}{(15)}-\dfrac{1}{(-30)}\\\\v=10\ cm[/tex]

Let h' is the size of the image. So,

[tex]\dfrac{h'}{h}=\dfrac{-v}{u}\\\\h'=\dfrac{-vh}{u}\\\\h'=\dfrac{-10\times 7.5}{-30}\\\\h'=2.5\ cm[/tex]

So, the image is located at a distance of 10 cm and the size of the image is 2.5 cm.

Answer:

we need it to work and without it we dont have strength to do anything

Answer:

energy is important to all living organisms. energy for producers comes from the sun, and energy for consumers comes from other living organisms. the abundance of energy available for organisms impacts the population.

Explanation:

The age of the universe is 13.8 billion years.

We know that,

1 billion years = 10⁹ years

So,

[tex]13.8\ \text{billion years}=13.8\times 10^9\ \text{years}[/tex]

We need to convert the age of the universe to the scientific notation.

Since, [tex]1\ \text{year}=3.154\times 10^7\ s[/tex]

So,

[tex]13.8\times 10^9\ \text{years}=13.8\times 10^9\times 3.154 \times 10^7\\\\=4.35\times 10^{17}\ s[/tex]

So, the age of the universe is [tex]4.35\times 10^{17}\ s[/tex].

Answer:

So the universe is approximately 1.34 x 10^10 years old

Explanation:

Edmentum

Answer:

        v = √k/m    x

Explanation:

We can solve this exercise using the energy conservation relationships

starting point. Fully compressed spring

        Em₀ = [tex]K_{e}[/tex] = ½ k x²

final point. Cart after leaving the spring

        [tex]Em_{f}[/tex] = K = ½ m v²

        Em₀ = Em_{f}

        ½ k x² = ½ m v²

        v = √k/m    x

Answer:

The answer is "Observe it in the infrared"

Explanation:

The infrared observatory is a device, which helps to track but also fix thermal light from sources out and the front of Earth's atmosphere like nebulae, young stars, and dust and gas in several other galaxies.

The very center of our galaxy is like a star cluster is situated in the central of our universe across the Milky Way supermassive, and the central black hole, around 4 million times our sun's mass.

The main super black hole named the Sagittarius A-star has been located in the nucleus of the galaxy, that's why in this question "Observe it in the infrared" is the correct answer.

Answer:

1) a    α,  m   I,  W=F.d    W =τ . θ,

2)  a = v²/r

Explanation:

1) The amounts of rotational and translational motion are related

acceleration is

        a = d²x / dt²

    linear displacement is equivalent to angular rotation, therefore angular acceleration is

      α = d²θ / dt²

force in linear motion is equivalent to moment in endowment motion

       F = m a

       τ = I α

the mass is the inertia of the translation, in rotational motion the moment of inertia is the rotational inertia

          I = m r²

Work is defined by W = F. d

in rotation it is defined by W = τ . θ

The linear momentum is p = mv

the angular momentum L = I w

momentum the linear motion is I = F dt

in the rotation it is I = τ dt

2) The velocity is a vector therefore it has modulus and direction, linear acceleration changes the modulus of velocity, whereas circular motion changes the direction (the other element of the vector).

      [tex]a_{c}[/tex]Ac = v²/r

Answer:

3.94

Explanation:

divide total mass by the number of blocks since they are identical

Answer:

3.94

Explanation:

You want to find the mass of one block. Since we know there is 8 blocks with the same mass, you can divide the total mass by 8 since the mass is equally distributed within the 8 blocks

Answer:

C: You must know force and distance to calculate both.

Explanation:

A force is said to have done work, when it has succeeded in causing displacement in an object from its starting point.

Power can be defined as the rate of doing work.

The work can be calculated using the following formula:

Work done = Force × Displacement

The power can be calculated using the following formula:

Power = Work \ Time

In order to calculate the work and power, the force and distance should be known. The force and distance will help in the calculating the work, which will in turn help in the calculation of the power of the object.

Hence, the answer is you must know force and distance to calculate both.

Answer:

B. 122N

Explanation:

f = 0.15 x 83 x 9.8 = 122.01

f = 122N

Explanation:

let us use the explanation below to get the intuition so desired;

According to Faraday's law of electro magnetic induction, when ever a coil/conductor is made to rotate in a magnetic field, voltage or emf is created and current is produced, in the long run energy has be produced or converted.

The conversion of this energy is made possible by the motion of the coil/conductor is the magnetic field, just by the  motion of the conductor cutting through the magnetic field, thus creating electro motive force(E.M.F) hence producing current, and ultimately energy is created

Answer:

1.92 x 10⁻¹²J

Explanation:

The magnetic force from the magnetic field gives the circulating protons gives the particle the necessary centripetal acceleration to keep it orbiting round the circular path. And from Newton's second law of motion, the force(F) is equal to the product of the mass(m) of the proton and the centripetal acceleration(a). i.e

F = ma

Where;

a = [tex]\frac{v^2}{r}[/tex]             [v = linear velocity, r = radius of circular path]

=> F = m[tex]\frac{v^2}{r}[/tex]           ------------(i)

We also know that the magnitude of this magnetic force experienced by the moving charge (proton) in a magnetic field is given by;

F = q v B sin θ       ----------(ii)

Where;

q = charge of the particle

v = velocity of the particle

B = magnetic field

θ = the angle between the velocity and the magnetic field.

Combining equations (i) and (ii) gives

m[tex]\frac{v^2}{r}[/tex] = q v B sin θ           [θ = 90° since the proton is orbiting at the maximum orbital radius]

=> m[tex]\frac{v^2}{r}[/tex] = q v B sin 90°

=> m[tex]\frac{v^2}{r}[/tex] = q v B

Divide both side by v;

=> m[tex]\frac{v}{r}[/tex] = qB

Make v subject of the formula

v = [tex]\frac{qBr}{m}[/tex]

From the question;

B = 1.25T

m = mass of proton = 1.67 x 10⁻²⁷kg

r = 0.40m

q = charge of a proton = 1.6 x 10⁻¹⁹C

Substitute these values into equation(iii) as follows;

v = [tex]\frac{(1.6*10^{-19})(1.25)(0.4)}{(1.67*10^{-27})}[/tex]

v = 4.79 x 10⁷m/s

Now, the kinetic energy, K, is given by;

K = [tex]\frac{1}{2}[/tex]mv²

m = mass of proton

v = velocity of the proton as calculated above

K = [tex]\frac{1}{2}(1.67*10^{-27} * (4.79 * 10^7)^2 )[/tex]

K = 1.92 x 10⁻¹²J

The kinetic energy is 1.92 x 10⁻¹²J

Answer:

50.176m ; 31.36m/s ; 31.36m/s

Explanation:

Given the following :

Time (t) = 3.2s

Height(s) of the building

The initial velocity (u) will be 0

Using the equation:

S = ut + 0.5at^2

Acceleration due to gravity (a) = g = +9.8m/s^2 (downward)

S = 0*t + 0.5(9.8)(3.2^2)

S = 0 + 50.176

S = 50.176m

B.) speed impact when it touches the ground:

v^2 = u^2 + 2aS

where v is the final Velocity

v^2 = 0^2 + 2(9.8)(50.176)

v^2 = 0 + 983.4496

v = √983.4496

v = 31.36m/s

C) Speed of throw from the ground to reach the top of the building

Here we need the initial velocity

Height (s) = 50.176m

Acceleration due to gravity g = a = - 9.8m/s ( upward)

Using the third equation of motion:

S = ut + 0.5at^2

50.176 = u*3.2 + 0.5(-9.8)(3.2^2)

50.176 = 3.2u - 50.176

50.176 + 50.176 = 3.2u

100.352 = 3.2u

u = 100.352 / 3.2

u = 31.36m/s

Answer:

The total frictional force equals 60 N

Explanation:

We know that F - f = ma where F is the applied force, f is the frictional force, m the mass of the object and a its acceleration.

Now, since the cart moves with constant velocity, its acceleration is zero. So, a = 0.

So F - f = ma

F - f = m(0) = 0

F - f = 0

So, F = f

From the above, we see that the frictional force equals the applied force which is equal to 60 N.

So, f the total frictional force equals 60 N

I think any pendulum would swing slower on the moon than it would on Earth.

The time it takes a pendulum to go through a complete back and forth swing is:

Time period = 2 π √(length/gravity)

You can see that 'gravity' is in the denominator of the fraction, so the smaller gravity gets, the longer the period gets.

To be a little bit more technical, the period is inversely proportional to the square root of gravity.

So the period for a complete swing on the moon would be  √(9.8/1.6) times as long as the complete swing of the same pendulum on Earth.

That number is roughly 2.47 .

So, for every 1 second that a pendulum takes to swing back and forth once on Earth, the same pendulum would take 2.47 seconds to do it on the moon.

Answer:

based on my opinion....

as we know that gravity in moon are less than gravity in earth.. since the force of gravity is less on the moon, the pendulum would swing slower at the same length, angle

and the frequency would be less.

I hope this helps

Answer:

V = 268.8 m/s

Explanation:

The speed of a wave in general is given by the following formula:

V = fλ

where,

V = Speed of that wave

f = Frequency of the wave

λ = wavelength of the wave

In this case we have a sound wave, travelling across carbon dioxide. The properties of sound wave are as follows:

V = Speed of Sound in Carbon dioxide = ?

f = frequency of sound wave = 140 Hz

λ = wavelength of sound wave = 1.92 m

Therefore,

V = (140 Hz)(1.92 m)

V = 268.8 m/s

Answer:

(a) By small angle approximation, we have;

F = -2×T×Δy/l

(b) [tex]The \ frequency \ of \ oscillation, \ f = \dfrac{1}{2\cdot \pi }\cdot\sqrt{\dfrac{2 \cdot T}{l \cdot M} }[/tex]

Explanation:

(a) The diagram shows the mass, M, being restored by two equal tension, T acting on the elastic strings l, such the restoring force, F acts along the path of motion of the mass, with distance Δy

Therefore, the component of the tension T that form part of the restoring force is given as follows;

Let the angle between the line representing the extension of the elastic strings T and the initial position of the string = ∅

Then we have;

String force, [tex]F_{string}[/tex] = T×sin∅ + T×cos∅ + T×sin∅ - T×cos∅  = 2×T×sin∅

Whereby the angle is small, we have;

sin∅ ≈ tan∅ = Δy/l

Which gives;

[tex]F_{string}[/tex] = 2×T×sin∅ = 2×T×Δy/l (for small angles)

Restoring force F = [tex]-F_{string}[/tex] = -2×T×Δy/l

F = -2×T×Δy/l

(b) Given that the the tensions do not change appreciably as the mass, M, oscillates from Δy we have;

By Hooke's law, F = -k×x

Whereby Δy corresponds to the maximum displacement of the mass, M from the rest position, which gives;

Which gives;

F = M×a = -k×Δy

a =  -k×Δy/M

d²(Δy)/dt² =  -k×Δy/M

When we put angular frequency as follows;

ω² = k/M

We get;

d²(Δy)/dt² =  -ω²×Δy

Which gives;

Δy(t) = A×cos(ωt + Ф)

The angular frequency is thus, ω = √(k/M)

Period of oscillation = 2·π/ω = 2·π/√(k/M)

The frequency of oscillation, f = 1/T = √(k/M)/(2·π)

Where:

k = 2·T/l, we have;

f = √(k/M)/(2·π) = √(2·T/l)/m)/(2·π)

The frequency of oscillation is given as follows;

[tex]f = \dfrac{1}{2\cdot \pi }\cdot\sqrt{\dfrac{2 \cdot T}{l \cdot M} }[/tex]

Answer:

(a) 17.0eV

(b) 10.55W

Explanation:

(a) The amount of work done (W) on an electron by an ideal battery of emf value of V as it moves from the positive to the negative terminal is given by;

W = q x V                 --------(i)

Where;

q = charge on the electron = 1e

From the question;

V = 17.0 V

Substitute the values of q and V into equation (i) as follows;

W = 1e x 17.0

W = 17.0eV

Therefore, the work done in electron volts is 17.0

(b) The power (P) of the battery as some electrons (n) pass through it at time t, is given as;

P = (n q V) / t            --------------(ii)

Where;

n = number of electrons = 3.88 x 10¹⁸

t = 1s

q = 1.6 x 10⁻¹⁹C

V = 17.0V

Substitute these values into equation (ii) as follows;

P = (3.88 x 10¹⁸ x 1.6 x 10⁻¹⁹ x 17.0) / 1

P = 10.55W

Therefore the power of the battery is 10.55W

Answer:

pressure=force/area

Answer:

There are no gaps in space between the photons as they travel. If you look at light as a wave, then there no gaps unless specifically placed there on purpose. ... The light from a distance star indeed spreads out and weakens as it travels, but this just reduces the wave strength and does not introduce gaps.