Man has never directly reached the inner layers of the Earth. But what phenomena allow to determine its chemical composition?

A continuous length of wire is made into 10 coaxial loops, located in the plane of this page. Each loop has a cross sectional ar

AleksandrR [38]

Answer:

e = 10 V

Explanation:

given,

number of the coaxial loops = 10

Cross sectional area = 0.5 m²

magnitude of magnetic field =

B = 3 T + (2 T/s)*t.

B = ( 3+ 2 t ) T

induced potential difference = ?

At time = 2 s

we know,

induced emf

$e = - N\dfrac{d\phi}{dt}$

∅ = B . A

$e = - N\dfrac{d(BA)}{dt}$

$e = - NA\dfrac{dB}{dt}$

$e = - 10 \times 0.5 \times \dfrac{d}{dt}(3 + 2 t)$

$e = - 10 \times 0.5 \times 2$

e = -10 V

magnitude of induced emf

|e| = |-10 V|

e = 10 V

the induced potential difference in the loop = e = 10 V

7 0

3 years ago

What's a Blackhole ?

Vlad1618 [11]

Answer:

A black hole is a region of spacetime where gravity is so strong that nothing—no particles or even electromagnetic radiation such as light—can escape from it. The theory of general relativity predicts that a sufficiently compact mass can deform spacetime to form a black hole.

7 0

3 years ago

What is the correct sequence for light traveling through a microscope?

Alexandra [31]

The answer would be B.

Since the light source comes from the bottom, you just need to find the sequence from the bottom going up.

It will start with the light source, then go through the slide, which will then go through the objective lens and lastly, the eyepiece lens. In summary, the sequence is:

B. Light --- slide ---- objective lens --- eyepiece lens

3 0

4 years ago

The gravitational self potential energy of a solid ball of mass density ρ and radius R is E. What is the gravitational self pote

Alenkasestr [34]

It will be
E = mgh.
where h and g are constant thus
m can be written as 4/3πr^3*density
E = 4/3πr^3* density
E? = 4/3π(2R)^3* density
= 4/3π8r^3
thus the e will be 4/3π8r^3* density/4/3πr^3*density nd thus you get 8E ..

8 0

3 years ago

5) Why does the first hill/drop on a roller coaster have to be the highest? (think about

prohojiy [21]

5) Due to the conservation of energy / due to the presence of friction

6) The work done is 4000 J

7) The distance travelled by the object is 4 m

8) The potential energy of the bird is 2450 J

Explanation:

5)

For a roller coaster in motion along the track, in absence of friction, the mechanical energy remains constant:

$E=U+K = const.$

where

U is the potential energy

K is the kinetic energy

At the beginning, the roller coaster is at rest, therefore its kinetic energy is zero and its mechanical energy is just equal to the potential energy:

$E=U=mgh$

where m is the mass of the roller coaster, g the acceleration of gravity, and h the height of the roller coaster above the ground at the first hill.

Since this amount of energy remains constant, the roller coaster cannot go to a higher hill: in fact, in that case it would reach a height $h'>h$ , but this is not possible, because it would mean that its mechanical energy would be $mgh' > E$ , larger than its mechanical energy, and since energy cannot be created (law of conservation of energy), the height of the next hills cannot be higher than the first one. Moreover, if we consider friction, then friction does work on the roller coaster in motion, and as a result, part of its mechanical energy is wasted (converted into thermal energy): therefore, the height of the following hills must be lower than the height of the first hill.

6)

The work done when lifting an object is equal to the gravitational potential energy gained by the object:

$W=Fh$

where

F is the weight of the object

h is the change in height of the object

For the object in this problem, we have

F = 200 N (weight)

h = 20 m (change in height)

Substituting, we find the work done:

$W=(200)(20)=4000 J$

7)

The work done when pushing an object in a direction parallel to the motion of the object is given by:

$W=Fd$

where

F is the magnitude of the force

d is the distance through which the object has travelled through

In this problem, we have:

W = 300 J (work done)

F = 75 N (magnitude of the force)

Solving the equation for d, we find the distance travelled:

$d=\frac{W}{F}=\frac{300}{75}=4 m$

8)

The potential energy of an object is the energy possessed by an object due to its position in a gravitational field. Near the Earth's surface, it is given by

$PE=mgh$