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3 years ago

The pioneer in the technique of photographic collage was:

1 answer:

4 0

The correct answer is Raoul Hausmann. He was an Austrian artist and writer. And he was a forefather and dominant figure of the Dada movement in Berlin, who was known particularly for his ironic photomontages and his provocative writing on art. One of the important figures there also includes the experimental photographic collages, sound poetry and institutional reviews that has a profound influence on the European Avant-Garde in the aftereffects of World War I.

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In proof testing of circuit boards, the probability that any particular diode will fail is 0.01. Suppose a circuit board contain

Sergio039 [100]

Answer: 2 diodes

Explanation:

Given : In proof testing of circuit boards, the probability that any particular diode will fail is 0.01.

The number of diodes contained by circuit board = 200

Then , the expected number of diodes to fail is given by :-

$0.01\times200=2$

Therefore, there are 2 diodes that we will expect to fail.

7 0

3 years ago

Suppose the original segment of wire is stretched to 10 times its original length. How much charge must be added to the wire to

Debora [2.8K]

Here we want to study how the linear charge density changes as we change the measures of our body.

We will find that we need to add 9*Q of charge to keep the linear charge density unchanged.

I will take two assumptions:

The charge is homogeneous, so the density is constant all along the wire.

As we work with a linear charge density we work in one dimension, so the wire "has no radius"

Originally, the wire has a charge Q and a length L.

The linear charge density will be given by:

λ = Q/L

Now the length of the wire is stretched to 10 times the original length, so we have:

L' = 10*L

We want to find the value of Q' such that λ' (the linear density of the stretched wire) is still equal to λ.

Then we will have:

λ' = Q'/L' = Q'/(10*L) = λ = Q/L

Q'/(10*L) = Q/L

Q'/10 = Q

Q' = 10*Q

So the new charge must be 10 times the original charge, this means that we need to add 9*Q of charge to keep the linear charge density unchanged.

If you want to learn more, you can read:

brainly.com/question/14514975

6 0

2 years ago

Astronomers discover a bright X-ray source in the Milky Way. They see no counterpart in the optical bands, but it has another st

VladimirAG [237]

Answer:

a. a black hole

Explanation:

X-ray emission from the central degrees of the Milky Way Bright X-ray emission traces the coherent edge brightened shell-like feature, dubbed the northern chimney, located north of Sgr A* and characterized by a diameter of about 160 pc. On the opposite side, the southern chimney appears as a bright linear feature. Bright X-ray emission is observed at high latitude

8 0

3 years ago

A charge of 32.0 nC is placed in a uniform electric field that is directed vertically upward and has a magnitude of 4.30x 104 V/

hodyreva [135]

A) The work done by the electric field is zero

B) The work done by the electric field is $9.1\cdot 10^{-4} J$

C) The work done by the electric field is $-2.4\cdot 10^{-3} J$

Explanation:

The electric field applies a force on the charged particle: the direction of the force is the same as that of the electric field (for a positive charge).

The work done by a force is given by the equation

$W=Fd cos \theta$

where

F is the magnitude of the force

d is the displacement of the particle

$\theta$ is the angle between the direction of the force and the direction of the displacement

In this problem, we have:

The force is directed vertically upward (because the field is directed vertically upward)
The charge moves to the right, so its displacement is to the right

This means that force and displacement are perpendicular to each other, so

$\theta=90^{\circ}$

and $cos 90^{\circ}=0$ : therefore, the work done on the charge by the electric field is zero.

In this case, the charge move upward (same direction as the electric field), so

$\theta=0^{\circ}$

and

$cos 0^{\circ}=1$

Therefore, the work done by the electric force is

$W=Fd$

and we have:

$F=qE$ is the magnitude of the electric force. Since

$E=4.30\cdot 10^4 V/m$ is the magnitude of the electric field

$q=32.0 nC = 32.0\cdot 10^{-9}C$ is the charge

The electric force is

$F=(32.0\cdot 10^{-9})(4.30\cdot 10^4)=1.38\cdot 10^{-3} N$

The displacement of the particle is

d = 0.660 m

Therefore, the work done is

$W=Fd=(1.38\cdot 10^{-3})(0.660)=9.1\cdot 10^{-4} J$

In this case, the angle between the direction of the field (upward) and the displacement (45.0° downward from the horizontal) is

$\theta=90^{\circ}+45^{\circ}=135^{\circ}$

Moreover, we have:

$F=1.38\cdot 10^{-3} N$ (electric force calculated in part b)

While the displacement of the charge is

d = 2.50 m

Therefore, we can now calculate the work done by the electric force:

$W=Fdcos \theta = (1.38\cdot 10^{-3})(2.50)(cos 135.0^{\circ})=-2.4\cdot 10^{-3} J$

And the work is negative because the electric force is opposite direction to the displacement of the charge.

Learn more about work and electric force:

brainly.com/question/6763771

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#LearnwithBrainly

5 0

3 years ago

A fold is a _ in a rock , and a fault is a _in a rock?

777dan777 [17]

A geological fold occurs when one or a stack of originally flat and planar surfaces, such as sedimentary strata, are bent or curved as a result of permanent deformation.

So A fold is a Bend? in a rock. Maybe.

A fault is a planar fracture or discontinuity in a volume of rock, across which there has been significant displacement as a result of rock-mass movement.

3 0

3 years ago

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