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3 years ago

Macro shots of flowers will typically have a shallow depth of field. True or False?

Physics

2 answers:

SSSSS [86.1K]3 years ago

5 0

It is TRUE that macro shots of flowers will typically have a shallow depth of field.

Macro photography (or photomacrography or macrography, and sometimes macrophotography), is extreme close-up photography, usually of very small subjects, in which the size of the subject in the photograph is greater than life size (though macrophotography technically refers to the art of making very largephotographs).

I am hoping that this answer has satisfied your query and it will be able to help you in your endeavor, and if you would like, feel free to ask another question.

tigry1 [53]3 years ago

3 0

Answer:

true

Explanation:

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What is one property of a suspension that is different from that of a solution or a colloid?

Step2247 [10]

Answer:

Explanation:

If left to rest it will separate.

3 0

2 years ago

The body falls in a free fall 11s.Calculate: a) from what height the body of the paddle) what path did it fall during the last s

JulsSmile [24]

Answer:

a) h = 593.50 m

b) h₁₁ = 103 m

c) vf = 107.91 m/s

Explanation:

We will use second equation of motion to find the height:

$h = v_{i}t + (\frac{1}{2})gt^2$

where,

h = height = ?

vi = initial speed = 0 m/s

t = time taken = 11 s

g = 9.81 /s²

Therefore,

$h = (0\ m/s)(11\ s) + (\frac{1}{2})(9.8\ m/s^2)(11\ s)^2\\\\$

h = 593.50 m

For the distance travelled in last second, we first need to find velocity at 10th second by using first equation of motion:

$v_{f} = v_{i} + gt$

where,

vf = final velocity at tenth second = v₁₀ = ?

t = 10 s

vi = 0 m/s

Therefore,

$v_{10} = 0\ m/s + (9.81\ m/s^2)(10\ s)\\\\v_{10} = 98.1\ m/s$

Now, we use the 2nd equation of motion between 10 and 11 seconds to find the height covered during last second:

$h = v_{i}t + (\frac{1}{2})gt^2$

where,

h = height covered during last second = h₁₁ = ?

vi = v₁₀ = 98.1 m/s

t = 1 s

Therefore,

$h_{11} = (98.1\ m/s)(1\ s) + (\frac{1}{2})(9.8\ m/s^2)(1\ s)^2\\\\$

h₁₁ = 103 m

Now, we use first equation of motion for complete motion:

$v_{f} = v_{i} + gt$

where,

vf = final velocity at tenth second = ?

t = 11 s

vi = 0 m/s

Therefore,

$v_{f} = 0\ m/s + (9.81\ m/s^2)(11\ s)$

vf = 107.91 m/s

8 0

3 years ago

Which of the following statements about a metal wire in equilibrium are true? Select all that apply. There cannot be excess char

olga_2 [115]

Answer:

There may be excess charges in the interior of the wire

The net electric field everywhere inside the wire is zero

The interior of the metal wire is neutral.

There may be excess charges on the surface of the wire.

There is no net flow of mobile electrons inside the wire.

Explanation:

For any metal wire in equilibrium position, there may be excess charges in the interior of the wire and the net electric field everywhere inside the wire is zero. Additionally, the interior of the metal wire is always neutral and there is likely to be excess charges on the surface of the wire. Moreover, it's important to note that for a metal wire in equilibrium, there is no net flow of mobile electrons inside the wire.

3 0

3 years ago

Which marine habitats would have the least access to primary producers?

nikitadnepr [17]

Answer:

Explanation:

4 0

2 years ago

A wire of density p is tapered so that its cross-sectional area varies with x according to

DochEvi [55]

The wave speed at the origin is v = 8.31 m/s

Given data

A=1.00× 10⁻⁵ x + 1.00× 10⁻⁶

A is in meters squared and x is in meters

tension in the wire is T

T = 24.0

x = 10.0m

density of aluminum = 2700

<h3>calculating for the speed of the wave at the origin</h3>

wave speed = v = sqrt ( T / mass per unit length)

mass per unit length = m / L

density = m / v

volume = v = A * L

density * A = m / L

wave speed = v = sqrt ( T / density * A )

v = sqrt ( 24 / ( 2700 * .00× 10⁻⁵ x + 1.00× 10⁻⁶ ) )

v = 8.31 m/s

Read more on speed of wave here: brainly.com/question/12969690

#SPJ4

4 0

2 years ago