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1 year ago

9

The inventor of the photographic process in which a photograph produced without a negative by exposing objects to light on light

sensitive paper, is named _____________
a. Man Ray
c. Dorothea Lange
b. Richard Avedon
d. Pablo Picasso

2 answers:

Bad White [126]1 year ago

8 0

Answer:

it's A man ray as in one of spongebob's evil villains

i think

.w.

Nikolay [14]1 year ago

7 0

<span>A. Man Ray --------------------</span>

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To avoid an accident, a driver steps on the brakes to stop a 1000-kg car traveling at 65km/h. if the braking distance is 35 m, h

IgorLugansk [536]

First, find the needed acceleration needed for the car to stop from its initial velocity given the distance. This is calculated through the equation,
2ad = Vf² - Vi²
where a and d are acceleration and distance, respectively. Vf and Vi are final and initial velocities, respectively. Substituting the known values,
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1 year ago

12*8A hollow steel ball weighing 4 pounds is suspended from a spring. This stretches the spring 13 feet. The ball is started in

max2010maxim [7]

Answer:

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Explanation:

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1 year ago

A positive point charge Q1 = 2.5 x 10-5 C is fixed at the origin of coordinates, and a negative point charge Q2 = -5.0 x 10-6 C

mario62 [17]

Answer:

3.62 m and - 1.4 m

Explanation:

Consider a location towards the positive side of x-axis beyond the location of charge Q₂

x = distance of the location from charge Q₂

d = distance between the two charges = 2 m

For the electric field to be zero at the location

E₁ = Electric field by charge Q₁ at the location = E₂ = Electric field by charge Q₂ at the location

$\frac{kQ_{1}}{(2 + x)^{2}}= \frac{kQ_{2}}{x^{2}}$

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x = 1.62 m

So location is 2 + 1.62 = 3.62 m

Consider a location towards the negative side of x-axis beyond the location of charge Q₁

x = distance of the location from charge Q₁

d = distance between the two charges = 2 m

For the electric field to be zero at the location

E₁ = Electric field by charge Q₁ at the location = E₂ = Electric field by charge Q₂ at the location

$\frac{kQ_{1}}{(x)^{2}}= \frac{kQ_{2}}{ (2 + x)^{2}}$

$\frac{2.5\times 10^{-5}}{(x)^{2}}= \frac{5 \times 10^{-6}}{(2+x)^{2}}$

x = - 1.4 m

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1 year ago

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bearhunter [10]

Answer:

0.1 m

Explanation:

It is given that,

Mass of the object, m = 350 g = 0.35 kg

Spring constant of the spring, k = 5.2 N/m

Amplitude of the oscillation, A = 10 cm = 0.1 m

Frequency of a spring mass system is given by :

Time period:

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1 year ago

Before leaving the house in the morning, you plop some stew in your slow cooker and turn it on Low. The slow cooker has a 160 Oh

guajiro [1.7K]

Answer:

Total charge flow through the cooker is 21600 C

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here it is given as

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now the charge flow through it is given as

$Q = i t$

total time is t = 8 hours

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$Q = 21600 C$

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2 years ago