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Dafna11 [192]
2 years ago
10

Ill give brainliest! 1. Which item below describes a quick change to

Physics
2 answers:
Damm [24]2 years ago
8 0

Answer:

I believe it's <u>D. a glacier flattening the land below it​.</u>

stiks02 [169]2 years ago
7 0
The best answer to go with is b
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The scientific method. When conducting research, scientists use the scientific method to collect measurable, empirical evidence in an experiment related to a hypothesis (often in the form of an if/then statement), the results aiming to support or contradict a theory.
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A uniform electric field of strength E points to the right. An electron is fired with a velocity v0 to the right and travels a d
zvonat [6]

Answer:

<u />D_l=d<u />

Explanation:

From the question we are told that:

The Electric field of strength direction =Right

The Velocity of The First Electron=V_0

The Velocity of The Second Electron=V_0

Therefore

V_{e1}=V_{e2}

Generally, the equation for the Horizontal Displacement of electron is mathematically given by

D=\frac{at^2}{2}

Where

Acceleration is given as

a=\frac{V_o}{2d}

And

Time

T=\frac{d}{v_0}

Therefore horizontal displacement towards the left is

D_l=\frac{(\frac{V_o}{2d})(\frac{d}{v_0})^2}{2}

<u />D_l=d<u />

5 0
2 years ago
During World War I, the Germans had a gun called Big Bertha that was used to shell Paris. The shell had an initial speed of 2.61
bonufazy [111]

Answer:

The shell hit at a distance of 1.9 x 10² km

The time of flight of the shell was 5.3 x 10² s

Explanation:

The position of the shell is given by the vector "r":

r  = (x0 + v0 * t * cos α ; y0 + v0 * t * sin α + 1/2 g t²)

where:

x0 = initial horizontal position

v0 = magnitude of the initial velocity

t = time

α = launching angle

y0 = initial vertical position

g = acceleration of gravity

When the shell hit, the vertical component (ry) of the vector position r is 0. See figure.

Then:

ry = 0 =  y0 + v0 * t * sin α + 1/2 g t²

Since the gun is at the center of our system of reference, y0 and x0 = 0

0 = t (v0 sin α + 1/2 g t)

t= 0 is discarded as solution

v0 sin α + 1/2 g t = 0

t = -2v0 sin α / g

t = (-2 * 2610 m/s * sin 81.9°)/ (-9.8 m/s²) = 5.3 x 10² s. This is the time of flight of the shell until it hit.

Then, the distance at which the shell hit is:

Distance = Module of r = ( x0 + v0 * t * cos α; 0) = x0 + v0 * t * cos α  

Distance = 2.61 km/s * 5.3 x 10² s * cos 81.9 = 1.9 x 10² km

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3 years ago
Which of the following is a zone found in the open ocean
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