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3 years ago

SCIENCE whoever gets this first will get a brainlest

Physics

2 answers:

gulaghasi [49]3 years ago

6 0

Answer:

im not 100% sure i should be doing this but i found a website that will help a ton.

https://geo.libretexts.org/Courses/Gettysburg_College/Book%3A_An_Introduction_to_Geology_(Johnson_Affolter_Inkenbrandt_and_Mosher)/02%3A_Plate_Tectonics/2.01%3A_Alfred_Wegener%E2%80%99s_Continental_Drift_Hypothesis

Goshia [24]3 years ago

5 0

Well he did have a lot evidence because we wouldn’t believe him otherwise

I’m pretty sure it’s:
He knew that plant and animal fossils, as well as rock layers, matched on the two continent of Africa and South America

Hope this helps :)

You might be interested in

How far from the earth must a body be along a line toward the sun so that the sun’s

Ivenika [448]

Answer:

r = 1.63×10^5 mi

Explanation:

Let r = distance of object from earth

Rs = distance between earth and sun

Ms = mass of the sun

= 3.24×10^5 Me (Me = mass of earth)

At a distance R from earth, the force Fs exerted by the sun on the object is equal to the force Fe exerted by the earth on the object. Using Newton's universal law of gravitation,

Fs = Fe

GmMs/(Rs - r)^2 = GmMe/r^2

This simplifies to

Ms/(Rs - r)^2 = Me/r^2

(3.24×10^5 Me)/(Rs - r)^2 = Me/r^2

Taking the reciprocal and then its square root, this simplifies further to

Rs - r = (569.2)r ----> Rs = 570.2r

r = Rs/570.2 = (9.3×10^7 mi)/570.2

= 1.63×10^5 mi

4 0

3 years ago

Vicky wanted to investigate water evaporation. She placed 50 mL of distilled water in three identical glass jars. She left one j

IgorC [24]

Answer:

D. the amount of water placed in jars

Explanation:

it was the same for all of the jars at the beginning

4 0

3 years ago

Read 2 more answers

What is the rate at which an object moves but doesn't include direction?

Juli2301 [7.4K]

Great question !

The rate at which an object covers distance, without worrying
about the direction it's moving, is the object's SPEED .

When the direction is also given, then you have the object's VELOCITY.

This question is important. It gives us a chance to point out that
"velocity" is not just a fancy word for speed that you use when you
want to sound smart. There's actually an important difference between
'speed' and 'velocity'.

6 0

3 years ago

A source for conducting formal research would include_

Wittaler [7]

C. written reports !

5 0

3 years ago

Read 2 more answers

Coulomb’s law and static point charge ensembles (15 points). A test charge of 2C is located at point (3, 3, 5) in Cartesian coor

fenix001 [56]

Answer:

a) $F_{r}= -583.72MN i + 183.47MN j + 6.05GN k$

b) $E=3.04 \frac{GN}{C}$

Step-by-step explanation.

In order to solve this problem, we mus start by plotting the given points and charges. That will help us visualize the problem better and determine the direction of the forces (see attached picture).

Once we drew the points, we can start calculating the forces:

$r_{AP}^{2}=(3-0)^{2}+(3-0)^{2}+(5+0)^{2}$

which yields:

$r_{AP}^{2}= 43 m^{2}$

(I will assume the positions are in meters)

Next, we can make use of the force formula:

$F=k_{e}\frac{q_{1}q_{2}}{r^{2}}$

so we substitute the values:

$F_{AP}=(8.99x10^{9})\frac{(1C)(2C)}{43m^{2}}$

which yields:

$F_{AP}=418.14 MN$

Now we can find its components:

$F_{APx}=418.14 MN*\frac{3}{\sqrt{43}}i$

$F_{APx}=191.30 MNi$

$F_{APy}=418.14 MN*\frac{3}{\sqrt{43}}j$

$F_{APy}=191.30MN j$

$F_{APz}=418.14 MN*\frac{5}{\sqrt{43}}k$

$F_{APz}=318.83 MN k$

And we can now write them together for the first force, so we get:

$F_{AP}=(191.30i+191.30j+318.83k)MN$

We continue with the next force. The procedure is the same so we get:

$r_{BP}^{2}=(3-1)^{2}+(3-1)^{2}+(5+0)^{2}$

which yields:

$r_{BP}^{2}= 33 m^{2}$

Next, we can make use of the force formula:

$F_{BP}=(8.99x10^{9})\frac{(4C)(2C)}{33m^{2}}$

which yields:

$F_{BP}=2.18 GN$

Now we can find its components:

$F_{BPx}=2.18 GN*\frac{2}{\sqrt{33}}i$

$F_{BPx}=758.98 MNi$

$F_{BPy}=2.18 GN*\frac{2}{\sqrt{33}}j$

$F_{BPy}=758.98MN j$

$F_{BPz}=2.18 GN*\frac{5}{\sqrt{33}}k$

$F_{BPz}=1.897 GN k$

And we can now write them together for the second, so we get:

$F_{BP}=(758.98i + 758.98j + 1897k)MN$

We continue with the next force. The procedure is the same so we get:

$r_{CP}^{2}=(3-5)^{2}+(3-4)^{2}+(5-0)^{2}$

which yields:

$r_{CP}^{2}= 30 m^{2}$

Next, we can make use of the force formula:

$F_{CP}=(8.99x10^{9})\frac{(7C)(2C)}{30m^{2}}$

which yields:

$F_{CP}=4.20 GN$

Now we can find its components:

$F_{CPx}=4.20 GN*\frac{-2}{\sqrt{30}}i$

$F_{CPx}=-1.534 GNi$

$F_{CPy}=4.20 GN*\frac{2}{\sqrt{30}}j$

$F_{CPy}=-766.81 MN j$

$F_{CPz}=4.20 GN*\frac{5}{\sqrt{30}}k$

$F_{CPz}=3.83 GN k$

And we can now write them together for the third force, so we get:

$F_{CP}=(-1.534i - 0.76681j +3.83k)GN$

So in order to find the resultant force, we need to add the forces together:

$F_{r}=F_{AP}+F_{BP}+F_{CP}$

so we get:

$F_{r}=(191.30i+191.30j+318.83k)MN + (758.98i + 758.98j + 1897k)MN + (-1.534i - 0.76681j +3.83k)GN$

So when adding the problem together we get that:

$F_{r}=(-0.583.72i + 0.18347j +6.05k)GN$

which is the answer to part a), now let's take a look at part b).

Basically, we need to find the magnitude of the force and divide it into the test charge, so we get:

$F_{r}=\sqrt{(-0.583.72)^{2} + (0.18347)^{2} +(6.05)^{2}}$

which yields:

$F_{r}=6.08 GN$

and now we take the formula for the electric field which is:

$E=\frac{F_{r}}{q}$

so we go ahead and substitute:

$E=\frac{6.08GN}{2C}$

$E=3.04\frac{GN}{C}$

7 0

4 years ago