1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
alexandr402 [8]
3 years ago
14

What are tree things that apply to relative dating

Physics
1 answer:
grigory [225]3 years ago
3 0
Rocks, earth aging, fossils
You might be interested in
1. A pumpkin with a mass of 2 kg accelerates 2 m/s/s when an unknown force is applied to it. What is the amount of the force? ​
Andreyy89

Answer:

<h3>The answer is 4 N</h3>

Explanation:

The force acting on an object given it's mass and acceleration can be found by using the formula

force = mass × acceleration

From the question we have

force = 2 × 2

We have the final answer as

<h3>4 N</h3>

Hope this helps you

8 0
3 years ago
Water is traveling into what sphere during infiltration?
o-na [289]
When precipitation hits the ground and hydrosphere
5 0
3 years ago
(15 points) :^|
mihalych1998 [28]
That is FALSE. The equation to calculate the charges has a distance component that is in the denominator which means that it is inversely proportional (as the distance os greater the force is smaller)
7 0
3 years ago
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).

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
3 years ago
If an asteroid were going to impact Earth, would you prefer to know in advance or let the NEO remain a mystery until impact? Why
ivanzaharov [21]

I would like to know in advance
6 0
3 years ago
Other questions:
  • This is may mastering physics homework and I just need help solving the question.
    8·1 answer
  • You throw a ball straight up with an initial velocity of 15.0 m/s. It passes a tree branch on the way up at a height of 7.00 m.
    14·1 answer
  • 1. Explain what dumping is, giving some examples. Does dumping raise any moral issues? What are they? What would an ethical rela
    7·1 answer
  • What is an example of velocity?
    7·1 answer
  • A light ray traveling through medium 1, index of refraction n1n1n_1 = 1.75, reaches the interface between medium 1 and medium 2,
    13·1 answer
  • A river flows at a rate of 2 km divided by h. A patrol boat travels 54 km upriver and returns in a total time of 9 hr. What is t
    9·1 answer
  • True or false the sun produces energy through nuclear fusion
    13·2 answers
  • Length of table is 1.0 metre, 1.00 metre and 1.000 metre. Which one is more accurate?​
    15·1 answer
  • Give an example of a vertical motion with a positive velocity and a negative acceleration. Give an example of a vertical motion
    11·1 answer
  • Margaret walks to the store using the following path 0.630 mi west, 0.370 mi north, 0.180 mi east. assume north to be along the
    6·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!