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777dan777 [17]
3 years ago
12

Which word is most closely associated with uniformitarianism

Physics
1 answer:
stellarik [79]3 years ago
5 0
Can you give me the choices.<span />
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Calculate the height from from which a body is released from rest if its velocity just before hitting the ground is30m\s
Kamila [148]

Answer:

height = 45 m

Explanation:

Initial velocity = 0 m/s

Final velocity = 30 m/s

Acceleration due to gravity = +10m/s^2 ( because ball is going down )

Using the eqn of motion

{v}^{2}  -  {u}^{2}  = 2gh

where v= final velocity ; u = initial velocity ; g = acceleration due to gravity ; h = height

So,

{30}^{2}  -  {0}^{2}  = 2 \times 10 \times h

=  > 20h = 900

=  > h =  \frac{900}{20}  = 45

5 0
3 years ago
Which of the following is not in the form of waves?
Nady [450]
C. Seismic energy
This is energy that is released in earthquakes.
6 0
3 years ago
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An ant travels 2.78 cm (West) and then turns and travels 6.25 cm (South 40 degrees East). What is the ant's total displacement?
andrey2020 [161]
From\ cosine\ theorem:\\\\&#10;c^2=a^2+b^2-2abcos(\angle between\ a\ and\ b)\\\\&#10;a=2,78\\b=6,25\\&#10;\angle between\ a\ and\ b=90+50=140^{\circ}\\\\&#10;c^2=2,78^2+6,25^2-2*2,78*6,25cos(140^{\circ})\\\\&#10;c^2=7,7284+39,0625-34,75*(-0,77)\\\\c^2=46,7909+26,7575\\\\c^2=735484\\\\c=8,58cm\\\\Total\ displacement\ is\ equal\ to\ 8,58cm.

5 0
2 years ago
Calculate curls of the following vector functions (a) AG) 4x3 - 2x2-yy + xz2 2
aleksandr82 [10.1K]

Answer:

The curl is 0 \hat x -z^2 \hat y -4xy \hat z

Explanation:

Given the vector function

\vec A (\vec r) =4x^3 \hat{x}-2x^2y \hat y+xz^2 \hat z

We can calculate the curl using the definition

\nabla \times \vec A (\vec r ) = \left|\begin{array}{ccc}\hat x&\hat y&\hat z\\\partial/\partial x&\partial/\partial y&\partial/\partial z\\A_x&X_y&A_z\end{array}\right|

Thus for the exercise we will have

\nabla \times \vec A (\vec r ) = \left|\begin{array}{ccc}\hat x&\hat y&\hat z\\\partial/\partial x&\partial/\partial y&\partial/\partial z\\4x^3&-2x^2y&xz^2\end{array}\right|

So we will get

\nabla  \times \vec A (\vec r )= \left( \cfrac{\partial}{\partial y}(xz^2)-\cfrac{\partial}{\partial z}(-2x^2y)\right) \hat x - \left(\cfrac{\partial}{\partial x}(xz^2)-\cfrac{\partial}{\partial z}(4x^3) \right) \hat y + \left(\cfrac{\partial}{\partial x}(-2x^2y)-\cfrac{\partial}{\partial y}(4x^3) \right) \hat z

Working with the partial derivatives we get the curl

\nabla  \times \vec A (\vec r )=0 \hat x -z^2 \hat y -4xy \hat z

6 0
3 years ago
PLEASE HURRY 20 PTS
Crank

Answer:

A. The electric field points to the left because the force on a negative charge is opposite to the direction of the field.

Explanation:

The electric force exerted on a charge by an electric field is given by:

where

F is the force

q is the charge

E is the electric field

We see that if the charge is negative, q contains a negative sign, so the force F and the electric field E will have opposite signs (which means they have opposite directions). This is due to the fact that the direction of the lines of an electric field shows the direction of the electric force experienced by a positive charge in that electric field: therefore, a negative charge will experience a force into opposite direction.

5 0
3 years ago
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