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irina1246 [14]
3 years ago
9

The majority of geologists believe that the sedimentary rock require 25,000 years to accumulate. true or false

Physics
2 answers:
nevsk [136]3 years ago
6 0
I believe False it takes a lot longer
Zarrin [17]3 years ago
6 0
True. The sedimentary rock takes the least amount of time to form out of all of the different types of rock
You might be interested in
If the coefficient of kinetic friction between a 22 kg kg crate and the floor is 0.27, what horizontal force is required to move
alina1380 [7]

Answer:

58.27 N

Explanation:

the data we have is:

mass: m=22kg

coefficient of friction: \mu =0.27

and we also know the acceleration of gravity is g=9.81m/s^2

We need to do an analysis of horizontal and vertical forces acting on the object:

-------

Vertically the forces acting on the object:

  • Normal force N (acting up from the object)
  • weight: w=mg (acting down from)

so the sum of forces in the vertical axis "y" are:

F_{y}=N-w\\F_{y}=N-mg

from Newton's second Law we know that F=ma, so:

ma_{y}=N-mg

and since the object is not accelerating in the vertical direction (the movement is only horizontal) a_{y}=0, and:

0=N-mg\\N=mg

-----------

now let's analyze the horizontal forces

  • frictional force: f= \mu N and since N=mg  --> f=\mu mg
  • force to move the object: F

and the two forces just mentioned must be opposite, thus the sum of forces in the "x" axis is:

F=ma_{x}=F-f\\ma_{x}=F-\mu mg

and we are told that the crate moves at a steady speed, thus there is no acceleration: a_{x}=0

and we get:

0=F-\mu mg\\F=\mu mg

substituting known values:

F=(0.27)(22kg)(9.81m/s^2)\\F=58.27N

3 0
3 years ago
10 The magnitude J of the current density in a certain lab wire with a circular cross section of radius R = 2.00 mm is given by
rewona [7]

Answer:

I = 0.002593 A = 2.593 mA

Explanation:

Current density = J = (3.00 × 10⁸)r² = Br²

B = (3.00 × 10⁸) (for ease of calculations)

The current through outer section is given by

I = ∫ J dA

The elemental Area for the wire,

dA = 2πr dr

I = ∫ Br² (2πr dr)

I = ∫ 2Bπ r³ dr

I = 2Bπ ∫ r³ dr

I = 2Bπ [r⁴/4] (evaluating this integral from r = 0.900R to r = R]

I = (Bπ/2) [R⁴ - (0.9R)⁴]

I = (Bπ/2) [R⁴ - 0.6561R⁴]

I = (Bπ/2) (0.3439R⁴)

I = (Bπ) (0.17195R⁴)

Recall B = (3.00 × 10⁸)

R = 2.00 mm = 0.002 m

I = (3.00 × 10⁸ × π) [0.17195 × (0.002⁴)]

I = 0.0025929449 A = 0.002593 A = 2.593 mA

Hope this Helps!!!

4 0
3 years ago
As a football player moves in a straight line [displacement (3.00 mm)i^i^ - (6.50 mm)j^j^], an opponent exerts a constant force
Brilliant_brown [7]

Answer:

Work done, W=(0.378i-1.092j)\ J

Explanation:

Displacement,

d=(3i-6.5j)\ mm\\\\d=(0.003i-0.0065j)\ m

Force, F=(126i+168j)\ N

Work done by the opponent do on the football player is given by :

W=F{\cdot} d\\\\W=(126i+168j){\cdot} (0.003i-0.0065j)\ m\\\\W=(0.378i-1.092j)\ J

So, the work done by the opponent do on the football player is  (0.378i-1.092j)\ J.

4 0
3 years ago
Read 2 more answers
Two trains approach each other on parallel tracks. Each has a speed of 95 kph relative to the ground. If they are initially 8.5
aliya0001 [1]

Answer: 2.83 minutes

Explanation:

It is understood that trains are approaching. That is, they have speeds of equal magnitude but opposite. When train A travels x meters northbound, then train B travels the same distance southbound.

Therefore trains approach at a speed of:

V = 95 +95 = 190\ km / h

Then:

V = \frac{x}{t}\\\\t = \frac{x}{V}

Where x is the distance between the trains

x = 8.5\ km

So the time in which both trains meet is:

t =\frac{8.5\ km}{180\ \frac{km}{h}}\\\\t=0.04722\ hours

This is:

0.04722*\frac{60\ minutes}{1\ hour} = 2.83\ minutes

<em />

<em>How long will it be before they reach one another ?</em>

<h3>2.83 minutes</h3>
3 0
3 years ago
If the period of a spring is 5 seconds what is the frequency
Scilla [17]

Answer:

C. 0.2 Hertz

Explanation:

The frequency of a spring is equal to the reciprocal of the period:

f=\frac{1}{T}

where

f is the frequency

T is the period

For the spring in this problem,

T = 5 s

therefore, the frequency is

f=\frac{1}{5 s}=0.2 Hz

7 0
3 years ago
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