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Marina86 [1]
3 years ago
6

Name five ways that the order of rock layers can be disturbed

Physics
1 answer:
neonofarm [45]3 years ago
5 0
Tilting,intrusions,folding,faults and if its uncomfortable.
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2. Four resistors of 2 ohm each are connected first in series and then in parallel with a battery.Find the ratio of electric cur
Serjik [45]

Answer:

Ratio of series current to parallel

= 1 : 8

Explanation:

Total resistance Rt

For series, Rt = 2+2+2+2 = 4ohms

For parallel, 1/Rt = 1/2 + 1/2 + 1/2 + 1/2

1/Rt = 4/2, Rt = 2/4 ohms.

If we use a 1V battery, then,

I = V/Rt

I = 1/4 = 0.25 ampere for series arrangement.

I = 1/0.5 = 2 ohms.

Ratio of current of series to parallel = 0.25 : 2

= 1 : 8

4 0
4 years ago
Read 2 more answers
Infrared waves from the sun are what make our skin feel warm on a sunny day. If an infrared wave has a frequency of 3.0 x 1012 H
djyliett [7]

Answer:

The wavelength of the infrared wave is <u>0.0001 m</u>.

Explanation:

Given:

Frequency of an infrared wave is, f=3.0\times 10^{12}\ Hz

We know that, infrared waves are electromagnetic waves. All electromagnetic waves travel with the same speed and their magnitude is equal to the speed of light in air.

So, speed of infrared waves coming from the Sun travels with the speed of light and thus its magnitude is given as:

v=c=3.0\times 10^8\ m/s

Where, 'v' is the speed of infrared waves and 'c' is the speed of light.

Now, we have a formula for the speed of any wave and is given as:

v=f\lambda

Where, \lambda \to \textrm{Wavelength of infrared wave}

Now, rewriting the above formula in terms of wavelength, \lambda, we get:

\lambda=\dfrac{v}{f}

Now, plug in 3.0\times 10^8 for 'v', 3.0\times 10^{12} for 'f' and solve for  \lambda. This gives,

\lambda=\frac{3.0\times 10^8}{3.0\times 10^{12}}\\\\\lambda=0.0001\ m

Therefore, the wavelength of the infrared wave is 0.0001 m.

5 0
3 years ago
An ore car of mass 39000 kg starts from rest and rolls downhill on tracks from a mine. At the end of the tracks, 19 m lower vert
cupoosta [38]

Answer:

The compression in the spring is 5.88 meters.                

Explanation:

Given that,

Mass of the car, m = 39000 kg

Height of the car, h = 19 m

Spring constant of the spring, k=4.2\times 10^5\ N/m

We need to find the compression in the spring in stopping the ore car. It can be done by balancing loss in gravitational potential energy and the increase in elastic energy. So,

mgh=\dfrac{1}{2}kx^2

x is the compression in spring

x=\sqrt{\dfrac{2mgh}{k}} \\\\x=\sqrt{\dfrac{2\times 39000\times 19\times 9.8}{4.2\times 10^5}} \\\\x=5.88\ m

So, the compression in the spring is 5.88 meters.                                                                                                                  

6 0
3 years ago
Glycerin (C3H8O3) is a nonvolatile liquid. What is the vapor pressure of a solution made by adding 154 g of glycerin to 316 mL o
garri49 [273]

Answer:

P_{sol}=50.4\ mm.Hg

Explanation:

According to given:

  • molecular mass of glycerin, M_g=3\times 12+8+3\times 16=92\ g.mol^{-1}
  • molecular mass of water, M_w=2+16=18\ g.mol^{-1}
  • ∵Density of water is 0.992\ g.cm^{-3}= 0.992\ g.mL^{-1}
  • ∴mass of water in 316 mL, m_w=316\times 0.992=313.5 g
  • mass of glycerin, m_g=154\ g
  • pressure of mixture, P_x=55.32\ torr= 55.32\ mm.Hg
  • temperature of mixture, T_x=40^{\circ}C

<em>Upon the formation of solution the vapour pressure will be reduced since we have one component of solution as non-volatile.</em>

<u>moles of water in the given quantity:</u>

n_w=\frac{m_w}{M_w}

n_w=\frac{313.5}{18}

n_w=17.42 moles

<u>moles of glycerin in the given quantity:</u>

n_g=\frac{m_g}{M_g}

n_g=\frac{154}{92}

n_g=1.674 moles

<u>Now the mole fraction of water:</u>

X_w=\frac{n_w}{n_w+n_g}

X_w=\frac{17.42}{17.42+1.674}

X_w=0.912

<em>Since glycerin is non-volatile in nature so the vapor pressure of the resulting solution will be due to water only.</em>

\therefore P_{sol}=X_w\times P_x

\therefore P_{sol}=0.912\times 55.32

P_{sol}=50.4\ mm.Hg

7 0
3 years ago
A ball A of mass 0.5 kg moving with a Velacity of 10 m/s a head on Collision with a ball B of mass 2kg moving with a Velocity of
Nesterboy [21]

Answer:

The common velocity v after collision is 2.8m/s²

Explanation:

look at the attachment above ☝️

3 0
2 years ago
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