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adelina 88 [10]
4 years ago
5

What are anticlines and synclines?

Physics
1 answer:
vesna_86 [32]4 years ago
3 0
<span>d. They are types of folds created by compressional stresses.</span>. 
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I have these questions I need to find the acceleration, Please explain how to do them, Don't have to explain all xD
34kurt
1. a=(v-v0)/t
a=14 m/s / 2s
a=7 m/s^2
2. <span>a=(v-v0)/t
</span>a=(30 m/s - 0 m/s) / 12 s
a=2.5 m/s^2
3. <span>a=(v-v0)/t
a=(37 m/s - 22 m/s) / 2 s
a= 7.5 m/s^2
4. </span><span>a=(v-v0)/t
a=(12 km/s - 0 km/s)/8 s
a=1.5 km/s^2
etc</span>
7 0
4 years ago
Two harmonic sound waves reach an overseveer simulatenouslt. the obsever hears the sound intensity rise and fall with a time of
irakobra [83]

Answer:

dF=2.5Hz

Explanation:

From the question we are told that:

Time T=0.2sec

Generally the Period is given as

 T= 2 * 0.2 = 0.4

Therefore difference in frequency dF

 dF=\frac{1}{T}

 dF=\frac{1}{0.4}

 dF=2.5Hz

4 0
3 years ago
A gas has a volume of 8.21 mL and exerts a pressure of 2.9 atm. What new volume will the gas have if the pressure is changed to
labwork [276]

Answer:

5.95 ml

Explanation:

Given info

P1=2.9 atm

V1=8.21 ml

P2=4 atm

From Boyle's law we know that p1v1=p2v2 where p and v are pressure and volume respectively. This is at a constant temperature. Making v2 the subject of formula then

V2=p1v1/p2

V2= 2.9*8.21/4=5.95 ml

3 0
3 years ago
List the metric symbols from largest to smallest
Ray Of Light [21]

Answer:

nano.  

micro.  

milli.

centi.  

kilo.  

mega.

giga.

tera.

Explanation:

7 0
3 years ago
Read 2 more answers
A 68.0-kg person jumps from rest off a 2.20-m-high tower straight down into the water. Neglect air resistance during the descent
Margarita [4]

Answer:

F= 1333.767\,N

Explanation:

The velocity of the swimmer just before touching the water is:

v = -\sqrt{2\cdot (9.807\,\frac{m}{s^{2}} )\cdot (2.20\,m)}

v \approx 6.569\,\frac{m}{s}

The average force exerted on the diver by the water is determined by the use of the Principle of Energy Conservation and the Work-Energy Theorem:

(68\,kg)\cdot (9.807\,\frac{m}{s^{2}})\cdot (2.20\,m) +\frac{1}{2}\cdot (68\,kg)\cdot (6.569\,\frac{m}{s} )^{2}-F\cdot(2.20\,m) = 0\,J

F= 1333.767\,N

6 0
3 years ago
Read 2 more answers
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