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Colt1911 [192]
3 years ago
12

Atmospheric pressure on the peak of Mt. Everest can be as low as 0.197 atm, which is why

Physics
1 answer:
lbvjy [14]3 years ago
7 0

Answer:2,030

Explanation:

40 atm x 10.0 L = 400

400/0.197 atm = 2,030

You might be interested in
Please fast answerrr thank you​
Vedmedyk [2.9K]

Answer:

50J

Explanation:

At the top you have(A)

KE_a = O

PE_a = 100J

KE + PE = 100J

At the bottom you have (C)

KE_c= 100J

PE_c=0J

KE+PE = 100J

At point C:

You are at half the height.

We know that at H, PE =100J

PE_c = mgH

At C,

PE_c= mg (H/2) *at half the height

*m and g stay the same

Intuitively, the higher you are, the more potential energy you have.

If you decrease the height by a half, your PE will also decrease

At A:

PE_a / (mg) = H

At B:

PE_b / (mg) = H/2

to also get H on the right hand side, multiply by 2

2 (PE_b/ (mg))= H

2PE_b / (mg) = H

Ok, now that we have set up 2 equations (where H is isolated), find PE at B

AT A = AT B *This way you are saying that H = H (you compare both equations)

PE_a / (mg) = 2x PE_b / (mg)

*mg are the same for both cancel them (you can do that because of the = sign)

PE_a =  2PE_b

We know that PE_a = 100J

100J/2 = PE_b

PE at b = 50J

**FIND KE at b

We know that

KE_b + PE_b is always 100J

100J = 50J + KE_b

KE_b = 50J

4 0
2 years ago
Can someone please answer this, ill give you brainliest and your getting 100 points.
dsp73
It’s c hopefully that helps
5 0
2 years ago
Read 2 more answers
Please help!<br>The question is attached!!​
Law Incorporation [45]

Answer:

the pic is blur,btw can u type sweetie

7 0
2 years ago
A negative ion of charge -2e is located at the origin and a second negative ion of charge -3e is located nearby at x = 3.8 nm ,
Rus_ich [418]

Answer:

\vec{F}_{21}=-5.63\times 10^{-11}N\\\\\vec{F}_{21}=\\

Explanation:

Given that

Q_1 = -2e\, C\\\\Q_2=-3e\,C\\\\x= 3.8 \times 10^{-9}\,m\\\\y= 3.2 \times 10^{-9}\,m\\\\r=\sqrt{x^2+y^2}\\\\r= 4.96\times 10^{-9} m\\

As both charges are negative so there exist force of repulsion in direction as shown in figure.

F_{12}=\frac{kQ_1Q_2}{r^2}\\\\F_{12}= \frac{(9\times 10^9)(6)(1.602\times 10^{-19})^2}{(4.96\times 10^{-9})^2}\\\\F_{12}=5.63\times 10^{-11}N

Angle at which force F12 is acting is

\theta=tan^{-1}\frac{3.2}{3.8}\\\\\theta=tan^{-1}\frac{y}{x}\\\\\theta= 40.1^o

F_{x}=F_{12}cos\theta\\\\F_{x}=(5.63\times 10^{-11})cos(40.1)\\\\F_{x}=4.306\times 10^{-11}N\\\\F_{y}=F_{12}sin\theta\\\\F_{y}=(5.63\times 10^{-11})sin(40.1)\\\\F_{y}=3.62\times 10^{-11}N\\\\

\vec{F}_{12}=\vec{F}_{x}+\vec{F}_y\\\\\vec{F}_{12}=4.30\times 10^{-11}\,\hat{i} + 3.62\times 10^{-11}\,\hat{j}\\\\\vec{F}_{12}=

Force exerted on charge -2e is equal in magnitude to F12 but is in opposite direction

F_{21}=-5.63\times 10^{-11}N

\vec{F}_{21}=

7 0
3 years ago
A _____ map provides information about how and when rocks formed in a particular area.A.contourB.geologicC.topographic
givi [52]

I believe the answer is B

6 0
3 years ago
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