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shusha [124]
3 years ago
15

Intermolecular forces hold

Physics
1 answer:
blagie [28]3 years ago
3 0
I believe Intermolecular forces hold, <span>molecules, ions, and atoms? But I would see if that doesn't sound familiar check it with a site or something?</span>
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Which of the following is not a way to make exercising on a cold day safer?
IRISSAK [1]

Answer:

D.

Explanation:

I think its D. because Tea during a cold workout isnt safe at all. Drinking tea while working out adds weight and you just simply dont need tea when working out.. Hope this helps man :)

4 0
3 years ago
Read 2 more answers
Which example represents the impact of religion on culture in the United States? A. children receiving counseling at school B. f
dmitriy555 [2]

Answer:

D. Banks and offices closing on sunday.

Explanation:

Church is usually on sunday and people need a day off to do it because service is about 3 hours long. When banks and offices close on this day it shows how religion has affected our lives and our culture.

5 0
3 years ago
Two point charges are fixed on the y axis: a negative point charge q1 = -25 μC at y1 = +0.18 m and a positive point charge q2 at
dedylja [7]

Answer:

50.91 \mu C

Explanation:

The magnitude of the net force exerted on q is known, we have the values and positions for q_{1} and q. So, making use of coulomb's law, we can calculate the magnitude of the force exerted byq_{1} on q. Then we can know the magnitude of the force exerted by q_{2} about q, finally this will allow us to know the magnitude of q_{2}

q_{1} exerts a force on q in +y direction, and q_{2} exerts a force on q in -y direction.

F_{1}=\frac{kq_{1} q }{d^2}\\F_{1}=\frac{(8.99*10^9)(25*10^{-6}C)(8.4*10^{-6}C)}{(0.18m)^2}=58.26 N\\

The net force on q is:

F_{T}=F_{1} - F_{2}\\25N=58.26N-F_{2}\\F_{2}=58.26N-25N=33.26N\\\mid F_{2} \mid=\frac{kq_{2}q}{d^2}

Rewriting for q_{2}:

q_{2}=\frac{F_{2}d^2}{kq}\\q_{2}=\frac{33.26N(0.34m)^2}{8.99*10^9\frac{Nm^2}{C^2}(8.4*10^{-6}C)}=50.91*10^{-6}C=50.91 \mu C

8 0
3 years ago
A balloon is filled to a volume of 7.00*10^2 mL at a temperature of 20.0°C. The balloon is then cooled at constant pressure to a
wolverine [178]

The final volume of the gas is 238.9 mL

Explanation:

We can solve this problem by using Charle's law, which states that for a gas kept at constant pressure, the volume of the gas (V) is proportional to its absolute temperature (T):

\frac{V}{T}=const.

Which can be also re-written as

\frac{V_1}{T_1}=\frac{V_2}{T_2}

where

V_1, V_2 are the initial and final volumes of the gas

T_1, T_2 are the initial and final temperature of the gas

For the gas in the balloon in this problem, we have:

V_1 = 7.00\cdot 10^2 mL = 700 mL is the initial volume

T_1=20.0^{\circ}C+273=293 K is the initial absolute temperature

V_2 is the final volume

T_2 = 1.00\cdot 10^2 K = 100 K is the final temperature

Solving for V_2,

V_2 = \frac{V_1 T_2}{T_1}=\frac{(700)(100)}{293}=238.9 mL

Learn more about ideal gases:

brainly.com/question/9321544

brainly.com/question/7316997

brainly.com/question/3658563

#LearnwithBrainly

6 0
3 years ago
Which statement about magnetic poles is accurate?
Mademuasel [1]
I see I'm late but I believe it's A. , opposites attract. I'm taking the test right now lol
6 0
3 years ago
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