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JulijaS [17]
3 years ago
13

A example of a substance that has low pH is

Physics
1 answer:
iren2701 [21]3 years ago
6 0

Answer:

lemons, which have a pH of 2 choice a

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A 2 kg ball if at rest. If the ball accelerates to 20 m/s, what is the change in momentum?
katrin [286]

Answer:

change in momentum

20m/s x2=40kg/m

4 0
3 years ago
What potential difference is needed to give a helium nucleus (q=2e) 85.0 kev of kinetic energy?
Vilka [71]
The kinetic energy K given to the helium nucleus is equal to its potential energy, which is 
E=q \Delta V
where q=2e is the charge of the helium nucleus, and \Delta V is the potential difference applied to it.
Since we know the kinetic energy, we have
E=K=85~keV=q \Delta V
and from this we can find the potential difference:
\Delta V =  \frac{K}{q}= \frac{85~keV}{2e}=42.5~kV

6 0
3 years ago
Which of the following is not a use for a weather radar?
Elodia [21]
Weather radar is used to locate precipitation<span>, calculate its motion, and estimate its type rain, snow, </span>hail etc

so, d.measuring temperature <span>is not a use for a weather radar

</span>effect of global warming-<span>heavier rainfall and flooding</span>
5 0
3 years ago
Read 2 more answers
What is water potential???​
melisa1 [442]

Answer:

Water potential is the potential energy of water per unit volume relative to pure water in reference conditions. Water potential quantifies the tendency of water to move from one area to another due to osmosis, gravity, mechanical pressure and matrix effects such as capillary action.

3 0
3 years ago
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You are designing a 108 cm3 right circular cylindrical can whose manufacture will take waste into account. There is no waste in
FinnZ [79.3K]

Explanation:

It is given that,

The volume of a right circular cylindrical, V=108\ cm^3

We know that the volume of the cylinder is given by :

V=\pi r^2 h

108=\pi r^2 h    

h=\dfrac{108}{\pi r^2}............(1)

The upper area is given by :

A=32r^2+2\pi rh

A=32r^2+2\pi r\times \dfrac{108}{\pi r^2}

A=32r^2+\dfrac{216}{r}

For maximum area, differentiate above equation wrt r such that, we get :

\dfrac{dA}{dr}=64r-\dfrac{216}{r^2}

64r-\dfrac{216}{r^2}=0

r^3=\dfrac{216}{64}

r = 1.83 m

Dividing equation (1) with r such that,

\dfrac{h}{r}=\dfrac{108}{\pi r}

\dfrac{h}{r}=\dfrac{108}{\pi 1.83}

\dfrac{h}{r}=59 \pi

Hence, this is the required solution.

8 0
3 years ago
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