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3 years ago

Which of the following will cause an increase in gas pressure in a closed container?

Physics

1 answer:

Volgvan3 years ago

5 0

D. both b. and c. Obviously adding more gas but reducing the volume because of the limited space in the container.

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PLEASE HELP ME 20 POINTS AND I WILL MARK BRAINLIEST!!!

Andreyy89

<span>B. a homogeneous mixture in which one substance is dissolved into another</span>

7 0

3 years ago

Read 2 more answers

Describe metallurgy. Check all that apply.

mylen [45]

Answer:

A, C, D

Explanation:

got it right

7 0

3 years ago

Compare the relative strengths of the nuclear force and the electric force

Serggg [28]

Answer:

To establish this relationship we must examine the potentials that these forces create. The electrical potential is described by

Ve = k q / r

The potential for strong nuclear force is

Vn (r) = - gs / 4pir exp (-mrc / h)

Where gs is the stacking constant and r the distance between the nucleons,

We can compare these potentials where the force is derived from the relationship

E = -dU / dr

F = q E

Explanation:

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3 years ago

You need to determine the density of a ceramic statue. if you suspend it from a spring scale, the scale reads 32.4 n . if you th

Mandarinka [93]

Well, the density of the water is
${1000 \frac{kg}{ {m}^{3} } }$
so i believe that is what the question is asking for :)

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3 years ago

A 560-g squirrel with a surface area of 930cm2 falls from a 5.0-m tree to the ground. Estimate its terminal velocity. (Use a dra

bija089 [108]

Explanation:

Terminal velocity is given by:

$v_t=\sqrt{\frac{2mg}{\rho C_dA}}$

Here, m is the mass of the falling object, g is the gravitational acceleration, $C_d$ is the drag coefficient, $\rho$ is the fluid density through which the object is falling, and A is the projected area of the object. in this case the projected area is given by:

$A=\frac{A_s}{2}=\frac{930cm^2}{2}=465cm^2\\465cm^2*\frac{1m^2}{10^4cm^2}=0.0465m^2\\560g*\frac{1kg}{10^3g}=0.56kg$

Recall that drag coefficient for a horizontal skydiver is equal to 1 and air density is $1.28\frac{kg}{m^3}$ .

$v_t=\sqrt{\frac{2(0.56kg)(9.8\frac{m}{s^2})}{(1.28\frac{kg}{m^3}(1)(0.0465m^2)}}\\v_t=13.58\frac{m}{s}$

Without drag contribution the motion of the person is an uniformly accelerated motion, thus:

$v_f^2=v_o^2+2gh\\v_f=\sqrt{2gh}\\v_f=\sqrt{2(9.8\frac{m}{s^2})(5m)}\\v_f=9.9\frac{m}{s}$

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3 years ago