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3 years ago

Consider the solubility graph above. According to the prompt, at 105oC, which of the solutes would still be unsaturated?

Physics

1 answer:

garik1379 [7]3 years ago

4 0

Answer: D. KNO3

Explanation:

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Calculate the displacement of a bee that flew out of the beehive due east for 4 km, went around a 0.52 m in diameter poplar tree

stiks02 [169]

Answer:

the only thing thats confusing me is no ?

Explanation:

please explain thank you :)

3 0

3 years ago

Jacki evaluated the expression below. 2 cubed (3 minus 1) + 4 (8 minus 12) = 2 cubed (2) + 4 (4) = 8 (2) + 16 = 16 + 16 = 32. Wh

Naddika [18.5K]

Answer:

Her error was that she did not subtract 12 from 8 correctly

Explanation:

Jackie did 8-12 instead of 12-8

8 0

3 years ago

Calculate the magnitude of the particle accelerator's magnetic field that causes an ionized helium atom (+2q e) with a momentum

trasher [3.6K]

Answer:

The magnetic field of the particle is 1.5 T.

(C) is correct option.

Explanation:

Given that,

Momentum of particle $p=4.8\times10^{-16}\ kg m/s$

Radius = 1.0 km

Charge of the particle $q=2\times1.6\times10^{-19}\ C$

We need to calculate the magnetic field

Using relation of radius of path in magnetic field

$r=\dfrac{mv}{qB}$

Here mv = p

$B=\dfrac{p}{qr}$

Put the value into the formula

$B=\dfrac{4.8\times10^{-16}}{1000\times2\times1.6\times10^{-19}}$

$B=1.5\ T$

Hence, The magnetic field of the particle is 1.5 T.

3 0

3 years ago

What is Plancks constant (h)?

Tatiana [17]

Planck's constant is 4.39048042 × 10-67 m4 kg2 / s2.

6 0

4 years ago

Suppose you wish to fabricate a uniform wire from a mass m of a metal with density rhom and resistivity rho. If the wire is to h

slamgirl [31]

Answer:

Explanation:

Resistivity and resistance are proportional and depends of the length and the cross-sectional area of the wire:

$R=\frac{L}{A}\rho$

furthermore, the density is the mass divided by the volume, and the volume can be written as the area multiplyed by the length:

$\rho_m=\frac{m}{A*L}$

Now you have tw equations and two variables, so you can solve for each of them.

first, solve for A in both equations and replace them:

$\frac{L}{R}\rho=\frac{m}{\rho_mL}$

$L^2=\frac{mR}{\rho_m \rho} \\L=$

now replace this into any of the previous equiations:

$R=\frac{\sqrt{\frac{mR}{\rho_m \rho} } \rho}{A} \\A=\sqrt{\frac{mR\rho^2}{\rho_m \rho R^2} }\\A=\sqrt{\frac{m\rho}{\rho_m R} }$

If you assume the wire has circular cross-sectional area, then the area is:

$A=\pi(\frac{d}{2} )^2$

solving for d:

$d=2\sqrt{\frac{A}{\pi} }$

replacing A and simplifying:

$d=2 \sqrt[4]{\frac{m\rho}{\rho_m R \pi^2} }$

5 0

3 years ago