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2 years ago

9

A substance has a freezing point of -36 degree C and a boiling point of 356 degree C. At what temperature would this substance b

e in its liquid state?

1 answer:

Travka [436]2 years ago

3 0

Answer:

Between -35 degrees C and 355 degrees C

Explanation:

At any temperature between, but not including the freezing and boiling points, the substance would be in its liquid state.

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Why is it, that no matter what we do in life, we die.

shutvik [7]

Answer:

the time comes eventually.

Explanation:

ur body just be giving up

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2 years ago

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3 years ago

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Find the magnitude of the electric field due to a charged ring of radius "a" and total charge "Q", at a point on the ring axis a

34kurt

Answer:

E= $\frac{KQ}{2\sqrt 2a^2}$

Explanation:

We are given that

Charge on ring= Q

Radius of ring=a

We have to find the magnitude of electric filed on the axis at distance a from the ring's center.

We know that the electric field at distance x from the center of ring of radius R is given by

$E=\frac{kQx}{(R^2+x^2)^{\frac{3}{2}}}$

Substitute x=a and R=a

Then, we get

$E=\frac{KQa}{(a^2+a^2)^{\frac{3}{2}}}$

$E=\frac{KQa}{(2a^2)^{\frac{3}{2}}}$

$E=\frac{KQa}{2\sqrt 2a^3}$

$E=\frac{KQ}{2\sqrt 2a^2}$

Where K= $9\times 10^9 Nm^2/C^2$

Hence, the magnitude of the electric filed due to charged ring on the axis of ring at distance a from the ring's center= $\frac{KQ}{2\sqrt 2a^2}$

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3 years ago

base your answer to this question on the information below and on your knowledge of physics. A toy launcher that is used to laun

Nadya [2.5K]

Answer: v = 2.24 m/s

Explanation: The <u>Law</u> <u>of</u> <u>Conservation</u> <u>of</u> <u>Energy</u> states that total energy is constant in any process and, it cannot be created nor destroyed, only transformed.

So, in the toy launcher, the energy of the compressed spring, called <u>Elastic</u> <u>Potential</u> <u>Energy (PE)</u>, transforms into the movement of the plastic sphere, called <u>Kinetic</u> <u>Energy (KE)</u>. Since total energy must be constant:

$KE_{i}+PE_{i}=KE_{f}+PE_{f}$

where the terms with subscript i are related to the initial of the process and the terms with subscript f relates to the final process.

The equation is calculated as:

$\frac{1}{2}kx^{2}+0=0+\frac{1}{2}mv^{2}$

$\frac{1}{2}kx^{2}=\frac{1}{2}mv^{2}$

$\frac{1}{2}50(0.1)^{2}=\frac{1}{2}(0.1)v^{2}$

$v^{2}=\frac{50(0.1)^{2}}{0.1}$

$v=\sqrt{50(0.1)}$

$v=\sqrt{5}$

v = 2.24

The maximum speed the plastic sphere will be launched is 2.24 m/s.

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3 years ago

The barometric pressure at sea level is 30 in of mercury when that on a mountain top is 29 in. If the specific weight of air is

stealth61 [152]

To solve this problem we will apply the concepts related to pressure, depending on the product between the density of the fluid, the gravity and the depth / height at which it is located.

For mercury, density, gravity and height are defined as

$\rho_m = 846lb/ft^3$

$g = 32.17405ft/s^2$

$h_1 = 1in = \frac{1}{12} ft$

For the air the defined properties would be

$\rho_a = 0.0075lb/ft^3$

$g = 32.17405ft/s^2$

$h_2 = ?$

We have for equilibrium that

$\text{Pressure change in Air}=\text{Pressure change in Mercury}$

$\rho_m g h_1 = \rho_a g h_2$

Replacing,

$(846)(32.17405)(\frac{1}{12}) = (0.0075)(32.17405)(h_2)$

Rearranging to find $h_2$

$h_2 = \frac{(846)(32.17405)(\frac{1}{12}) }{(0.0075)(32.17405)}$

$h = 9400ft$

Therefore the elevation of the mountain top is 9400ft

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3 years ago