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2 years ago

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How high would the level be in a gasoline barometer at normal atmospheric pressure?

1 answer:

Sergio [31]2 years ago

7 0

Answer:

h = 13.06 m

Explanation:

Given:

- Specific gravity of gasoline S.G = 0.739

- Density of water p_w = 997 kg/m^3

- The atmosphere pressure P_o = 101.325 KPa

- The change in height of the liquid is h m

Find:

How high would the level be in a gasoline barometer at normal atmospheric pressure?

Solution:

- When we consider a barometer setup. We dip the open mouth of an inverted test tube into a pool of fluid. Due to the pressure acting on the free surface of the pool, the fluid starts to rise into the test-tube to a height h.

- The relation with the pressure acting on the free surface and the height to which the fluid travels depends on the density of the fluid and gravitational acceleration as follows:

P = S.G*p_w*g*h

Where, h = P / S.G*p_w*g

- Input the values given:

h = 101.325 KPa / 0.739*9.81*997

h = 13.06 m

- Hence, the gasoline will rise up to the height of 13.06 m under normal atmospheric conditions at sea level.

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elena-s [515]

Answer:

the horizontal distance is 4.355 meters

Explanation:

The computation of the horizontal distance while travelling in the air is shown below:

Data provided in the question is as follows

Velocity = u = 7.70 m/s

H = 1.60 m

R = horizontal direction

Based on the above information

As we know that

R = u × time

where,

Time = $\sqrt{\frac{2H}{g} }$

So,

= $7.70\times \sqrt{\frac{2\times 1.60}{10} }$

= 4.355 meters

hence, the horizontal distance is 4.355 meters

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2 years ago

sweet-ann [11.9K]

Answer:

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2 years ago

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A charge 3q is at the origin, and a charge −2q is on the positive x axis at x=a. part a where on the x-axis would you place a th

Mila [183]

Answer:

$x = a + \frac{a\sqrt2}{\sqrt3 - \sqrt2}$

Explanation:

Position of charge 3q is x = 0

position of charge -2q is x = a

so here we know that

when two charges are of opposite nature then the electric field will be zero on the line joining the two charges at the position near to smaller magnitude charge

So here the electric field will be zero if the field due to 3q is counterbalanced by field due to -2q

so here we can say

$\frac{k(3q)}{(r+a)^2} + \frac{k(-2q)}{r^2} = 0$

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

$\frac{r+a}{r} = \sqrt{\frac{3}{2}}$

$1 + \frac{a}{r} =\sqrt{\frac{3}{2}}$

$\frac{a}{r} = \frac{\sqrt3 - \sqrt2}{\sqrt2}$

so we will have

$r = \frac{a\sqrt2}{\sqrt3 - \sqrt2}$

so the x coordinate of this position is given as

$x = a + \frac{a\sqrt2}{\sqrt3 - \sqrt2}$

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3 years ago

The device used to measure a masses of a body is kilogram . true or false

olga nikolaevna [1]

Answer: false

Explanation:

While kilograms are the unit used to measure body mass, the device used is a scale.

Hope it helps :)

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2 years ago

A 0.20-kg mass is oscillating on a spring over a horizontal frictionless surface. When it is at a displacement of 2.6 cm for equ

valentinak56 [21]

Explanation:

The given data is as follows.

mass = 0.20 kg

displacement = 2.6 cm

Kinetic energy = 1.4 J

Spring potential energy = 2.2 J

Now, we will calculate the total energy present present as follows.

Total energy = Kinetic energy + spring potential energy

= 1.4 J + 2.2 J

= 3.6 Joules

As maximum kinetic energy of the object will be equal to the total energy.

So, K.E = Total energy

= 3.6 J

Also, we know that

K.E = $\frac{1}{2}mv^{2}_{m}$

or, v = $\sqrt{\frac{2K.E}{m}}$

= $\sqrt{2 \times 3.6 J}{0.2 kg}$

= $\sqrt{36}$

= 6 m/s

thus, we can conclude that maximum speed of the mass during its oscillation is 6 m/s.

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2 years ago