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2 years ago

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3. Atmospheric pressure is often given in units of millimeters of mercury. This refers to the height of a mercury column above t

he mercury's surface at the base of a barometer . The force exerted by the atmosphere on the surface of the mercury in the reservoir equals the weight of the mercury in the column . If the mercury column extends 760 mm above the mercury's surface in the reservoir , what is the atmosphere's pressure over the mercury ? Use 13.6*10^ 3 kg/m^ 3 for the density of mercury

1 answer:

gregori [183]2 years ago

6 0

Answer:

101397.16 pa

Explanation:

The pressure recorded will be equal to pgh

Where p = density of mercury = 13.6x10^3 kg/m^ 3

g = acceleration due to gravity 9.81 m/s^2

h = height of mercury in the column = 760 mm = 760x10^-3 m

Pressure = 13.6x10^3 x 9.81 x 760x10^-3 = 101397.16 pa

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A 3-cm high object is in front of a thin lens. The object distance is 4 cm and the image distance is –8 cm. (a) What is the foca

xenn [34]

Answer:

a) Focal length of the lens is 8 cm which is a convex lens

b) 6 cm

c) The lens is a convex lens and produces a virtual image which is upright and two times larger than the object.

Explanation:

u = Object distance = 4 cm

v = Image distance = -8 cm

f = Focal length

Lens Equation

$\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\\\Rightarrow \frac{1}{f}=\frac{1}{4}+\frac{1}{-8}\\\Rightarrow \frac{1}{f}=\frac{1}{8}\\\Rightarrow f=\frac{8}{1}=-8\ cm$

a) Focal length of the lens is 8 cm which is a convex lens

Magnification

$m=-\frac{v}{u}\\\Rightarrow m=-\frac{-8}{4}\\\Rightarrow m=2$

b) Height of image is 2×3 = 6 cm

Since magnification is positive the image upright

c) The lens is a convex lens and produces a virtual image which is upright and two times larger than the object.

8 0

2 years ago

(a) Aircraft sometimes acquire small static charges. Suppose a supersonic jet has a 0.500 - μC charge and flies due west at a sp

12345 [234]

(a) $2.64\cdot 10^{-8} N$ north

We can treat the aircraft as a single point charge moving in a magnetic field. In this case, the magnetic force exerted on the plane is

$F=qvB sin \theta$

where

$q=0.500 \mu C = 0.500\cdot 10^{-6} C$ is the charge on the plane

v = 660 m/s is the velocity

$B=8.00\cdot 10^{-5} T$ is the magnitude of the magnetic field

$\theta=90^{\circ}$ is the angle between the direction of motion of the jet and of the magnetic field

Substituting,

$F=(0.5\cdot 10^{-6})(660)(8.0\cdot 10^{-5})=2.64\cdot 10^{-8} N$

The direction can be found by using Fleming's left hand rule. We have:

- index finger: magnetic field direction (straight up)

- middle finger: velocity of the plane (due west)

- force: thumb --> north

(b) Not negligible

As we can see from part (a), the magnitude of the force is not really big, so the effects are negligible.

For instance, we can compare this force with the weight of a plane. If we take a Boeing 737, its mass is about 80,000 kg, so its weight is

$W=mg=(80000)(9.8)=784,000 N$

As we can see, this is several orders of magnitude bigger than the magnetic force calculated at point (a), so the effects of the magnetic force are negligible.

8 0

2 years ago

A fisherman has caught a very large, 5.0kg fish from a dock that is 2.0m above the water. He is using lightweightfishing line th

agasfer [191]

Answer:

t = 2 s

Explanation:

As we know that fish is pulled upwards with uniform maximum acceleration

then we will have

$T - mg = ma$

here we know that maximum possible acceleration of so that string will not break is given as

$T = 54 N$

now we have

$54 - (5 \times 9.8) = 5 a$

$a = 1 m/s^2$

now for such acceleration we can use kinematics

$d = \frac{1}{2}at^2$

$2 = \frac{1}{2}(1) t^2$

t = 2 s

7 0

2 years ago

You have negotiated with the Omicronians for a base on the planet Omicron Persei 7. The architects working with you to plan the

steposvetlana [31]

Answer:

5.724 meters / second^2

Explanation:

We are given two pieces of information, 5.24 flurg = 1 meter, 1 grom = 0.493 second. If that is so, we can say that there are two possible conversion units, 5.25 flurg / meter, and 0.493 second / grom.

_____

We want to convert 7.29 flurg / grom^2 ( I believe? ) to the units meters / second^2. But, let's break this down into bits. It would be convenient to first convert 7.29 flurg / grom^2 to the units meters / grom^2, by dividing the conversion factors as to cancel out the appropriate things, which we will go into detail on a bit later ( using the first conversion factor ). Respectively we can convert meters / grom^2 to meters / grom * s, canceling out the flurg ( through the second conversion factor ). And now we would need to get rid of the grom, dividing similarly.

_____

( 1 ) ( flurg / grom^2 ) / ( flurg / meters ) - first conversion unit

= flurg / grom^2 * meters /flurg

= ( meters * flurg ) / ( grom^2 * flurg )

= meters /grom^2,

7.29 flurg / grom^2 / 5.24 flurg / meter = ( About ) 1.39 meter / grom^2

( 2 ) ( meter / grom^2 ) / ( second / grom ) - second conversion unit

= meter / grom^2 * grom / second

= ( meter * grom ) / ( grom^2 * second )

= meter / ( grom * second ),

( 1.39 meter / grom^2 ) / 0.493 second / grom = ( About ) 2.82195 meter / grom * second

( 3 ) ( 2.82195 meter / ( grom * second ) ) / 0.493 second / grom = 5.724 meter / second^2

( And thus, the value of gOP7 in the units the architects will use should be about 5.724 meters / second^2 )

8 0

2 years ago

Argon in the amount of 1.5 kg fills a 0.04-m3 piston cylinder device at 550 kPa. The piston is now moved by changing the weights

Arlecino [84]

Answer:

275 kPa

Explanation:

mass of the gas=m=1.5 kg

initial volume if the gas=V₁=0.04 m³

initial pressure of the gas= P₁=550 kPa

as the condition is given final volume is double the initial volume

V₂=final volume

V₂=2 V₁

As the temperature is constant

T₁=T₂=T

$\frac{P1V1}{T1}$ = $\frac{P2 V2}{T2}$

putting the values in the equation.

$\frac{P1V1}{T1}$ = $\frac{P2 *2V1}{T2}$

P₂= $\frac{P1}{2}$

P₂= $\frac{550}{2}$

P₂=275 kPa

So the final pressure of the gas is 275 kPa.

3 0

2 years ago