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Doss [256]
3 years ago
14

A container is filled to a depth of 22.0 cm with water. On top of the water floats a 28.0-cm-thick layer of oil with specific gr

avity 0.500. What is the absolute pressure at the bottom of the container? 1 Pa
Physics
1 answer:
Ilya [14]3 years ago
5 0

Answer:104.853 kPa

Explanation:

Given

Depth of water h_2=22 cm

on the top of water a layer of h_1=28 cm oil layer is present

specific gravity of oil \rho _o=0.5\rho _w

Assuming Atmospheric Pressure to be 101.325 KPa

We know Pressure due to Pressure Difference is given by

\Delta P=\rho \times g\times h

\Delta P=\rho _0\times g\times h_1+\rho _w\times g\times h_2

\Delta P=0.5\rho _w\times g\times h_1+\rho _w\times g\times h_2

\Delta P=0.5\times 1000 \times 9.8\times 0.28+1000\times 9.8\times 0.22

\Delta P=1.372\times 10^3+2.156\times 10^3

\Delta P=3.528 kPa

Absolute\ Pressure=Atmospheric\ Pressure+\Delta P=101.325+3.528

Absolute\ Pressure =101.325+3.528=104.853 kPa

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The answer is C sorry if it’s wrong
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3 years ago
A 3.6-volt battery is used to operate a cell phone
Marina CMI [18]

Explanation :

It is given that,

Potential energy, V=3.6\ V

Power dissipated,  P=0.064\ Watt

We know that the power dissipated is given by :

P=VI

I is the current passing through the phone.

I=\dfrac{P}{V}

I=\dfrac{0.064\ W}{3.6\ V}

I=0.017\ A

or

I = 0.018 A

Hence, the current that passes through the phone is (1) 0.018 A.

3 0
3 years ago
Read 2 more answers
Sam, whose mass is 75 kg , takes off across level snow on his jet-powered skis. The skis have a thrust of 160 N and a coefficien
Volgvan

Answer:

top speed = 17.25

Total height = 281.19 m

Explanation:

given data

mass = 75 kg

thrust = 160 N

coefficient of kinetic friction = 0.1

solution

we get here frictional force acting that is

frictional force = \mu *m*g   .............1

frictional force = 0.1 × 75 × 9.8

frictional force = 73.5 N

and

Net force acting will be F = 160 - 73.5  N

F = 86.5 N

so

Acceleration in the First 15 second  will be

F = ma .........2

86.5 = 75 × a

a = 1.15 m/s²

and

now After 15 second the velocity will be  as

v = u + at   ..........3

here u is 0

so v will be

V = 1.15 × 15

v = 17.25

and

now we get travels distance S  in 15 s

s = u × t + 0.5 × a × t²  

here u is 0

so distance s will be

s = 0.5 × a × t²  

s = 0.5 ×  1.15 × 15²  

s = 129.37 m

and

now  acceleration acting is

F =  \mu *m*g  

m a =  \mu *m*g

a = \mu* g

a = - 0.98

here it is negative it mean downward nature of acceleration

and

now we get distance s by this formula

V² - u² = 2 a s    

here v velocity is 0  and

u initial velocity is 17.25 m/s

put here value

0 - 17.25² = 2 × (-0.98) × s    

solve it we get

s = 151.82 m

so

Total height is

Total height = 129.37 m + 151.82 m

Total height = 281.19 m

7 0
3 years ago
an object on a planet has a mass of 243 Kg. what is the acceleration of the object, if the radius of the planet is 2.32 x 10^7m
Ksenya-84 [330]

Answer:

7.87x10^5m/s^2

Explanation:

3 0
3 years ago
A man pushes a 10kg box with a constant acceleration of 5m/s2. What force is applied to the box?
drek231 [11]
FORMULA

\boxed {F = m \times a}

F = force
m = mass
a = acceleration

Using the formula

F = 10 \times 5

Multiply

\boxed {\textsf {F = 10 N}}
8 0
3 years ago
Read 2 more answers
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