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butalik [34]
3 years ago
11

A student finds an unlabeled liquid container in his lab. He notices that the container has two liquids. Since the two liquids h

ave different density, there is a distinct separation between them. Water, which has a density of 1.00g/cm^3, fills the lower portion of the container to a depth of 20.0 cm.The fluid which is floating on top of the water is 30.7 cm deep. If the absolute pressure on the bottom of the container is 104900 Pa, what is the density ???? of the unknown fluid? The acceleration due to gravity is g=9.81 m/s^2 and atmospheric pressure is P0=101300 Pa.
Physics
1 answer:
raketka [301]3 years ago
5 0

Answer:

\rho = 1848.03 kg m^{-3}

Explanation:

given data:

density of water \rho = 1 gm/cm^3 = 1000 kg/m^3

height  of water  = 20 cm  =0.2 m

Pressure  p = 1.01300*10^5 Pa

pressure at bottom

P =  P_{fluid} + P_{h_2 o}

P   = P_{fluid}  + \rho g h

P_{fluid}  = P - \rho g h

                 = 1.01300*10^5 - 1000*0.2*9.8

                 = 99340 Pa

p_{fluid}  = P_{atm} + \rho g h_{fluid}                       h_[fluid} = 0.307m

99340 = 104900 + \rho *9.8*0.307

\rho = 1848.03 kg m^{-3}

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Answer:c

Explanation:

7 0
3 years ago
What is the resistance of : A) A 1.70 m long copper wire that is 0.700 mm in diameter? B) A 20.0 cm long piece of carbon with a
astra-53 [7]

Answer:

(I). The resistance of the copper wire is 0.0742 Ω.

(II). The resistance of the carbon piece is 1.75 Ω.

Explanation:

Given that,

Length of copper wire = 1.70 m

Diameter = 0.700 mm

Length of carbon piece = 20.0 cm

Cross section areaA = (2.00\times10^{-3})^2\ m

(I). We need to calculate the area of copper wire

Using formula of area

A=\pi r^2

A=3.14\times(\dfrac{0.700\times10^{-3}}{2})^2

We need to calculate the resistance

Using formula of resistance

R=\dfrac{\rho l}{A}

Put the value into the formula

R=\dfrac{1.68\times10^{-8}\times1.70}{3.14\times(\dfrac{0.700\times10^{-3}}{2})^2}

R=0.0742\ \Omega

(II). We need to calculate the resistance

Using formula of resistance

R=\dfrac{\rho l}{A}

Put the value into the formula

R=\dfrac{3.5\times10^{-5}\times20\times10^{-2}}{(2.00\times10^{-3})^2}

R=1.75\ \Omega

Hence, (I). The resistance of the copper wire is 0.0742 Ω.

(II). The resistance of the carbon piece is 1.75 Ω.

8 0
3 years ago
We have an Atwood device, two blocks connect by a string strung over a pulley, but the twist this time is that both blocks are o
Zanzabum

The Acceleration of the system is 6.41 m/s².

Given,

α= 15°, m₁ = 7kg

β= 65°, m₂ = 11 kg

Let, a be the acceleration and T is the tensions at the end it's the cord.

Let, the mass m₂ be coming down along the inclined plane along the inclined surface towards downward m₂g sin β and the tension in the upward direction,

Resultant force, m₂a=m₂g sin β -T

11a=((11) ×g sin 65°) -T  ...(i)

Now, considering the motion of m₁ which moves downwards, the forces are m₁g sinα, and T both are acting downwards.

Resultant force m₁a = m₁g sin α+T

7a =7g sin 15°+T  ...(ii)

Solving both the equations by adding them,

18a=11gsin 65°+7g sin 15°-T+T

18a=11gsin 65°+7g sin 15°=115.45

a=115.45/18=6.41 m/s²

Hence, the Acceleration of the system is 6.41 m/s².

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6 0
2 years ago
I would like help with this physics problem
Darina [25.2K]

(a) This is a freefall problem in disguise - when the ball returns to its original position, it will be going at the same speed but in the opposite direction. So the ball's final velocity is the negative of its initial velocity.

Recall that

v_f=v_i+at

We have v_f=-v_i, so that

-2v_i=at\implies-2\left(8\,\dfrac{\mathrm m}{\mathrm s}\right)=\left(-2\,\dfrac{\mathrm m}{\mathrm s^2}\right)t\implies t=8\,\mathrm s

(b) The speed of the ball at the start and at the end of the roll are the same 8 m/s, so the average speed is also 8 m/s.

(c) The ball's average velocity is 0. Average velocity is given by \dfrac{v_i+v_f}2, and we know that v_f=-v_i.

(d) The position of the ball x_f at time t is given by

x_f=x_i+v_it+\dfrac12at^2

Take the starting position to be the origin, x_i=0. Then after 6 seconds,

x_f=\left(8\,\dfrac{\mathrm m}{\mathrm s}\right)(6\,\mathrm s)+\dfrac12\left(-2\,\dfrac{\mathrm m}{\mathrm s^2}\right)(6\,\mathrm s)^2=42\,\mathrm m

so the ball is 42 m away from where it started.

We're not asked to say in which direction it's moving at this point, but just out of curiosity we can determine that too:

x_f-x_i=\dfrac{v_i+v_f}2t\implies42\,\mathrm m=\dfrac{8\,\frac{\mathrm m}{\mathrm s}+v_f}2(6\,\mathrm s)\implies v_f=6\,\dfrac{\mathrm m}{\mathrm s}

Since the velocity is positive, the ball is still moving up the incline.

8 0
3 years ago
How to solve this step by step
victus00 [196]
Well, it goes 60 miles in one hour......so set up a ratio.....
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7 0
2 years ago
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