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zavuch27 [327]
3 years ago
6

A U-tube with a cross-sectional area of 1.00 cm2 is open to the atmosphere at both ends. Water is poured into the tube until the

water rises part-way along the straight sides, and then 5.00 cm3 of oil is poured into one end. As a result, the top surface of the oil ends up 0.550 cm higher than the water surface on the other side of the U. What is the density of the oil
Physics
1 answer:
Doss [256]3 years ago
8 0

Answer:

0.89 g/cm^3 = 890 kg/m^3

Explanation:

Cross sectional area of U-tube ( A ) = 1.00 cm^2

volume of oil ( V )  = 5.00 cm^3

change between top surface = 0.550 cm

height of oil = 5 cm  ( volume / area )

height of water = 5 - 0.550 = 4.45 cm

pressure at the oil-water junction = Pressure on the second side of the U-tube at same level

Po * g * Hoil = Pw * g * Hwater

Po * 5 = 1 * 4.45

∴ Density of oil ( Po ) = 4.45 / 5  g/cm^3 = 0.89 g/cm^3

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Assertion(A):The distance moved by an object in unit time is called its speed. Reason (R):Faster vehicles have higher speeds. i)
ikadub [295]

Answer:

it's ii) R is correct while A is incorrect

Explanation:

cuz, both statements are correct but the reason is not the correct reason for the assertion,

8 0
3 years ago
Estimate the average rate of change in elevation from d to points i, ii, and iii, assuming the distance from d to i is 3500 m, t
Anit [1.1K]
Ummm i am not going to be able say i am high

8 0
3 years ago
Can i get help asap im taking a test
Ainat [17]

Answer:

4 seconds per meter

Explanation:

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5 0
4 years ago
If I wanted to generate a maximum emf of 20 V, what angular velocity (radians/sec, aka Hz) would be required given a circular co
sergij07 [2.7K]

Answer:

Angular velocity, \omega=35.36\ rad/s

Explanation:

It is given that,

Maximum emf generated in the coil, \epsilon=20\ V

Diameter of the coil, d = 40 cm

Radius of the coil, r = 20 cm = 0.2 m

Number of turns in the coil, N = 500

Magnetic field in the coil, B=9\times 10^{-3}\ T

The angle between the area vector and the magnet field vector varies from 0 to 2 π radians. The formula for the maximum emf generated in the coil is given by :

\epsilon=NBA\omega

\omega=\dfrac{\epsilon}{NBA}

\omega=\dfrac{20\ V}{500\times 9\times 10^{-3}\ T\times \pi (0.2\ m)^2}

\omega=35.36\ rad/s

So, the angular velocity of the circular coil is 35.36 rad/s. Hence, this is the required solution.

6 0
3 years ago
A brave child decides to grab onto an already spinning merry‑go‑round. The child is initially at rest and has a mass of 34.5 kg.
rodikova [14]

Answer:

the moment of inertia of the merry go round is 38.04 kg.m²

Explanation:

We are given;

Initial angular velocity; ω_1 = 37 rpm

Final angular velocity; ω_2 = 19 rpm

mass of child; m = 15.5 kg

distance from the centre; r = 1.55 m

Now, let the moment of inertia of the merry go round be I.

Using the principle of conservation of angular momentum, we have;

I_1 = I_2

Thus,

Iω_1 = I'ω_2

where I' is the moment of inertia of the merry go round and child which is given as I' = mr²

Thus,

I x 37 = ( I + mr²)19

37I = ( I + (15.5 x 1.55²))19

37I = 19I + 684.7125

37I - 19 I = 684.7125

18I = 684.7125

I = 684.7125/18

I = 38.04 kg.m²

Thus, the moment of inertia of the merry go round is 38.04 kg.m²

7 0
4 years ago
Read 2 more answers
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