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Liula [17]
3 years ago
14

Suppose you pour water into a container until it reaches a depth of 12 cm. Next, you carefully pour in an 8.5 cm thickness of ol

ive oil so that it floats on top of the water.
What is the pressure at the bottom of the container?
Physics
1 answer:
Phantasy [73]3 years ago
3 0

Answer:

the pressure at the bottom is approximately 103264 Pa

Explanation:

From Pascal's law, the pressure at the bottom of the container is the pressure from the atmosphere and the columns of water and olive , therefore

Pressure at the bottom = Pressure at the surface of the liquid ( atmospheric pressure) + pressure of the column of water + pressure of the column of olive oil

since

- Pressure at the surface of the liquid = atmospheric pressure = 101325 Pa

- pressure of the column of water = density of water * gravity * level of water

= 1000 kg/m³ * 9.8 m/s² * 0.12 m = 1176 Pa

- pressure of the column of olive oil=  density of olive oil* gravity * level of olive oil = 916 kg/m³ * 9.8 m/s² * 0.085 m = 763 Pa

therefore

Pressure at the bottom = 101325 Pa + 1176 Pa + 763 Pa = 103264 Pa

density of olive oil was taken from internet sources ( we can check that is lower than the one of water, and thus it floats )

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A body moving with an initial velocity of 30m/s accelerates uniformly at the rate of 10m/s . what is the distance covered during
nikdorinn [45]

Answer:

The distance covered by the body is, S = 800 m

Explanation:

Given data,

The initial velocity of the body, u = 30 m/s

The acceleration of the body, a = 10 m/s²

Let the time period of travel be, t = 10 s

Using the II equations of motion,

                       S = ut + ½ at²

Substituting the given values,

                        S = 30 x 10 + ½ x 10 x 10²

                         S = 800 m

Hence, the distance covered by the body is, S = 800 m

5 0
3 years ago
The energy equivalent of the rest mass of an electron is
DerKrebs [107]

Answer:energy times mass

Explanation: yeh

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3 years ago
A 90-meter train travels at a constant speed of 10 meters per second. How long will it take to cross a 0.06 bridge?
salantis [7]

Answer:

The time taken for the train to cross the bridge is 9.01 s

Explanation:

Given;

length of the train, L₁ = 90 m

length of the bridge, L₂ = 0.06 m

speed of the train, v = 10 m/s

Total distance to be traveled, = L₁ + L₂

                                                  = 90 m + 0.06 m

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Time of motion = Distance / speed

Time of motion = 90.06 / 10

Time of motion = 9.006 s ≅ 9.01 s

Therefore, the time taken for the train to cross the bridge is 9.01 s

5 0
3 years ago
What is the momentum of a 50 at a speed of 5m/s
aleksandr82 [10.1K]
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A resistor and a capacitor are connected in series across an ideal battery having a constant voltage across its terminals. (a) A
mariarad [96]

Answer:

The question is not complete. see the complete question in the explanation section. The correct option is highlighted in bold

Explanation:

(a)A resistor and a capacitor are connected in series across an ideal battery having a constant voltage across its terminals. At the moment contact is made with the battery, the voltage across the resistor is

I.     greater than the battery's terminal voltage.  

II.    equal to the battery's terminal voltage.  

III.  less than the battery's terminal voltage, but greater than zero.  

IV.  zero.

<em>Option (i) is not correct as the voltage across the resistor cannot be greater than the terminal voltage since the current is yet to flow through the resistor. Option (ii) is correct as both the resistor voltage and the terminal voltage will just equal at the instance of connection. Option (ii) can only be possible after the current must have passed through the resistor for a while not immediately after contact. Option (iv) is not correct, as this can only be possible is the contact is open. </em>

(b)A resistor and a capacitor are connected in series across an ideal battery having a constant voltage across its terminals. At the moment contact is made with the battery the voltage across the capacitor is  

I.   greater than the battery's terminal voltage.  

II.  equal to the battery's terminal voltage.  

III. less than the battery's terminal voltage, but greater than zero.  

IV. zero.

<em>Option (i) is not correct as the capacitor is yet to charge talk less of the its voltage exceeding that of the battery. Option (ii) can only be correct if the capacitor is fully charged not when it has just been connected. Option (iii) can only occur if the capacitor is discharging. Option (iv) is the correct answer as the capacitor is about to start charging </em>

8 0
4 years ago
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