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shtirl [24]
3 years ago
6

A large bottle contains 150 L of water, and is open to the atmosphere. If the bottle has a flat bottom with an area of 2 ft, cal

culate the absolute pressure at the bottom of the bottle. Give your answer in Pa. Assume the water has a density of 1000 kg/m, and assume the system is at sea level on Earth. Answer: over 100,000 Pa
Physics
1 answer:
Yuri [45]3 years ago
6 0

Answer:

Total pressure exerted at bottom =  119785.71 N/m^2

Explanation:

given data:

volume of water in bottle = 150 L = 0.35 m^3

Area of bottle = 2 ft^2

density of water = 1000 kg/m

Absolute pressure on bottom of bottle will be sum of atmospheric pressure and pressure due to water

Pressure due to water P = F/A

F, force exerted by water = mg

m, mass of water = density * volume

                             =  1000*0.350 = 350 kg

F  = 350*9.8 = 3430 N

A = 2 ft^2 = 0.1858 m^2  

so, pressure P = 3430/ 0.1858 = 18460.71 N/m^2

Atmospheric pressure

At sea level atmospheric pressure is 101325 Pa

Total pressure exerted at bottom  = 18460.71 + 101325 = 119785.71 N/m^2

Total pressure exerted at bottom =  119785.71 N/m^2

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erastova [34]

A scientist would write that number as 1.49 x 10⁸ kilometers .

(Or, if the scientist is in France or the UK, he might write it as  1.49 x 10⁸ kilometres .)

6 0
3 years ago
A constant torque of 3 Nm is applied to an unloaded motor at rest at time t = 0. The motor reaches a speed of 1,393 rpm in 4 s.
irakobra [83]

Answer:

The moment of inertia of the motor is 0.0823 Newton-meter-square seconds.

Explanation:

From Newton's Laws of Motion and Principle of Motion of D'Alembert, the net torque of a system (\tau), measured in Newton-meters, is:

\tau = I\cdot \alpha (1)

Where:

I - Moment of inertia, measured in Newton-meter-square seconds.

\alpha - Angular acceleration, measured in radians per square second.

If motor have an uniform acceleration, then we can calculate acceleration by this formula:

\alpha = \frac{\omega - \omega_{o}}{t} (2)

Where:

\omega_{o} - Initial angular speed, measured in radians per second.

\omega - Final angular speed, measured in radians per second.

t - Time, measured in seconds.

If we know that \tau = 3\,N\cdot m, \omega_{o} = 0\,\frac{rad}{s }, \omega = 145.875\,\frac{rad}{s} and t = 4\,s, then the moment of inertia of the motor is:

\alpha = \frac{145.875\,\frac{rad}{s}-0\,\frac{rad}{s}}{4\,s}

\alpha = 36.469\,\frac{rad}{s^{2}}

I = \frac{\tau}{\alpha}

I = \frac{3\,N\cdot m}{36.469\,\frac{rad}{s^{2}} }

I = 0.0823\,N\cdot m\cdot s^{2}

The moment of inertia of the motor is 0.0823 Newton-meter-square seconds.

5 0
3 years ago
A load of mass 5kg is raised through a height of 2m. calculate the work done against (g=10mls)​
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The work done against gravity is 100 J

Explanation:

The work done against gravity in order to lift an object is equal to the change in gravitational potential energy of the object:

W=mg\Delta h

where

m is the mass of the object

g is the acceleration of gravity

\Delta h is the change in height of the object

For the object in this problem, we have:

m = 5 kg

g=10 m/s^2

\Delta h = 2 m

Substituting into the equation,

W=(5)(10)(2)=100 J

Learn more about work:

brainly.com/question/6763771

brainly.com/question/6443626

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3 0
3 years ago
The Hall effect can be used to determine the density of mobile electrons in a conductor. A thin strip of the material being inve
solmaris [256]

Answer:

the density of mobile electrons in the material is 3.4716 × 10²⁵ m⁻³

Explanation:

Given the data in the question;

we make use of the following expression;

hall Voltage VH = IB / ned

where I = 2.25 A

B = 0.685 T

d =  0.107 mm =  0.107 × 10⁻³ m

e = 1.602×10⁻¹⁹ C

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so from the form; VH = IB / ned

VHned = IB

n = IB / VHed

so we substitute

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n = 3.4716 × 10²⁵ m⁻³

Therefore, the density of mobile electrons in the material is 3.4716 × 10²⁵ m⁻³

5 0
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kati45 [8]

Answer:

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6 0
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