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2 years ago

E. Describe in short the structure of a mercury barometer

Physics

1 answer:

Readme [11.4K]2 years ago

6 0

Answer:

A mercury barometer has a glass tube that is closed at the top and open at the bottom.At the bottom of the tube is a pool of mercury.The mercury sits in a circular,shallow dish surrounding the tube.The mercury in the tube will adjust itself to match the atmospheric pressure above the dish.

hope this helps you

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1. A liquid of mass 250g is heated with an electric heater. Its temperature rises from 30°C to 80°C, the specific heat capacity

statuscvo [17]

Answer:

1) 50 seconds 2) 100°C

Explanation:

(Follows formula of Power=Energy/Time)

1) 500W x X = 2000J/kg°C x .25kg x 50°C

X = 50 seconds.

2) 2000W x 300s = 1000J/kg°C x 2kg x X

X = 300

Initial temperature => 400°C-300°C = 100°C

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2 years ago

The amount of pressure required to move a 6800 lb force with a 6" d piston is ___ psi.

Katena32 [7]

The pressure needed in PSI = Pounds of force needed divided by the cylinder Area
The Cylinder rod Area is 21.19 sq inches
Thus, the pressure= 6800/21.19
= 320.91 PSI

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2 years ago

The internal shear force V at a certain section of a steel beam is 80 kN, and the moment of inertia is 64,900,000 . Determine th

Luba_88 [7]

Here is the complete question

The internal shear force V at a certain section of a steel beam is 80 kN, and the moment of inertia is 64,900,000 . Determine the horizontal shear stress at point H, which is located L = 20 mm below the centriod

The missing image which is the remaining part of this question is attached in the image below.

Answer:

The horizontal shear stress at point H is $\mathbf{\tau_H \approx 42.604 \ N/mm^2}$

Explanation:

Given that :

The internal shear force V = 80 kN = 80 × 10³ N

The moment of inertia = 64,900,000

The length = 20 mm below the centriod

The horizontal shear stress $\tau$ can be calculated by using the equation:

$\tau = \dfrac{VQ}{Ib}$

where;

Q = moment of area above or below the point H

b = thickness of the beam = 10 mm

From the centroid ;

Q = $Q_1 + Q_{2}$

Q = $A_1y_1 + A_{2}y_{2}$

Q = ( ( 70 × 10) × (55) + ( 210 × 15) (90 + 15/2) ) mm³

Q = ( ( 700) × (55) + ( 3150 ) ( 97.5) ) mm³

Q = ( 38500 + 307125 ) mm³

Q = 345625 mm³

$\tau_H = \dfrac{VQ}{Ib}$

$\tau_H = \dfrac{80*10^3 * 345625}{64900000*10 }$

$\tau_H = \dfrac{2.765*10^{10}}{649000000 }$

$\tau_H = 42.60400616 \ N/mm^2$

$\mathbf{\tau_H \approx 42.604 \ N/mm^2}$

The horizontal shear stress at point H is $\mathbf{\tau_H \approx 42.604 \ N/mm^2}$

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2 years ago

A heavy flywheel is accelerated (rotationally) by a motor that provides constant torque and therefore a constant angular acceler

aleksandrvk [35]

Answer:

Part a)

$t_1 = \frac{\omega_1}{\alpha}$

Part b)

$\theta = \frac{1}{2}\alpha t_1^2$

Part c)

$t = \frac{t_1}{5}$

Explanation:

Part a)

As we know that it is having constant torque so here the time taken by it to accelerate is given as

$\omega_f = \omega_i + \alpha t$

$\omega_1 = 0 + \alpha t_1$

$t_1 = \frac{\omega_1}{\alpha}$

Part b)

angular displacement is given as

$\theta = \omega_i t_1 + \frac{1}{2}(\alpha) t_1^2$

$\theta = 0 + \frac{1}{2}\alpha t_1^2$

$\theta = \frac{1}{2}\alpha t_1^2$

Part c)

As we know that the angular deceleration produced by the brakes is given as

$\alpha_d = - 5\alpha$

now we have

$\omega_f = \omega _i + \alpha t$

$0 = \omega_1 - 5 \alpha_1 t$

$t = \frac{\omega_1}{5 \alpha}$

As we know that

$t_1 = \frac{\omega_1}{\alpha}$

so we have

$t = \frac{t_1}{5}$

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1 year ago

An airplane travels horizontally at a constant velocity v. An object is dropped from the plane and one second later another obje

Delvig [45]

Answer:

the vertical distance between the two object will increase uniformly when they are dropped after a fixed interval of time

Explanation:

Since airplane is moving horizontally with constant speed v

so when object is dropped from the plane then the speed of the object will be same as that of the speed of the airplane

so we can say that two object when dropped after some interval of time then they always lie in same vertical line

now we know that they both have same acceleration in vertical line so the motion of two objects relative to each other in vertical direction is always uniform motion because they have no acceleration with respect to each other

So the vertical distance between the two object will increase uniformly when they are dropped after a fixed interval of time

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2 years ago

E. Describe in short the structure of a mercury barometer​

E. Describe in short the structure of a mercury barometer