A glass rod is charged by the process of _______

The aqueous humor in a person's eye is exerting a force of 0.242 N on the 1.21 cm2 area of the cornea. What pressure is this in

Virty [35]

Answer:

Explanation:

Pressure is defined as the ratio of force applied to an object to its area.

Pressure = Force/Area

Given parameters

Force = 0.242N

Area = 1.21cm²

Required parameters

Pressure = 0.242/1.21

Pressure = 0.2N/cm²

Using the conversion to convert the pressure to mmHg

1N/cm² = 75.0062mmHg

0.2N/cm² = y

y = 0.2 * 75.0062

y = 15.00124mmHg

<em>Hence the pressure in mmHg is 15.00124mmHg</em>

4 0

2 years ago

At a certain place, Earth's magnetic field has magnitude B =0.703 gauss and is inclined downward at an angle of 75.4° to the hor

Irina18 [472]

Answer:

The charge flows in coulombs is

$dq=1.843x10^{-5}C$

Explanation:

The current magnitude of current is given by the resistance and the induced Emf as:

$I=N*\frac{dF}{Rdt}$

$\frac{dq}{dt}=\frac{dF}{Rdt}=dq=N*\frac{dF}{R}$

$dq=\frac{N*\beta*A*(Cos(\alpha_f)-Cos(\alpha_i}{R}$

$N=1300$ , $\beta=0.703$ , $A=\pi*r^2=\pi*0.10^2=0.01\pi m^2$ , $R=99.4+202=301.4$ Ω

$\alpha_f=14.6$ , $\alpha_i=165.4$

Replacing :

$dq=\frac{1300*0.703x10^{-4}*0.01\pi*(0.9667-(-0.9667))}{202+99.4}$

$dq=1.843x10^{-5}C$

5 0

3 years ago

A 1.50-kg iron horseshoe initially at 550°C is dropped into a bucket containing 25.0 kg of water at 20.0°C. What is the final te

Ber [7]

Answer:

Te = 23.4 °C

Explanation:

Given:-

- The mass of iron horseshoe, m = 1.50 kg

- The initial temperature of horseshoe, Ti_h = 550°C

- The specific heat capacity of iron, ci = 448 J/kgC

- The mass of water, M = 25 kg

- The initial temperature of water, Ti_w = 20°C

- The specific heat capacity of water, cw = 4186 J/kgC

Find:-

What is the final temperature of the water–horseshoe system?

Solution:-

- The interaction of horseshoe and water at their respective initial temperatures will obey the Zeroth and First Law of thermodynamics. The horseshoe at higher temperature comes in thermal equilibrium with the water at lower temperature. We denote the equilibrium temperature as (Te) and apply the First Law of thermodynamics on the system:

m*ci*( Ti_h - Te) = M*cw*( Te - Ti_w )

- Solve for (Te):

m*ci*( Ti_h ) + M*cw*( Ti_w ) = Te* (m*ci + M*cw )

Te = [ m*ci*( Ti_h ) + M*cw*( Ti_w ) ] / [ m*ci + M*cw ]

- Plug in the values and evaluate (Te):

Te = [1.5*448*550 + 25*4186*20 ] / [ 1.5*448 + 25*4186 ]

Te = 2462600 / 105322

Te = 23.4 °C

7 0

3 years ago

Read 2 more answers

If an automobile had a 100%-efficient engine, transferring all of the fuel's energy to work, would the engine be warm to your to

svetlana [45]

Answer:

The engine would be warm to touch, and the exhaust gases would be at ambient temperature. The engine would not vibrate nor make any noise. None of the fuel entering the engine would go unused.

Explanation:

In this ideal engine, none of these events would happen due to the nature of the efficiency.

We can define efficiency as the ratio between the used energy and the potential generable energy in the fuel.

n=W, total/(E, available).

However, in real engines the energy generated in the combustion of the fuel transforms into heat (which heates the exhost gases, and the engine therefore transfering some of this heat to the environment). Also, there are some mechanical energy loss due to vibrations and sound, which are also energy that comes from the fuel combustion.

5 0

3 years ago

The two masses in the Atwood's machine shown in the figure are initially at rest at the same height. After they are released, th

Inga [223]

According to the description given in the photo, the attached figure represents the problem graphically for the Atwood machine.

To solve this problem we must apply the concept related to the conservation of energy theorem.

PART A ) For energy conservation the initial kinetic and potential energy will be the same as the final kinetic and potential energy, so

$E_i = E_f$