A current is always developed in the secondary of a transformer. true or false

At a depth of 1030 m in Lake Baikal (a fresh water lake in Siberia), the pressure has increased by 100 atmospheres (to about 107

dangina [55]

Answer:

A volume of a cubic meter of water from the surface of the lake has been compressed in 0.004 cubic meters.

Explanation:

The bulk modulus is represented by the following differential equation:

$K = - V\cdot \frac{dP}{dV}$

Where:

$K$ - Bulk module, measured in pascals.

$V$ - Sample volume, measured in cubic meters.

$P$ - Local pressure, measured in pascals.

Now, let suppose that bulk remains constant, so that differential equation can be reduced into a first-order linear non-homogeneous differential equation with separable variables:

$-\frac{K \,dV}{V} = dP$

This resultant expression is solved by definite integration and algebraic handling:

$-K\int\limits^{V_{f}}_{V_{o}} {\frac{dV}{V} } = \int\limits^{P_{f}}_{P_{o}}\, dP$

$-K\cdot \ln \left |\frac{V_{f}}{V_{o}} \right| = P_{f} - P_{o}$

$\ln \left| \frac{V_{f}}{V_{o}} \right| = \frac{P_{o}-P_{f}}{K}$

$\frac{V_{f}}{V_{o}} = e^{\frac{P_{o}-P_{f}}{K} }$

The final volume is predicted by:

$V_{f} = V_{o}\cdot e^{\frac{P_{o}-P_{f}}{K} }$

If $V_{o} = 1\,m^{3}$ , $P_{o} - P_{f} = -10132500\,Pa$ and $K = 2.3\times 10^{9}\,Pa$ , then:

$V_{f} = (1\,m^{3}) \cdot e^{\frac{-10.1325\times 10^{6}\,Pa}{2.3 \times 10^{9}\,Pa} }$

$V_{f} \approx 0.996\,m^{3}$

Change in volume due to increasure on pressure is:

$\Delta V = V_{o} - V_{f}$

$\Delta V = 1\,m^{3} - 0.996\,m^{3}$

$\Delta V = 0.004\,m^{3}$

A volume of a cubic meter of water from the surface of the lake has been compressed in 0.004 cubic meters.

8 0

3 years ago

A body is at aheight of 81m and is ascending upwards with a velocity of 12m/s .a body of 2 kg weight is dropped from it.if g=10m

butalik [34]

First the velocity drops to zero in 1.2 secs. In those seconds it went upwards for 7.2 m, then it went from 87.2 to 0m. x-x0=v0*t+1/2*g*t^2 ergo t=sqrt(2x/g) that is 4.1761 s. Finally the total time required is 5.3761 s

7 0

3 years ago

A block of mass m is attached to a rope wound around the outer rim of a disk of radius R and moment of inertia I, which is free

Hoochie [10]

Answer:

Explanation:

I is the moment of inertia of the pulley, α is the angular acceleration of the pulley and T is the tension in the rope. Let a is the linear acceleration.

The relation between the linear acceleration and the angular acceleration is

a = R α .... (1)

According to the diagram,

T x R = I x α

T x R = I x a / R from equation (1)

T = I x a / R² .... (2)

mg - T = ma .... (3)

Substitute the value of T from equation (2) in equation (3)

$mg - \frac{Ia}{R^{2}}=ma$

$a=\frac{mg}{m+\frac{I}{R^{2}}}$

T is the acceleration in the system

Substitute the value of a in equation (2)

$T = \frac{I}{R^{2}}\times \frac{mg}{m+\frac{I}{R^{2}}}$

$T=\frac{I\times mg}{I+mR^{2}}$

This is the tension in the string.

4 0

3 years ago

Establishing a potential difference The deflection plates in an oscilloscope are 10 cm by 2 cm with a gap distance of 1 mm. A 10

Nezavi [6.7K]

Answer:

t = 23.9nS

Explanation:

given :

Area A= 10 cm by 2 cm => 2 x 10^-2m x 10 x 10^-2m

distance d= 1mm=> 0.001

resistor R= 975 ohm

Capacitance can be calculated through the following formula,

C = (ε0 x A )/d

C = (8.85 x 10^-12 x (2 x 10^-2 x 10 x 10^-2))/0.001

C = 17.7 x 10^-12 (pico 'p' = 10^-12)

C = 17.7pF

the voltage between two plates is related to time, There we use the following formula of the final voltage

Vc = Vx (1-e^-(t/CR))

75 = 100 x (1-e^-(t/CR))

75/100 = (1-e^-(t/CR))

.75 = (1-e^-(t/CR))

.75 -1 = -e^-(t/CR)

-0.25 = -e^-(t/CR) --->(cancelling out the negative sign)

e^-(t/CR) = 0.25

in order to remove the exponent, take logs on both sides

-t/CR = ln (0.25)

t/CR = -ln(0.25)

t = -CR x ln (0.25)

t = -(17.7 x 10^-12 x 975) x (-1.38629)

t = 23.9 x $10^{-9$

t = 23.9ns

Thus, it took 23.9ns for the potential difference between the deflection plates to reach 75 volts

6 0

3 years ago

Read 2 more answers

Astronauts need special tools to work in space. One specially designed tool that astronauts use is a screwdriver. Using Newton’s

Softa [21]

Due to the fact that for every action there is an equal reaction for every turn on the screw the screw will exert energy equal to the amount you gave

3 0

3 years ago