A circuit is built based on the circuit diagram shown.

An early submersible craft for deep-sea exploration was raised and lowered by a cable from a ship. When the craft was stationary

Talja [164]

Answer:

A) 5.2 x 10³ N

B) 8.8 x 10³ N

Explanation:

Part A)

$F_{g}$ = weight of the craft in downward direction = tension force in the cable when stationary = 7000 N

$T$ = Tension force in upward direction

$F_{d}$ = Drag force in upward direction = 1800 N

Force equation for the motion of craft is given as

$F_{g}$ - $F_{d}$ - $T$ = 0

7000 - 1800 - $T$ = 0

$T$ = 5200 N

$T$ = 5.2 x 10³ N

Part B)

$F_{g}$ = weight of the craft in downward direction = tension force in the cable when stationary = 7000 N

$T$ = Tension force in upward direction

$F_{d}$ = Drag force in downward direction = 1800 N

Force equation for the motion of craft is given as

$T$ - $F_{g}$ - $F_{d}$ = 0

$T$ - 7000 - 1800 = 0

$T$ = 8800 N

$T$ = 8.8 x 10³ N

4 0

3 years ago

Define about the different types of silk found around the world

Sunny_sXe [5.5K]

Answer:

In short, there are four types of natural silk produced around the world: Mulberry silk, Eri silk, Tasar silk and Muga silk. Mulberry silk contributes around as much as 90% of silk production, with the mulberry silkworm generally being regarded as the most important.

7 0

2 years ago

Diferencia entre transferencia y transformación de energía. Da ejemplos de cada caso.

Firdavs [7]

Answer:

La transformación de energía es un proceso en el que la energía se intercambia entre un sistema y el medio ambiente en al menos dos formas de energía diferentes entre sí. Por ejemplo, un panel solar convierte la energía lumínica en energía eléctrica.

En cambio, en la transferencia de energía, esta no cambia su forma sino que es transmitida de un cuerpo a otro. El ejemplo más claro es el de la fogata, que transmite calor al medio ambiente a través de radiación.

6 0

3 years ago

Mechanical energy is conserved in the presence of which of the following types of forces?magnetic

Tju [1.3M]

<h2>Answer: electrostatic and gravitational force </h2><h2 />

Mechanical energy remains constant (conserved) if only <u>conservative forces</u> act on the particles.

In this sense, the following forces are conservative:

-Gravitational

-Elastic

-Electrostatics

While the Friction Force and the Magnetic Force are not conservative.

According to this, mechanical energy is conserved in the presence of electrostatic and gravitational forces.

7 0

3 years ago

An infinite line of charge with linear density λ1 = 8.2 μC/m is positioned along the axis of a thick insulating shell of inner r

bixtya [17]

1) Linear charge density of the shell: $-2.6\mu C/m$

2) x-component of the electric field at r = 8.7 cm: $1.16\cdot 10^6 N/C$ outward

3) y-component of the electric field at r =8.7 cm: 0

4) x-component of the electric field at r = 1.15 cm: $1.28\cdot 10^7 N/C$ outward

5) y-component of the electric field at r = 1.15 cm: 0

Explanation:

1)

The linear charge density of the cylindrical insulating shell can be found by using

$\lambda_2 = \rho A$

where

$\rho = -567\mu C/m^3$ is charge volumetric density

A is the area of the cylindrical shell, which can be written as

$A=\pi(b^2-a^2)$

where

$b=4.7 cm=0.047 m$ is the outer radius

$a=2.7 cm=0.027 m$ is the inner radius

Therefore, we have :

$\lambda_2=\rho \pi (b^2-a^2)=(-567)\pi(0.047^2-0.027^2)=-2.6\mu C/m$

2)

Here we want to find the x-component of the electric field at a point at a distance of 8.7 cm from the central axis.

The electric field outside the shell is the superposition of the fields produced by the line of charge and the field produced by the shell:

$E=E_1+E_2$

where:

$E_1=\frac{\lambda_1}{2\pi r \epsilon_0}$

where

$\lambda_1=8.2\mu C/m = 8.2\cdot 10^{-6} C/m$ is the linear charge density of the wire

r = 8.7 cm = 0.087 m is the distance from the axis

And this field points radially outward, since the charge is positive .

And

$E_2=\frac{\lambda_2}{2\pi r \epsilon_0}$

where

$\lambda_2=-2.6\mu C/m = -2.6\cdot 10^{-6} C/m$

And this field points radially inward, because the charge is negative.

Therefore, the net field is

$E=\frac{\lambda_1}{2\pi \epsilon_0 r}+\frac{\lambda_2}{2\pi \epsilon_0r}=\frac{1}{2\pi \epsilon_0 r}(\lambda_1 - \lambda_2)=\frac{1}{2\pi (8.85\cdot 10^{-12})(0.087)}(8.2\cdot 10^{-6}-2.6\cdot 10^{-6})=1.16\cdot 10^6 N/C$

in the outward direction.

3)

To find the net electric field along the y-direction, we have to sum the y-component of the electric field of the wire and of the shell.

However, we notice that since the wire is infinite, for the element of electric field $dE_y$ produced by a certain amount of charge $dq$ along the wire there exist always another piece of charge $dq$ on the opposite side of the wire that produce an element of electric field $-dE_y$ , equal and opposite to $dE_y$ .

Therefore, this means that the net field produced by the wire along the y-direction is zero at any point.

We can apply the same argument to the cylindrical shell (which is also infinite), and therefore we find that also the field generated by the cylindrical shell has no component along the y-direction. Therefore,

$E_y=0$

4)

Here we want to find the x-component of the electric field at a point at

r = 1.15 cm

from the central axis.

We notice that in this case, the cylindrical shell does not contribute to the electric field at r = 1.15 cm, because the inner radius of the shell is at 2.7 cm from the axis.

Therefore, the electric field at r = 1.15 cm is only given by the electric field produced by the infinite wire:

$E=\frac{\lambda_1}{2\pi \epsilon_0 r}$

where:

$\lambda_1=8.2\mu C/m = 8.2\cdot 10^{-6} C/m$ is the linear charge density of the wire

r = 1.15 cm = 0.0115 m is the distance from the axis

This field points radially outward, since the charge is positive . Therefore,

$E=\frac{8.2\cdot 10^{-6}}{2\pi (8.85\cdot 10^{-12})(0.0115)}=1.28\cdot 10^7 N/C$

5)

For this last part we can use the same argument used in part 4): since the wire is infinite, for the element of electric field $dE_y$ produced by a certain amount of charge $dq$ along the wire there exist always another piece of charge $dq$ on the opposite side of the wire that produce an element of electric field $-dE_y$ , equal and opposite to $dE_y$ .

Therefore, the y-component of the electric field is zero.

Learn more about electric field:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

4 0

3 years ago