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NikAS [45]
3 years ago
10

3- For given three vectors a, b and c, c = a x b, then the vector c is:​

Physics
1 answer:
o-na [289]3 years ago
8 0

Answer:

VB

Explanation:

You might be interested in
Which graph accurately shows the relationship between kinetic energy and speed as speed increases?
mixer [17]

Answer:

B

Explanation:

kinetic energy (KE) is the energy possessed by moving bodies. It can be expressed as:

KE = \frac{1}{2}mv^{2}

Where: m is the mass of the object, and v its speed.

For example, a stone of mass 2kg was thrown and moves with a speed of 3 m/s. Determine the kinetic energy of the stone.

Thus,

KE =  \frac{1}{2} x 2 x 3^{2}

     = 9

KE = 9.0 Joules

Assume that the speed of the stone was 4 m/s, then its KE would be:

KE =  \frac{1}{2} x 2 x 4^{2}

     = 16

KE = 16.0 Joules

Therefore, it can be observed that as speed increases, the kinetic energy increases. Thus option B is appropriate.

3 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
Alan rents a private jet for the weekend. He flies 400 km south to New York, then flies 700 km west to Chicago, then 1200 km to
Ratling [72]

Answer:

Total distance = 400+700+1200= 2300km

Explanation:

the resultant of d 1st right angle triangle + 1200

= 806.2 + 1200 = 2006.2km

7 0
3 years ago
8. A rectangle is measured to be 6.4 +0.2 cm by 8.3 $0.2 cm.
mamaluj [8]

Answer:

a) The perimeter of the rectangle is 29.4 centimeters.

b) The uncertainty in its perimeter is 0.8 centimeters.

Explanation:

a) From Geometry we remember that the perimeter of the rectangle (p), measured in centimeters, is represented by the following formula:

p = 2\cdot (w+l) (1)

Where:

w - Width, measured in centimeters.

l - Length, measured in centimeters.

If we know that w = 6.4\,cm and l = 8.3\,cm, then the perimeter of the rectangle is:

p = 2\cdot (6.4\,cm+8.3\,cm)

p = 29.4\,cm

The perimeter of the rectangle is 29.4 centimeters.

b) The uncertainty of the perimeter (\Delta p), measured in centimeters, is estimated by differences. That is:

\Delta p = 2\cdot (\Delta w + \Delta l)  (2)

Where:

\Delta w - Uncertainty in width, measured in centimeters.

\Delta l - Uncertainty in length, measured in centimeters.

If we know that \Delta w = 0.2\,cm and \Delta l = 0.2\,cm, then the uncertainty in perimeter is:

\Delta p = 2\cdot (0.2\,cm+0.2\,cm)

\Delta p = 0.8\,cm

The uncertainty in its perimeter is 0.8 centimeters.

5 0
3 years ago
Which quantities appear in the equation for the magnetic field component of an electromagnetic wave?
tankabanditka [31]

Answer:

The speed of light (c)

Explanation:

The equation that relates the magnetic field component of an electromagnetic wave the the electric field component of the wave is:

E=cB

where

E is the magnitude of the electric field component

B is the magnitude of the magnetic field component

c is the speed of light in a vacuum, whose value is

c=3.0 \cdot 10^8 m/s

Re-arranging the equation to solve for B, we find:

B=\frac{E}{c}

5 0
3 years ago
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