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pishuonlain [190]
4 years ago
8

Do you think most other people participate in these measures?

Physics
1 answer:
Fynjy0 [20]4 years ago
3 0

what?theres no picture. what does this mean?

You might be interested in
Please simplify and write the below paragraph.
NikAS [45]

Field lines become straight and perpendicular because every point of circular loop the concentric circles become larger and larger as we move away from the wire.

3 0
3 years ago
Gravity is a <br><br> A.pushing force<br> B. not a force at all<br> C. pulling force
melamori03 [73]

Answer:

C

Explanation:

gravity is a pulling force according to Newton

4 0
3 years ago
The capacitance of the variable capacitor of a radio can be changed from 100 to 350 pF by turning the dial from 0° to 180°. With
tangare [24]

Answer:

0.7 mJ

Explanation:

<u>Identify the unknown:  </u>

The work required to turn the dial from 180° to 0°  

<u>List the Knowns:  </u>

Capacitance when the dial is set at 180°: C = 350 pF = 350 x 10^-12 F Capacitance when the dial is set at 0°: C = 100 pF = 100 x 10^-12 F  

Voltage of the battery: V = 130 V  

<u>Set Up the Problem:</u>  

<em><u>Energy stored in a capacitor: </u></em>

U_c=1/2*V^2*C

      =1/2*Q^2/C

<em><u>When the dial is set at 180°:</u></em><em>  </em>

U_c=1/2*(130)^2*350*10^-12=10^-4

Q=√2*U_c*C=4*10^-7

<u><em>When the dial is set at 0°:</em></u>  

U_c=1/2*(4*10^-7)^2/100*10^-12

      =8*10^-4 J

<u><em>Solve the Problem:  </em></u>

ΔU_c=7*10^-4 J

        =0.7 mJ

note:

there maybe error in calculation but method is correct

4 0
3 years ago
Read 2 more answers
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
Write down the conservation of momentum?​
otez555 [7]
Law of conservation of momentum states that when two objects collide with each other , the sum of their linear momentum always remains same or we can say conserved and is not effected by any action, reaction only in case is no external unbalanced force is applied on the bodies.
Let,
m
A
​
= Mass of ball A
m
B
​
= Mass of ball B
u
A
​
= initial velocity of ball A
u
B
​
= initial velocity of ball B
v
A
​
= Velocity after the collision of ball A
v
B
​
= Velocity after the collision of ball B
F
ab
​
= Force exerted by A on B
F
ba
​
= Force exerted by B on A
Now,
Change in the momentum of A= momentum of A after the collision - the momentum of A before the collision
= m
A
​
v
A
​
−m
A
​
u
A
​

Rate of change of momentum A= Change in momentum of A/ time taken
=
t
m
A
​
v
A
​
−m
A
​
u
A
​

​

Force exerted by B on A (F
ba
​
);
F
ba
​
=
t
m
A
​
v
A
​
−m
A
​
u
A
​

​
........ [i]
In the same way,
Rate of change of momentum of B=
t
m
b
​
v
B
​
−m
B
​
u
B
​

​

Force exerted by A on B (F
ab
​
)=
F
ab
​
=
t
m
B
​
v
B
​
−m
B
​
u
B
​

​
.......... [ii]
Newton's third law of motion states that every action has an equal and opposite reaction, then,
F
a
​
b=−F
b
​
a [ ' -- ' sign is used to indicate that 1 object is moving in opposite direction after collision]

Using [i] and [ii] , we have
t
m
B
​
v
B
​
−m
B
​
u
B
​

​
=−
t
m
A
​
v
A
​
−m
A
​
u
A
​

​

m
B
​
v
B
​
−m
B
​
u
B
​
=−m
A
​
v
A
​
+m
A
​
u
A
​

Finally we get,
m
B
​
v
B
​
+m
A
​
v
A
​
=m
B
​
u
B
​
+m
A
​
u
A
​

This is the derivation of conservation of linear momentum.
5 0
3 years ago
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