A student increased the resister of the voltmeter year not the circuit

An electric dipole is formed from ± 5.0 nC point charges spaced 3.0 mm apart. The dipole is centered at the origin, oriented alo

Ymorist [56]

Answer:

The electric field strength at point (x,y) = ( 20 mm ,0cm) is =<u>16321.0769 N/C</u>

The electric field strength at point (x,y) = (0cm, 20 mm) is =<u>35321.58999 N/C</u>

Explanation:

Question: What is the electric field strength at point (x,y) = ( 20 mm ,0cm)?

Answer:

The electric field at any given point of the dipole is given as:

E= (KP) ÷ (r^2 + a^2)^3/2

Where:

K = 9x10^9 Nm^2/c^2 (coloumb constant)

P = (0.003) (5x10^-9c) which is the movement of the dipole

(0.003) is arrived at when mm is converted to m. 3.0 mm space apart was converted to a meter.

r= the point, in the question above is 20mm = 0.02m

Now, the electric field, E can be calculated by putting the values in the formula above:

E = (KP) ÷ (r^2 + a^2)^3/2

= (9x10^9 Nm^2/c^2) (0.003 m) (5x10^-9c) ÷ [ (0.02m)^2 + (0.003)^2]^3/2

= 0.135 ÷ (8.271513x10^-6)

=<u>16321.0769 N/C</u>

Question: What is the electric field strength at point (x,y) = (0cm, 20 mm )?

Answer:

Here, the electric field, E= 2krp ÷ (r^2 - a^2)^2

E= 2 (9x10^9 Nm^2/c^2) (0.02m) (0.003 m) (5x10^-9c) ÷ [(0.02m)^2 - (0.003)^2]^2

= 0.0054 ÷ 0.000000152881

=<u>35321.58999 N/C</u>

8 0

3 years ago

Suppose you are in an elevator. As the elevator starts upward, its speed will increase. During this time when the elevator is mo

kkurt [141]

Answer:increased

Explanation:

It is given that elevator speed is increasing while moving upward i.e.its acceleration is increasing .

This causes the apparent to be increased if measured using weighing machine.

considering upward direction to be positive

N-mg=ma

N=m(g+a)

where N=Normal reaction=Apparent weight

a=acceleration of Elevator

thus you feel as if your weight is increased.

6 0

4 years ago

A driver drove for 3 hours at a certain speed. if he drove 12 more miles at the same speed, the distance would be 132 miles. at

Zepler [3.9K]

Answer:

40mph

Explanation:

1st leg DATA:

time = 3 hrs ; speed = r mph ; distance = 3r miles

------------

2nd leg DATA:

speed = r mph ; distance = 12 miles

--------------------------------

3r + 12 = 132

3r = 120

rate = 40 mph

4 0

3 years ago

What has a larger capacitance, an aluminum sphere with a 10 cm diameter or one with a 100 cm diameter? Question 16 options: 10 c

creativ13 [48]

$\boxed{\sf C=\dfrac{Q}{V}}$

But

$\boxed{\sf \Delta V_{R_2\to R_1}={\displaystyle{\int}^{R_1}_{R_2}}dV=-{\displaystyle{\int}^{R_1}_{R_2}}\dfrac{kQ}{r^2}dR}$

Hence higher the radius lower the voltage
Lower the voltage higher the capacitance .

<h3>100cm diameter having aluminium sphere has a larger capacitance</h3>

3 0

3 years ago

You move a 2.5 kg book from a shelf that is 1.2 m above the ground to a shelf that is 2.6 m above the ground. What is the change

Sophie [7]

The change in potential energy of an object is given by
$U=mg \Delta h$
where
m is the mass of the object
g is the gravitational acceleration
$\delta h$ is the increase in altitude of the object

In our problem, $m=2.5 kg$ is the mass of the book, $g=9.81 m/s^2$ and
$\Delta h=2.6 m -1.2 m=1.4 m$ is the increase in altitude of the book, so its variation of potential energy is
$U=mg\Delta h=(2.5 kg)(9.81 m/s^2)(1.4 m)=34.3 J$

8 0

4 years ago