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dem82 [27]
3 years ago
13

Please Help.............................

Physics
2 answers:
Natalka [10]3 years ago
7 0
Neutrons don't have any charge
kramer3 years ago
4 0
Protons!! hope i helped :)
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Objects 1 and 2 attract each other with a electrostatic force of 18.0 units. If the charge
xxMikexx [17]

Answer:

new force is 6 times of the initial force.

Explanation:

Let the charges on two objects is q₁ and q₂. The electric force between charges is given by :

F=\dfrac{kq_1q_2}{r^2}

Objects 1 and 2 attract each other with a electrostatic force of 18.0 units

If the charge  of Object 1 is doubled and the charge of object 2 is tripled, it means, q_1'=2q_1 and q_2'=3q_2. New force is given by :

F'=\dfrac{kq_1'q_2'}{r^2}\\\\F'=\dfrac{k(2q_1)(3q_2)}{r^2}\\\\F'=6\dfrac{kq_1q_2}{r^2}\\\\F'=6F

So, the new electrostatic force between objects will become 6 times of the initial force.

5 0
3 years ago
8. The fact that voltage can be created by exerting force on a crystal is used in which type of sensor?
AfilCa [17]

Option C

The fact that voltage can be created by exerting force on a crystal is used in Knock sensor

<u>Explanation:</u>

Any knock to an engine exhibits as a little shake that is distinguished by the knock sensor. This sensor acts by altering the fluctuation to an electrical sign, which is later transferred to the processor mastering the ignition system.

There the variation in quake to the voltage sign modifies the timing improvements on the kindling. The knock sensor is placed on the engine base, cylinder cap or consumption manifold. This is because its purpose is to sense fluctuations affected by engine knock or explosion.

5 0
3 years ago
A uniform electric field with a magnitude of 5750 N/C points in the positive x direction. Find the change in electric potential
castortr0y [4]

Explanation:

Given that,

Electric field = 5750 N/C

Charge q=+10.5\times10^{-6}\ C

Distance = 5.50 cm

(a). When the charge is moved in the positive x- direction

We need to calculate the change in electric potential energy

Using formula of electric potential energy

\Delta U=-W

\Delta U=-F\cdot d

\Delta U=-q(E\cdot d)

Put the value into the formula

\Delta U=-10.5\times10^{-6}\times5750\times5.50\times10^{-2}

\Delta U=-3.32\times10^{-3}\ J

The change in electric potential energy  is -3.32\times10^{-3}\ J

(b). When the charge is moved in the negative x- direction

We need to calculate the change in electric potential energy

Using formula of electric potential energy

\Delta U=-W

\Delta U=-F\cdot (-d)

\Delta U=-q(E\cdot (-d))

Put the value into the formula

\Delta U=-10.5\times10^{-6}\times5750\times(-5.50\times10^{-2})

\Delta U=3.32\times10^{-3}\ J

The change in electric potential energy  is 3.32\times10^{-3}\ J

Hence, This is the required solution.

3 0
3 years ago
2. Four resistors of 2 ohm each are connected first in series and then in parallel with a battery.Find the ratio of electric cur
Serjik [45]

Answer:

Ratio of series current to parallel

= 1 : 8

Explanation:

Total resistance Rt

For series, Rt = 2+2+2+2 = 4ohms

For parallel, 1/Rt = 1/2 + 1/2 + 1/2 + 1/2

1/Rt = 4/2, Rt = 2/4 ohms.

If we use a 1V battery, then,

I = V/Rt

I = 1/4 = 0.25 ampere for series arrangement.

I = 1/0.5 = 2 ohms.

Ratio of current of series to parallel = 0.25 : 2

= 1 : 8

4 0
3 years ago
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What is the moon phase today
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Waxxing Gibbous Phase
4 0
3 years ago
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