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V125BC [204]
3 years ago
12

When resistors are connected in series, we can be certain that:______

Physics
1 answer:
Debora [2.8K]3 years ago
7 0

Answer:

A) the same current flows in each one.

Explanation:

When resistors are connected in series, Ohm's law states that the current flowing through each resistor is the same. Since it is similar to Ohm's law for a single resistor, except that the series resistors appear as a sum. Then, the total resistance of a series circuit is equal to the sum of the individual resistors.

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if a force is applied to an object is not greater than the starting friction what will happen to the object
MArishka [77]

If you match the amount of static friction that can be generated when the object is at rest, it will not move because there is zero force the force has to be greater that the static friction in order to have it include motion. But if there was any more force will cause acceleration otherwise if it was lower force it will deceleration.

4 0
3 years ago
The electric field on the surface of an irregularly shaped conductor varies from 74.0 kN/C to 14.0 kN/C.
mrs_skeptik [129]

Answer:

(a). The local surface charge density at the point on the surface where the radius of curvature of the surface is greatest is 123.9 nC/m².

(b). The local surface charge density at the point on the surface where the radius of curvature of the surface is greatest is 654.9 nC/m².

Explanation:

Given that,

Electric field E_{1}=74.0\ kN/C

Electric field E_{2}=14.0\ kN/C

When the radius of curvature is greatest, the electric field at the surface will be smaller.

Where the radius of curvature is greatest

(a). We need to calculate the local surface charge density at the point on the surface

Using formula of charge density

\sigma=\epsilon_{0}E_{2}

Put the value into the formula

\sigma=8.85\times10^{-12}\times14\times10^{3}

\sigma=1.239\times10^{-7}\ C/m^2

\sigma=123.9\times10^{-9}\ C/m^2

\sigma=123.9\ nC/m^2

The local surface charge density at the point on the surface where the radius of curvature of the surface is greatest is 123.9 nC/m².

(b). We need to calculate the local surface charge density at the point on the surface where the radius of curvature of the surface is smallest

Using formula of charge density

\sigma=\epsilon_{0}E_{1}

Put the value into the formula

\sigma=8.85\times10^{-12}\times74\times10^{3}

\sigma=6.549\times10^{-7}\ C/m^2

\sigma=654.9\times10^{-9}\ C/m^2

\sigma=654.9\ nC/m^2

The local surface charge density at the point on the surface where the radius of curvature of the surface is greatest is 654.9 nC/m².

Hence, (a). The local surface charge density at the point on the surface where the radius of curvature of the surface is greatest is 123.9 nC/m².

(b). The local surface charge density at the point on the surface where the radius of curvature of the surface is greatest is 654.9 nC/m².

7 0
3 years ago
When the mass of an object increases, it gravitational pull(increases)(decreases)(stay the same
Sonbull [250]

I'm pretty sure the answer is the gravitational pull  increases

7 0
4 years ago
Read 2 more answers
When Kevin pulls his cotton shirt off his body, the electrons get transferred from the (shirt or body) to the (shirt or body) .
Masja [62]
<span>When Kevin pulls his cotton shirt off his body, the electrons get transferred from the shirt (in form of static charges i.e. electrons to the body. So, the shirt becomes positively charged and Kevin’s body becomes negatively charged.

As a result of charge transfer from the shirt to the body, we can hear a crackling sound. or if observed in dark, a sparkle can be seen.</span>
6 0
3 years ago
A 7.5 nC point charge and a - 2.9 nC point charge are 3.2 cm apart. What is the electric field strength at the midpoint between
Oduvanchick [21]

Answer:

Net electric field, E_{net}=91406.24\ N/C

Explanation:

Given that,

Charge 1, q_1=7.5\ nC=7.5\times 10^{-9}\ C

Charge 2, q_2=-2.9\ nC=-2.9\times 10^{-9}\ C

distance, d = 3.2 cm = 0.032 m

Electric field due to charge 1 is given by :

E_1=\dfrac{kq_1}{r^2}

E_1=\dfrac{9\times 10^9\times 7.5\times 10^{-9}}{(0.032)^2}

E_1=65917.96\ N/C

Electric field due to charge 2 is given by :

E_2=\dfrac{kq_2}{r^2}

E_2=\dfrac{9\times 10^9\times 2.9\times 10^{-9}}{(0.032)^2}

E_2=25488.28\ N/C

The point charges have opposite charge. So, the net electric field is given by the sum of electric field due to both charges as :

E_{net}=E_1+E_2

E_{net}=65917.96+25488.28

E_{net}=91406.24\ N/C

So, the electric field strength at the midpoint between the two charges is 91406.24 N/C. Hence, this is the required solution.

3 0
4 years ago
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