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Galina-37 [17]
2 years ago
12

The lamp has a resistance of 10 ohms each resistor has a resistance of 10 ohms what is the total resistance of the circuit

Physics
1 answer:
prisoha [69]2 years ago
7 0
Mark Brainliest please

Answer : if connected series, 20 ohms
And if connected parallel, answer will be less than 20 ohms

When resistors are wired in series, the total circuit resistance increases because each resistor contributes opposition to the circuit's current flow. Therefore if a 10 ohm resistor is placed in series with another 10 ohm resistor, the total resistance contributed by the two resistors is 20 ohms.
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Glycerin is poured into an open U-shaped tube until the height in both sides is 20 cm. Ethyl alcohol is then poured into one arm
cestrela7 [59]

Answer:

0.043 m

Explanation:

From the attachment, the shaded part is the ethyl alcohol. The crossed part on the other hand, is that of glycerin.

The height of the Ethyl Alcohol is h2 = 0.25 m, it's density is ρ2 = 790 kg/m³. The density of glycerin is ρ1 = 1260 kg/m³

If we assume pressure at the two points to be the same, then

P1 = P2

ρ1.g.V1 = ρ2.g.V2

ρ1.A.h1 = ρ2.A.h2

ρ1.h1 = ρ2.h2, making h1 subject of formula

h1 = ρ2.h2 / ρ1, so that

h1 = 790 * 0.25 / 1260

h1 = 197.5 / 1260

h1 = 0.157 m

Δh = 0.2 - 0.157

Δh = 0.043 m or 4.3 cm

8 0
3 years ago
Find the charge on the capacitor in an LRC-series circuit when L = 1 2 h, R = 10 Ω, C = 0.01 f, E(t) = 50 V, q(0) = 1 C, and i(0
alex41 [277]

Answer:

Explanation:

Given that,

In an LRC circuit

L = 1/2h

R = 10 Ω,

C = 0.01 f

E(t) = 50 V,

q(0) = 1 C, and

i(0) = 0 A.

q(t) = C

We can to fine the charge after a long time, let say t→∞

The Kirchoff second law for the system is

L•dq²/dt + R•dq/dt + q/C  = E(t)

Divide through by L

dq²/dt + R/L •dq/dt + q/LC = E(t)/L

Now inserting the values of R, L, C and E

dq²/dt+10/½ •dq/dt +q/½×0.01=50/½

dq²/dt + 20•dq/dt + 200q  =  100

Let solve the differential equation

First : homogenous solution

Using D operator

D² + 20D +200 = 0

Solving the quadratic equation using formula method

D = (-b±√b²-4ac)/2a

D = (-20±√20²-4×1×200) /2

D = (-20±√400-800)/2

D = (-20±20•i)/2

D = -10±10•i

So we have a complex solution

Then, the complementary solution is

q(t) = e(-10t)[ Acos10t + BSin10t]

A and B are constant

Let find the particular solution using the method of undetermine coefficient

Let assume particular solution of

q(t) = C, I.e q(t) Is a constant

So, inserting this into the equation below

dq²/dt + 20•dq/dt + 200q  =  100

200q = 100

q = 100/200

q = ½

Then, the particular solution is ½

So, the total solution is the sum of particular solution and complementary solution

q = e(-10t)[ Acos10t + BSin10t] + ½

Using the initial conditions

q(0) = 1

1 = e(0) [ACos0 + BSin0] +½

1 = A+½

A = ½

Also i(0) = 0

I(t) = q'(t)

Then,

q'(t) = -10•e(-10t)[ Acos10t + BSin10t] + e(-10t)[ -10Asin10t + 10BCos10t]

0 = -10e(0) [ ACos0 + BSin0] + e(0)[-10ASin0 +10BCos0]

0 = -10(A) + 10B

A=B=½

So the general equation becomes

q(t) = e(-10t)[ ½cos10t + ½Sin10t] + ½

So, as t→∞, the aspect of e(-10t) become zero

So the charge stabilizes at q = ½C after a long time

q = ½C as t→∞

6 0
3 years ago
Read 2 more answers
Our textbook describes a common model for friction (both static and kinetic) in which the area of contact between interacting su
Mkey [24]

Contact area is not a factor of the amount of friction force acting between two bodies in contact, preventing rolling or sliding

Reasons:

  • Friction force is the product of the normal reaction and the coefficient of friction.
  • The coefficient of friction is higher in softer tires than hard tires which are required to be wider to reduce the pressure that can cause deformation withstood by hard tires.
  • The wide tires in drag racing cars can be thought of being made from the soft compound tire material having a higher coefficient of friction that increases the friction force.

Other advantages of wide tires are;

  • Increased traction
  • Shorter stopping distance

Learn more here:

brainly.com/question/17209968

7 0
2 years ago
What are the magnitudes of (a) the angular velocity, (b) the radial acceleration, and (c) the tangential acceleration of a space
KonstantinChe [14]

Answer:

(a)w=2.485*10^{-3}m/s\\(b)a=26.92m/s^2\\(c)\alpha =0rad/s^2

Explanation:

Given data

Linear speed of spaceship V=39000 km/h =10833.3 m/s

The radius of circular motion r=4360 km=4360000 m

For Part (a) angular velocity

The angular velocity is given by:

w=\frac{v_{speed}}{r_{radius}}

Substitute the given values

So

w=\frac{10833.3m/s}{4360000m} \\w=2.485*10^{-3}m/s

For Part(b) The radial acceleration of spaceship

The radial acceleration is given by:

a_{r}=w^2 r

Substitute given values and value of ω

So

a_{r}=(2.485*10^{-3}m/s)^2(4360000m)\\a_{r}=26.92m/s^2

For Part (c) The tangential acceleration

The spaceship is rotating with constant angular velocity,then the  tangential acceleration of spaceship is zero

So

\alpha =0rad/s^2

4 0
3 years ago
Consider formula A to be v = and formula B to be v2 = G. Write the letter of the appropriate formula to use in each scenario. De
Fofino [41]

Answer:

Situation one:

The moon will be experiencing gravitational force in the form of centripetal force, so we equate the two formulas.

Gravitational force = GMm /r²

Centripetal force = mv²/r

Equating,

GMm/r² = mv²/r

v² = GM/r

The first scenario will use the formula v² = GM/r

Situation 2:

The second situation will use the simple distance over time formula for velocity, where the distance will be the circumference and the time will be in seconds.

6 0
3 years ago
Read 2 more answers
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