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never [62]
3 years ago
5

Two resistors, of R1 = 3.11 Ω and R2 = 6.15 Ω, are connected in series to a battery with an EMF of 24.0 V and negligible interna

l resistance. Find the current I1 through the first resistor and the potential difference ΔV2 between the ends of the second resistor.
Physics
1 answer:
Firlakuza [10]3 years ago
7 0
Part (a):
1- Since the resistors are in series, therefore, the total resistance is the summation of the two resistors.
Therefore:
Rtotal = R1 + R2 = 3.11 + 6.15 = 9.26 ohm

2- Since the two resistors are in series, therefore, the current flowing in both is the same. We will use ohm's law to get the current as follows:
V = I*R
V is the voltage of the battery = 24 v
I is the current we want to get
R is the total resistance = 9.26 ohm
Therefore:
24 = 9.26*I
I = 24 / 9.26
I = 2.59 ampere

Part (b):
To get the voltage across the second resistor, we will again use Ohm's law as follows:
V = I*R
V is the voltage we want to get
I is the current in the second resistor = 2.59 ampere
R is the value of the second resistor = 6.15 ohm
Therefore:
V = I*R
V = 2.59 * 6.15
V = 15.9285 volts

Hope this helps :)
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Answer:

2.25in³

Explanation:

For a 12 awg conductor the minimum volume allowance as stated by the NEC is 2.25in³

See attached Table 314.16(B) from NEC 2011

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3 years ago
The speed of the 2 kg cart after 5 seconds is ? cm/s.
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The answer to this question is 1cm/s
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2 years ago
Whats the force of gravitation of a 10kg rock and 100kg boulder which are 5 meters apart​
spayn [35]

Answer:

F = 2.6692 x 10⁻⁹ N

Explanation:

Given,

The mass of the rock, m = 10 kg

The mass of the boulder, M = 100 kg

The distance between them, d = 5 m

The gravitational force between the two bodies is proportional to the product of their masses and inversely proportional to the square of the distance between them. It is given by the formula

                                   <em> F = GMm/d²  newton</em>

Where,

                                 G - Universal gravitational constant

Substituting the given values,

                                 F = 6.673 x 10⁻¹¹ x 100 x 10 / 5²

                                 F = 2.6692 X 10⁻⁹ N

Hence, the force between the two bodies is, F = 2.6692 X 10⁻⁹ N

6 0
3 years ago
To determine the muzzle velocity of a bullet fired from a rifle, you shoot the 2.47-g bullet into a 2.43-kg wooden block. The bl
Elza [17]

Answer:

The velocity of the bullet on leaving the gun's barrel is 236.36 m/s.

Explanation:

Given;

mass of the bullet, m₁ = 2.47 g = 0.00247 kg

mass of the wooden block, m₂ = 2.43 kg

initial velocity of the wooden block, u₂ = 0

height reached by the bullet-block system after collision = 0.295 cm = 0.00295 m

let the initial velocity of the bullet on leaving the gun's barrel = v₁

let final velocity of the bullet-wooden block system after collision = v₂

Apply the principle of conservation of linear momentum;

Total initial momentum = Total final momentum

m₁v₁ + m₂u₂ = v₂(m₁ + m₂)

0.00247v₁  + 2.43 x 0  =  v₂(2.43 + 0.00247)

0.00247v₁ = 2.4325v₂ -------(1)

The kinetic energy of the bullet-block system after collision;

K.E = ¹/₂(m₁ + m₂)v₂²

K.E = ¹/₂ (2.4325)v₂²

The potential energy of the bullet-block system after collision;

P.E = mgh

P.E = (2.4325)(9.8)(0.00295)

P.E = 0.07032

Apply the principle of conservation of mechanical energy;

K.E = P.E

¹/₂ (2.4325)v₂² = 0.07032

1.21625 v₂²  = 0.07032

v₂²  = 0.07032  / 1.21625

v₂² = 0.0578

v₂ = √0.0578

v₂ = 0.24 m/s

Substitute v₂ in equation (1), to obtain the initial velocity of the bullet;

0.00247v₁ = 2.4325v₂

0.00247v₁ = 2.4325 (0.24)

0.00247v₁ = 0.5838

v₁ = 0.5838 / 0.00247

v₁ = 236.36 m/s

Therefore, the velocity of the bullet on leaving the gun's barrel is 236.36 m/s.

5 0
2 years ago
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algol13

Answer:

Chemical elements, a dense lustrous yellow precious metal of group.

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