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notsponge [240]
2 years ago
13

Showing results for a 4 olm and 8 ohm resistor are connected in series. The current through the 4 olm resistor is 2 amps. What i

s the current through the 8 olm resister?
Plzzzz help 20 points
Physics
1 answer:
kirill115 [55]2 years ago
7 0

Answer:

Let R1= resistance of 4ohms

R2= resistance of 8ohms

Equivalent resistance R will be

R=R1 + R2

=> 4+8=12ohms

The current through the two resistors will be the same since they are connected in series. Notwithstanding, the voltage will drop to appreciate the change.

Explanation:

You might be interested in
The normal eye, myopic eye and old age
yanalaym [24]

Answer:

1)    f’₀ / f = 1.10, the relationship between the focal length (f'₀) and the distance to the retina (image) is given by the constructor's equation

2) the two diameters have the same order of magnitude and are very close to each other

Explanation:

You have some problems in the writing of your exercise, we will try to answer.

1) The equation to be used in geometric optics is the constructor equation

          \frac{1}{f} = \frac{1}{p} + \frac{1}{q}

where p and q are the distance to the object and the image, respectively, f is the focal length

* For the normal eye and with presbyopia

the object is at infinity (p = inf) and the image is on the retina (q = 15 mm = 1.5 cm)

        \frac{1}{f'_o} = 1/ inf + \frac{1}{1.5}

        f'₀ = 1.5 cm

this is the focal length for this type of eye

* Eye with myopia

the distance to the object is p = 15 cm the distance to the image that is on the retina is q = 1.5 cm

           1 / f = 1/15 + 1 / 1.5

           1 / f = 0.733

            f = 1.36 cm

this is the focal length for the myopic eye.

In general, the two focal lengths are related

         f’₀ / f = 1.5 / 1.36

         f’₀ / f = 1.10

The question of the relationship between the focal length (f'₀) and the distance to the retina (image) is given by the constructor's equation

2) For this second part we have a diffraction problem, the point diameter corresponds to the first zero of the diffraction pattern that is given by the expression for a linear slit

          a sin θ= m λ

the first zero occurs for m = 1, as the angles are very small

          tan θ = y / f = sin θ / cos θ

for some very small the cosine is 1

          sin θ = y / f

where f is the distance of the lens (eye)

           y / f = lam / a

in the case of the eye we have a circular slit, therefore the system must be solved in polar coordinates, giving a numerical factor

           y / f = 1.22 λ / D

           y = 1.22 λ f / D

where D is the diameter of the eye

          D = 2R₀

          D = 2 0.1

          D = 0.2 cm

           

the eye has its highest sensitivity for lam = 550 10⁻⁹ m (green light), let's use this wavelength for the calculation

         

* normal eye

the focal length of the normal eye can be accommodated to give a focus on the immobile retian, so let's use the constructor equation

      \frac{1}{f} = \frac{1}{p} + \frac{1}{q}

sustitute

       \frac{1}{f} = \frac{1}{25} + \frac{1}{1.5}

       \frac{1}{f}= 0.7066

        f = 1.415 cm

therefore the diffraction is

        y = 1.22  550 10⁻⁹  1.415  / 0.2

        y = 4.75 10⁻⁶ m

this is the radius, the diffraction diameter is

       d = 2y

       d_normal = 9.49 10⁻⁶ m

* myopic eye

In the statement they indicate that the distance to the object is p = 15 cm, the retina is at the same distance, it does not move, q = 1.5 cm

       \frac{1}{f} = \frac{1}{15} + \frac{1}{ 1.5}

        \frac{1}{f}= 0.733

         f = 1.36 cm

diffraction is

        y = 1.22 550 10-9 1.36 10-2 / 0.2 10--2

        y = 4.56 10-6 m

the diffraction diameter is

        d_myope = 2y

         d_myope = 9.16 10-6 m

         \frac{d_{normal}}{d_{myope}} = 9.49 /9.16

        \frac{d_{normal}}{d_{myope}} =  1.04

we can see that the two diameters have the same order of magnitude and are very close to each other

8 0
3 years ago
If an astronaut can throw a certain wrench 15.0 m vertically upward on earth, how high could he throw it on our moon if he gives
Sever21 [200]

Answer:

hm= 18.2 m

Explanation:

Solution is attached

5 0
3 years ago
A jet airliner moving initially at 889 mph
Eduardwww [97]

Answer:

1500 mph

Explanation:

Take east to be +x and north to be +y.

The x component of the velocity is:

vₓ = 889 cos 0° + 830 cos 59°

vₓ = 1316.5 mph

The y component of the velocity is:

vᵧ = 889 sin 0° + 830 sin 59°

vᵧ = 711.4 mph

The speed is found with Pythagorean theorem:

v² = vₓ² + vᵧ²

v² = (1316.5 mph)² + (711.4 mph)²

v = 1496 mph

Rounded to two significant figures, the jet's speed relative to the ground is 1500 mph.

8 0
3 years ago
(please help i gotta turn this in a few minutes 10 points!)
pogonyaev

Answer:

3a, 2b,4c,1d

Explanation:

what do I need to explain just something you know

7 0
3 years ago
A 15g football was kicked to an acceleration of 55m/s2. What was the force applied to it? 40N 3.7N 825N
Pepsi [2]

Answer:

f =ma = 0.015 * 55 = 0.825 N

so yeah that's ur ans

8 0
2 years ago
Read 2 more answers
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