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Mice21 [21]
2 years ago
4

During the experiment, you adjusted the length of the rheostat LL6 so that 1 glow flowed through bulb H. Now that we have seen t

hat the (conduction) current measured by a DMM corresponds to the same physical concept as ‘glow’, it makes sense to ask "what current goes through bulb H when 1 glow goes through bulb H?" The consensus in the past has been about 30 mA but you might have been using a slightly larger or smaller current. Remember that you assumed that the bulbs in circuit 7 were identical. If the resistance of rheostat LL6 has been adjusted so that 30 mA flows through bulb H then what current flows through bulb B? What current flows through bulb C?

Physics
1 answer:
riadik2000 [5.3K]2 years ago
4 0

Answer:

IB = IC =IH = 30 mA

Explanation:

please look at the solution in the attached Word file

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A 0.20 kg mass on a horizontal spring is pulled back 2.0 cm and released. If, instead, a 0.40 kg mass were used in this same exp
KIM [24]

Answer:

The total mechanical energy does not change if the value of the mass is changed. That is, remain the same

Explanation:

The total mechanical energy of a spring-mass system is equal to the elastic potential energy where the object is at the amplitude of the motion. That is:

E=U=\frac{1}{2}kA^2       (1)

k: spring constant

A: amplitude of the motion = 2.0cm

As you can notice in the equation (1), the total mechanical energy of the system does not depend of the mass of the object. It only depends of the amplitude A and the spring constant.

Hence, if you use a mass of 0.40kg the total mechanical energy is the same as the obtained with a mas 0.20kg

Remain the same

8 0
2 years ago
Before you start taking measurements though, we’ll first make sure you understand the underlying concepts involved. By what meth
Svetradugi [14.3K]

Answer:

If they are metallic spheres  they are connected to earth and a charged body approaches

non- metallic (insulating) spheres in this case are charged by rubbing

Explanation:

For fillers, there are two fundamental methods, depending on the type of material.

If they are metallic spheres, they are connected to earth and a charged body approaches, this induces a charge of opposite sign and of equal magnitude, then it removes the contact to earth and the sphere is charged.

If the non- metallic (insulating) spheres in this case are charged by rubbing with some material or touching with another charged material, in this case the sphere takes half the charge and when separated each sphere has half the charge and with equal sign.

8 0
2 years ago
The two hot-air balloons in the drawing are 48.2m and 61.0 m above the ground.A person in the left balloon observes that the rig
mafiozo [28]

Answer:

The horizontal distance x between the two balloons is 54.15 m

Explanation:

The diagram described as obtained online is presented in the image attached to this solution.

Let the horizontal distance between the two balloons be x

Difference in height (vertical distance) between the two balloons = 61 - 48.2 = 12.8 m

Using trigonometric relations, it is evident that

Tan 13.3° = 12.8/x

x = 12.8/tan 13.3° = 12.8/0.2364 = 54.15 m

4 0
2 years ago
Stu wanted to calculate the resistance of a light bulb connected to a 4.0 V battery, with a resulting current of 0.5 A. He used
Dimas [21]
His answer was incorrect because according to ohm's law the formula used should have been R=V/I instead of multiplying and the answer should be 8ohms
4 0
2 years ago
Read 2 more answers
It has been proposed that extending a long conducting wire from a spacecraft (a "tether") could be used for a variety of applica
denis23 [38]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The angle between shuttle's velocity and the Earth's field is  \theta =   24.2^o

Explanation:

From the question we are told that

     The length of eire let out is  L = 250 \ m

      The emf generated is \epsilon = 40 V

      The earth magnetic field is B = 5.0 *10^{-5} T

     The speed of the shuttle and tether is v =  7.80 * 10^3 \  m/s

The emf generated is mathematically represented as

                             \epsilon = L\ v\ B\ sin \ \theta

making \theta  the subject of the formula

                        \theta =   sin ^{-1}[ \frac{\epsilon}{L  * B  *v} ]

substituting values

                        \theta =   sin ^{-1}[ \frac{40}{250  * (5*10^{-5})  *(7.80 *10^{3})} ]

                        \theta =   24.2^o

6 0
2 years ago
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