11.1. Is ammeter A1 connected in series or in parallel in this diagram?​

Answers

Answer 1

Ammeter is always connected in series with the circuit in which the current is to be measured.

An ammeter is connected, to gauge the current flowing through a part or circuit. Since current in a series connection stays constant and an ammeter's resistance is relatively low, the current being measured is unaffected. For the purpose of measuring current, an ammeter is linked in series.

A shunt running parallel to the metre carries the majority of the current at high current values, hence an ammeter can measure a wide range of current values. An ammeter is represented by a circle with a capital A inside it on circuit diagrams.

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Related Questions

(a) Calculate the force (in N) the woman in the figure below exerts to do a push-up at constant speed, taking all data to be known to three digits. (You may need to use torque methods from a later chapter.) 401.15

(b)How much work (in J) does she do if her center of mass rises 0.260 m?

(c) What is her useful power output (in W) if she does 30 push-ups in 1 min? (Should work done lowering her body be included? See the discussion of useful work in Work, Energy, and Power in Humans.)

Answers

The force is 400.2 N

The work done is 120 J

The power is 48W

What is Force?

Force is a physical concept that describes the influence that one object has on another object, causing it to accelerate or deform. Force can be defined as any influence that changes the motion of an object, such as a push or a pull.

How to solve:

under equilibrium condition

F * 1.45 m =68 kg * 9.81 m/s^2 *0.87 m

F =400.2 N

b)

work done = m*g*h =68 kg*9.81 m/s^2*0.180 m =120.0744 J =120 J

c)

power =120.0744 J *(24 /60 s) =48.02976 W = 48 W

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Which of the following was one of Hubble's conclusions due to red shift?
A. There are millions of galaxies in the universe, not just ours.
B. The universe is contracting.
C. Background radiation shows the Big Bang occurring.
D. The universe is expanding.

Answers

D. The universe is expanding. Hubble's observation of red shift in the light from distant galaxies led him to conclude that these galaxies were moving away from us and that the universe was expanding.

What was Hubble's observation?

Hubble's most famous observation was his discovery of the relationship between the redshift of light from distant galaxies and their distance from Earth. He observed that the light from distant galaxies was shifted toward longer, redder wavelengths, which indicated that the galaxies were moving away from us. By analyzing the degree of redshift, Hubble was able to calculate the distance of these galaxies from Earth and found that they were much farther away than previously thought. This led him to conclude that the universe was expanding and laid the foundation for the Big Bang theory.

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NO USE OF AI BOTS TO answer question pls answer thanks

Answers

Answer:

19.2 centimeters or 0.192 meters. Both are the same.

Explanation:

T = 2π√(L/g)

T = 60.0 s / 14 ≈ 4.29 s

L = (gT²) / (4π²) ≈ 0.384 m

h + 0.384 m = height above the field

H = h + 0.192 m

H = h

h ≈ 0.192 m or 19.2 cm.

Nuclear fusion in a star produces elements up to, but no heavier than, ________.
A. iron
B. lead
C. carbon
D. nitrogen

Answers

Answer:

A. iron.

In the process of nuclear fusion, lighter elements are fused together to form heavier elements. This process releases energy and is what powers the star. However, the fusion of elements heavier than iron requires energy, rather than releasing it. Therefore, once a star has produced iron in its core, it is no longer able to sustain nuclear fusion and will eventually undergo a supernova explosion.

1. Which characteristic of a substance is constant?
a phase
b mass
Ospecific heat
d kinetic energy

Answers

B: mass will be the correct answer

Macmillan Learning
When a massive star reaches the end of its life, it is possible for a supernova to occur. This may result in the formation of a very
small, but very dense, neutron star, the density of which is about the same as a neutron. A neutron has a mass of 1.7 x 10-27 kg
and an approximate radius of 1.2 x 10-15 m. The mass of the sun is 2.0 x 1030 kg.

Answers

Okay, let's break this down step-by-step:

1) A neutron has a mass of 1.7 x 10-27 kg and an approximate radius of 1.2 x 10-15 m.

So we know the mass and radius of a single neutron.

2) The mass of the sun is 2.0 x 1030 kg.

So we know the total mass of the sun, which is much greater than a neutron.

3) When a massive star reaches the end of its life, it can explode as a supernova.

This supernova can form a neutron star.

4) A neutron star has a density about the same as a neutron.

So we can conclude that a neutron star has a density of:

Density = Mass / Volume

= (1.7 x 10-27 kg) / (4/3 * pi * (1.2 x 10-15 m)3)

= 1.6 x 1017 kg/m3

5) A neutron star forms from the core collapse of a massive star during supernova.

So it has a mass on the order of 1-2 times that of the sun (2 x 1030 kg),

but compressed into a sphere only about 10-20 km in radius.

So its mass would be huge, around 2 x 1030 kg, but confined to a tiny volume,

giving it an immense density, around 1.6 x 1017 kg/m3, the same as a neutron.

Does this help explain the concepts and walk through the calculations? Let me know if you have any other questions!

A charge of −0.0004 C is a distance of 3 meters from a charge of 0.0003 C. What is the magnitude of the force between them?

Answers

The magnitude of the force is 3.6 x 10^-6 N, rounded to two significant figures.

The magnitude of the force between two point charges can be calculated using Coulomb's law:

F = k * (q1 * q2) /[tex]r^2[/tex]

where F is the magnitude of the force in Newtons (N), k is Coulomb's constant (9 x 10^9 N*m^2/C^2), q1 and q2 are the magnitudes of the charges in Coulombs (C), and r is the distance between the charges in meters (m).

Plugging in the given values, we get:

F =[tex](9 * 10^{9} N*m^{2}/C^2) * (-0.0004\ C) * (0.0003 C) / (3 m)^{2}[/tex]

Simplifying the expression, we get:

F = [tex]-3.6 * 10^{-6} N[/tex]

Note that the negative sign in the result indicates that the force is attractive. The magnitude of the force is [tex]3.6 * 10^{-6} N[/tex], rounded to two significant figures.

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A magnifying glass has a converging lens of focal length of 10.3 cm. At what distance from a nickel should you hold this lens to get an image with a magnification of +1.53?

Answers

The magnifying glass should be held 16.4 cm away from the nickel to obtain an image with a magnification of +1.53.

Using the thin lens equation, 1/f = 1/o + 1/i, where f is the focal length, o is the object distance, and i is the image distance, and the magnification equation, M = -i/o, where M is the magnification, we can solve for the object distance.

First, solve for the image distance i:

M = -i/o

1.53 = -i/o

i = -1.53o

Then, substitute i into the thin lens equation:

1/0.103 = 1/o + 1/(-1.53o)

Solving for o, we get:

o = 16.4 cm

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