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Softa [21]
2 years ago
14

Q Explain the following terms. a) A source of electric current b) Electrical appliances

Physics
1 answer:
exis [7]2 years ago
4 0

Answer:a)battery

B)ironer

Explanation:

You might be interested in
A cannonball with a mass of 1.0 kilogram is fired horizontally from a 500.-kilogram cannon, initially at rest, on a horizontal,
vfiekz [6]
The impulse J is equal to the magnitude of the force applied to the cannonball times the time it is applied:
J=F \Delta t
But the impulse is also equal to the change in momentum of the cannonball:
J=\Delta p
If we put the two equations together, we find
F \Delta t= \Delta p
And since we know the magnitude of the average force and the time, we can calculate the change in momentum:
\Delta p= F \Delta t=(8.0 \cdot 10^3 N)(1.0 \cdot 10^{-1} s)=800 kg m/s
7 0
3 years ago
The brakes application to a car produce an acceleration of 6ms2 in the opposite direction to the motion .If the car takes 2 seco
Anna [14]

Answer:12 meter

Explanation:

acceleration(a)=6m/s^2

Time(t)=2 seconds

Distance =(a x t^2)/2

Distance =(6 x 2^2)/2

Distance=(6 x 2 x 2)/2

Distance=24/2

Distance =12

Distance is 12 meters

7 0
2 years ago
Cesium-137 undergoes beta decay and has a half-life of 30.0 years. How many beta particles are emitted by a 14.0-g sample of ces
Mandarinka [93]

Answer: 0.81\times 10^{16} beta particles

Explanation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Given mass = 14.0 g

Molar mass = 137 g/mol

\text{Number of moles of cesium}=\frac{14.0g}{137g/mol}=0.102moles

According to avogadro's law, 1 mole of every substance weighs equal to its molecular mass and contains avogadro's number 6.023\times 10^{23} of particles.

1 mole of cesium contains atoms =  6.023\times 10^{23}

0.102 moles of cesium contains atoms =  \frac{6.023\times 10^{23}}{1}\times 0.102=0.614\times 10^{23}

The relation of atoms with time for radioactivbe decay is:

N_t=N_0\times \frac{1}{2}^{\frac{t}{t_{\frac{1}{2}}}}

Where N_t =atoms left undecayed

N_0 = initial atoms

t = time taken for decay = 3 minutes

{t_{\frac{1}{2}}} = half life = 30.0 years = 1.577\times 10^7 minutes

The fraction that decays  :  1-(\frac{1}{2})^{\frac{3}{1.577\times 10^7}}=1.32\times 10^{-7}

Amount of particles that decay is  = 0.614\times 10^{23}\times 1.32\times 10^{-7}=0.81\times 10^{16}

Thus 0.81\times 10^{16} beta particles are emitted by a 14.0-g sample of cesium-137 in three minutes.

7 0
3 years ago
Why is pseudoscience bad?
USPshnik [31]

Answer:

It is quite difficult to picture a pseudoscientist—really picture him or her over the course of a day, a year, or a whole career. What kind or research does he or she actually do, what differentiates him or her from a carpenter, or a historian, or a working scientist? In short, what do such people think they are up to?

… it is a significant point for reflection that all individuals who have been called “pseudoscientists” have considered themselves to be “scientists”, with no prefix.

The answer might surprise you. When they find time after the obligation of supporting themselves, they read papers in specific areas, propose theories, gather data, write articles, and, maybe, publish them. What they imagine they are doing is, in a word, “science”. They might be wrong about that—many of us hold incorrect judgments about the true nature of our activities—but surely it is a significant point for reflection that all individuals who have been called “pseudoscientists” have considered themselves to be “scientists”, with no prefix.

What is pseudoscience?

“Pseudoscience” is a bad category for analysis. It exists entirely as a negative attribution that scientists and non‐scientists hurl at others but never apply to themselves. Not only do they apply the term exclusively as a discrediting slur, they do so inconsistently. Over the past two‐and‐a‐quarter centuries since the term popped into the Western European languages, a great number of disparate doctrines have been categorized as sharing a core quality—pseudoscientificity, if you will—when in fact they do not. It is based on this diversity that I refer to such beliefs and theories as “fringe” rather than as “pseudo”: Their defining characteristic is the distance from the center of the mainstream scientific consensus in whichever direction, not some essential property they share.

Scholars have by and large tended to ignore fringe science as regrettable sideshows to the main narrative of the history of science, but there is a good deal to be learned by applying the same tools of analysis that have been used to understand mainstream science. This is not, I stress, to imply that there is no difference between hollow‐Earth theories and geophysics; on the contrary, the differences are the point of the analysis. Focusing on the historical and conceptual relationship between the fringe and the core of the various sciences as that blurry border has fluctuated over the centuries provides powerful analytical leverage for understanding where contemporary anti‐science movements come from and how mainstream scientists might address them.

As soon as professionalization blossomed, tagging competing theories as pseudoscientific became an important tool for scientists to define what they understood science to be

The central claim of this essay is that the concept of “pseudoscience” was called into being as the shadow of professional science. Before science became a profession—with formalized training, credentialing, publishing venues, careers—the category of pseudoscience did not exist. As soon as professionalization blossomed, tagging competing theories as pseudoscientific became an important tool for scientists to define what they understood science to be. In fact, despite many decades of strenuous effort by philosophers and historians, a precise definition of “science” remains elusive. It should be noted however that the absence of such definitional clarity has not seriously inhibited the ability of scientists to deepen our understanding of nature tremendously.

Explanation:

8 0
2 years ago
Two identical 0.400 kg masses are pressed against opposite ends of a light spring of force constant 1.75 N/cm, compressing the s
nlexa [21]

Answer:

0.853 m/s

Explanation:

Total energy stored in the spring = Total kinetic energy of the masses.

1/2ke² = 1/2m'v².................... Equation 1

Where k = spring constant of the spring, e = extension, m' = total mass, v = speed of the masses.

make v the subject of the equation,

v = e[√(k/m')].................... Equation 2

Given: e = 39 cm = 0.39 m, m' = 0.4+0.4 = 0.8 kg, k = 1.75 N/cm = 175 N/m.

Substitute into equation 2

v = 0.39[√(1.75/0.8)

v = 0.39[2.1875]

v = 0.853 m/s

Hence the speed of each mass = 0.853 m/s

7 0
3 years ago
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