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Dmitry [639]
3 years ago
10

Which of the following is an example of work being done on an object?

Physics
2 answers:
slega [8]3 years ago
6 0

the answer is d. all of these

tensa zangetsu [6.8K]3 years ago
4 0

Answer:

d. All of these

Explanation:

work is said to be done when a force is applied to an object through a certain distance. the SI unit of workdone is joules or newton per meter

mathematically

workdone = force x distance.

from the answers,  work is being done because there is force applied in a certain distance.

  • from wagon is used to carry vegetables from a garden.
  • pulley is used to get water from a well.
  • hammer is used to remove a nail from a wall.

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The big bang theory has finally answered one of the biggest questions of science—the origin of the universe.
ss7ja [257]

Answer:

False

Explanation:

According to the big bang theory, matter was an infinitely small and very high density point which at one point exploded and expanded in all directions, creating what we know as our Universe, which also includes space and time . This happened about 13.8 billion years ago. Theoretical physicists have managed to reconstruct this chronology of events from 1/100 of a second after the Big Bang. After the explosion, while the Universe expanded, it cooled sufficiently and the first subatomic particles were formed: Electrons, Positrons, Mesons, Barions, Neutrinos, Photons among others. Today more than 90 particles are known. This theory solves many unknowns and is very well received by the scientific community, however there is still much to solve, for example, one of the great unsolved scientific problems in the expanding Universe model is whether the Universe is open or closed.

An attempt to solve this problem is to determine if the average density of matter in the Universe is greater than the critical value in Friedmann's model. The mass of a galaxy can be measured by observing the movement of its stars; multiplying the mass of each galaxy by the number of galaxies, it is seen that the density is only 5 to 10% of the critical value.

6 0
3 years ago
The asteroid 2007 VK184 has one chance in 2940 of hitting the Earth between the years 2048 and 2057. The asteroid is 130 meters
N76 [4]

Answer:

 The average death rate for this type of event is closest to 350 people per event.

Explanation:

Between the years 2048 and 2057

Is 10 years. Since there is only one chance of event between these years, there is no point to consider it.

The total number of events = 2940 chances.

average death rate per event = total number of dead people divided by total number of events of occurrence.

1000000/2940 = 340.134

The average death rate for this type of event is therefore closest to 350 people per event

4 0
2 years ago
The circumference of a sphere was measured to be
professor190 [17]

To solve this problem we will apply the concepts related to the calculation of the surface, volume and error through the differentiation of the formulas given for the calculation of these values in a circle. Our values given at the beginning are

\phi = 76cm

Error (dr) = 0.5cm

The radius then would be

\phi = 2\pi r \\76cm = 2\pi r\\r = \frac{38}{\pi} cm

And

\frac{d\phi}{dr} = 2\pi \\d\phi = 2\pi dr \\0.5 = 2\pi dr

PART A ) For the Surface Area we have that,

A = 4\pi r^2 \\A = 4\pi (\frac{38}{\pi})^2\\A = \frac{5776}{\pi}

Deriving we have that the change in the Area is equivalent to the maximum error, therefore

\frac{dA}{dr} = 4\pi (2r) \\dA = 4r (2\pi dr)

Maximum error:

dA = 4(\frac{38}{\pi})(0.5)

dA = \frac{76}{\pi}cm^2

The relative error is that between the value of the Area and the maximum error, therefore:

\frac{dA}{A} = \frac{\frac{76}{\pi}}{\frac{5776}{\pi}}

\frac{dA}{A} = 0.01315 = 1.31\%

PART B) For the volume we repeat the same process but now with the formula for the calculation of the volume in a sphere, so

V = \frac{4}{3} \pi r^3

V = \frac{4}{3} \pi (\frac{38}{\pi})^3

V = \frac{219488}{3\pi^2}

Therefore the Maximum Error would be,

\frac{dV}{dr} = \frac{4}{3} 3\pi r^2

dV = 2r^2 (2\pi dr)

dV = 4r^2 (\pi dr)

Replacing the value for the radius

dV = 4(\frac{38}{\pi})^2(0.5)

dV = \frac{2888}{\pi^2} cm^3

And the relative Error

\frac{dV}{V} = \frac{ \frac{2888}{\pi^2}}{ \frac{219488}{3\pi^2} }

\frac{dV}{V} = 0.03947

\frac{dV}{V} = 3.947\%

3 0
2 years ago
Which image illustrates refraction?
maks197457 [2]

Answer:

B illustrates refraction

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3 years ago
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