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Amiraneli [1.4K]
2 years ago
13

Which of the following quantities can be measured in the same units as Work?

Physics
1 answer:
arlik [135]2 years ago
3 0

Answer:

Kinetic energy and potential energy.

Explanation:

Kinetic energy is directly proportional to the mass of the object and to the square of its velocity: K.E. = 1/2 m v^2. If the mass has units of kilograms and the velocity of meters per second, the kinetic energy has units of kilograms-meters squared per second squared. Kinetic energy is usually measured in units of Joules (J); one Joule is equal to 1 kg m^2 / s^2.

The formula for potential energy depends on the force acting on the two objects. For the gravitational force, the formula is P.E. = mgh, where m is the mass in kilograms, g is the acceleration due to gravity (9.8 m/s^2 at the surface of the earth) and h is the height in meters. Notice that gravitational potential energy has the same units as kinetic energy, kg m^2/s^2. In fact, all energy has the same units, kg m^2/s^2, and is measured using the unit Joule (J).

Work is also measured with the unit Joule (J).

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They will subtract to form a combined wave with a lower amplitude

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Which of the following scenarios would be optimal for obtaining a date from radioactive decay using these isotopes: 87Rb, 147Sm,
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Answer:

a) 238U, 40K and 87Rb, b)   235U and to a lesser extent 40K , c)  he 235U,

d) possibility is 14C , e)this period would be ideal for 14C , f) 14C should be used since it is the one with the least average life time, even though the measurements must be very careful

Explanation:

One of the applications of radioactive decay is the dating of different systems.

To do this, the quantity of radioactive material in a meter is determined and with the average life time, the time of the sample is found.

Let's write the half-life times of the given materials

87Rb T ½ = 4.75 1010 years

147Sm T ½ = 1.06 1011 years

235U = 7,038 108 years

238U = 4.47 109 years

40K = 1,248 109 years

14C = 5,568 103 years

we already have the half-life of the different elements given

a) meteors. As these decomposed in the formation of the solar system, their life time is around 3 109 to 5 109 years, so it is necessary to look for elements that have a life time of this order, among the candidates we have 238U, 40K and 87Rb if these elements were at the moment of the formation of these meteors, there must still be rations in them, instead elements 14C already completely adequate

b) rock. The formation period is 4.20-108 years, therefore one of the most promising elements is 235U and to a lesser extent 40K since it is more abundant in rocks. The other elements with higher life times have not decayed and therefore will not give a true value and the 14C is completely decayed

c) volcanic ash. Formation time 6107 years, the only element that has the possibility of having a count is the 235U, the others have a life time so long that they have not decayed and the 14C is complete, unbent

d) scarp of an earthquake formation time 5 101 years, The only one that has any possibility is 14C even when it has declined very little, all the others, you have time to long that has not decayed

e) INCA excavation. The time of this civilization is about 10000 to 500 years (104 to 5 102 years), we see that this period would be ideal for 14C since it has some period of cementation, the others have not decayed

f) Tree in Blepharitis. 14C should be used since it is the one with the least average life time, even though the measurements must be very careful because of a period of disintegration. We have such a long time that they have not decayed

8 0
3 years ago
The spectra for elements are _____.
mina [271]

Answer:

B. Same for a chemical family

4 0
3 years ago
Assuming the pick-up trucks, trailers and road conditions are exactly the same, which vehicle will take a longer distance to sto
mars1129 [50]

I’ve answered this before so I know the question is missing an important given and that given is: <span>1 has an empty trailer and the other has a fully loaded one. 

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6 0
3 years ago
A small but measurable current of 5.8 × 10-10 A exists in a copper wire whose diameter is 3.0 mm. The number of charge carriers
swat32

Answer:

a) The current density ,J = 2.05×10^-5

b) The drift velocity Vd= 1.51×10^-15

Explanation:

The equation for the current density and drift velocity is given by:

J = i/A = (ne)×Vd

Where i= current

A = Are

Vd = drift velocity

e = charge ,q= 1.602 ×10^-19C

n = volume

Given: i = 5.8×10^-10A

Raduis,r = 3mm= 3.0×10^-3m

n = 8.49×10^28m^3

a) Current density, J =( 5.8×10^-10)/[3.142(3.0×10^-3)^2]

J = (5.8×10^-10) /(2.83×10^-5)

J = 2.05 ×10^-5

b) Drift velocity, Vd = J/ (ne)

Vd = (2.05×10^-5)/ (8.49×10^28)(1.602×10^-19)

Vd = (2.05×10^-5)/(1.36 ×10^10)

Vd = 1.51× 10^-5

8 0
3 years ago
Read 2 more answers
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