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What are the various components of \(E=mc^2\)?
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What are the various components of \(E=mc^2\)?What are the various components of \(E=mc^2\)? ...
by MathsGee Platinum
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asked in Mathematics Jan 11
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If you start with 24 grams of a radioactive element with a half-life of one month, how many grams of the radioactive element will be left after 2 months?

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Two bodies of equal mass are moving in circular paths at equal speed.
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Two bodies of equal mass are moving in circular paths at equal speed.Two bodies of equal mass are moving in circular paths at equal speed. The first body moves in a circle whose radius is twice as great as that of the s ...
by Digitized Wooden
(1,150 points)
asked in Physics Oct 5, 2021
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After a projectile is fired horizontally near the earth's surface, the horizontal component of its velocity, neglecting any air resistance:
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Lenz's law is an example of the law of conservation of one of the following. Is it conservation of:
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Lenz's law is an example of the law of conservation of one of the following. Is it conservation of:Lenz's law is an example of the law of conservation of one of the following. Is it conservation of: w) mass x) charge y) momentum z) energy ...
by Digitized Wooden
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asked in Physics Oct 5, 2021
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A ball falling vertically from rest for 3 seconds travels very nearly:
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A ball falling vertically from rest for 3 seconds travels very nearly:A ball falling vertically from rest for 3 seconds travels very nearly: w) 14.7 meters x) 29.4 meters y) 44.1 meters z) 88.2 meters ...
by Digitized Wooden
(1,150 points)
asked in Physics Oct 5, 2021
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The half-life of an isotope of an element is 5 days. The mass of a 10 gram sample of this isotope remaining after 20 days is:
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If the distance between the earth and moon were halved, the force of the attraction between them would be:
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If the distance between the earth and moon were halved, the force of the attraction between them would be:If the distance between the earth and moon were halved, the force of the attraction between them would be: w) one fourth as great x) one half as gre ...
by Digitized Wooden
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asked in Physics Oct 5, 2021
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Two steel balls are at a distance S from one another. As the mass of ONE of the balls is doubled, the gravitational force of attraction between them is:

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The time needed for a net force of 10 newtons to change the velocity of a 5 kilograms mass by 3 meters/second is:
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The time needed for a net force of 10 newtons to change the velocity of a 5 kilograms mass by 3 meters/second is:The time needed for a net force of 10 newtons to change the velocity of a 5 kilograms mass by 3 meters/second is: w) 1.5 seconds x) 6 seconds y) 16 ...
by Digitized Wooden
(1,150 points)
asked in Physics Oct 5, 2021
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If the resultant force acting on a body of constant mass is zero, the body's momentum is:
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If the resultant force acting on a body of constant mass is zero, the body's momentum is:If the resultant force acting on a body of constant mass is zero, the body's momentum is: w) increasing x) decreasing y) always zero z) constant ...
by Digitized Wooden
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asked in Physics Oct 5, 2021
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The velocity, \(\mathrm{V}\), of a particle after t seconds, is \(V=3 t^{2}+2 t-1\). Find the acceleration of the particle after 2 seconds.
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A force ( \(10 i+4 j)\) N acts on a body of mass \(2 \mathrm{~kg}\) which is at rest. Find the velocity after 3 seconds.
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A force ( \(10 i+4 j)\) N acts on a body of mass \(2 \mathrm{~kg}\) which is at rest. Find the velocity after 3 seconds.A force ( \(10 i+4 j)\) N acts on a body of mass \(2 \mathrm{~kg}\) which is at rest. Find the velocity after 3 seconds. A. \(\left(\frac{5 i}{3}+\fr ...
by MathsGee Platinum
(103,030 points)
asked in Mathematics Sep 12, 2021
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A ball is kicked straight up in the air. If its height after \(t\) seconds is \(h(t)=276 t-6 t^{2}\), calculate the maximum height the ball will attain.
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If the displacement of a particle (in meter) after \(t\) seconds is \(s=t^{3}-4 t+6\), determine the velocity after 8 seconds.
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If the displacement of a particle (in meter) after \(t\) seconds is \(s=-3 t^{2}-2 t-3\), determine the acceleration after 6 seconds.
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If the displacement of a particle (in meters) after \(t\) seconds is \(s=-5 t^{2}+4 t+6\), determine the velocity after 4 seconds.
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The displacement of a particle (in meters) after \(t\) seconds is: \[ s=4 t+4 \]
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The displacement of a particle (in meters) after \(t\) seconds is: \[ s=4 t+4 \]The displacement of a particle (in meters) after \(t\) seconds is: \ s=4 t+4 \ (a) Determine the displacement after 4 seconds. (b) Determine the dis ...
by MathsGee Platinum
(103,030 points)
asked in Mathematics Aug 31, 2021
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