Answer:
7kgm/s
Explanation:
Using the law of conservation of momentum which states that the sum of momentum of bodies before collision is equal to the sum of the bodies after collision.
Let P1A and P1B be the initial momentum of the bodies A and B respectively
Let P2A and P2B be the final momentum of the bodies A and B respectively after collision.
Based on the law:
P1A+P2A = P1B + P2B
Given P1A = 5kgm/s
P2A = 0kgm/s(ball B at rest before collision)
P2A = -2.0kgm/s (negative because it moves in the negative x direction)
P2B = ?
Substituting the values in the equation gives;
5+0 = -2+P2B
5+2 = P2B
P2B = 7kgm/s
Harder. Not compressible(unless using an extremely strong force). Non-metal have more of a chance of breaking than metals.
Answer:
B
Explanation:
this is because the neutrons do not have a charge, the things that have charge in an atom are electrons and protons.
and in the nucleus of an atom, there are protons and neutrons so you can see that A is not the answer
if you see the periodic table, you will know that the number of electrons and protons are equal, so the charges cancel each other out, hence the charge of an atom will be neutral
let me give you a tip which I got from my teacher, never write there is no charge in the atom, this suggests that there is no protons or electrons.
instead, write, the it is neutral
hope it helps if not please report it so that someone else gets to try it out
Answer:
recall that heat absorbed released is given by
Q = mc*(T2 - T1)
where
m = mass (in g)
c = specific heat capacity (in J/g-k)
T = temperature (in C or K)
*note: Q is (+) when heat is absorbed and (-) when heat is released.
substituting,
Q = (480)*(0.97)*(234 - 22)
Q = 98707 J = 98.7 kJ
Explanation:
<span>D. sill
A sill forms when hot magma squeezes between layers of rock to form a hardened sheet of rock. A sill is basically a tabular sheet intrusion between old layers of sedimentary rock, volcanic beds, or in the direction of foliation inherent in metamorphic rock. The word sill means a concordant intrusive sheet.</span>