Two identical balls collide<span> head on. The </span>initial velocity<span> of </span>one<span> is 0.75 </span>m/s<span> east, while that of the </span>other one<span> is 0.43 </span>m/s west<span>.</span>
In order to predict whether a star will eventually fuse oxygen into a heavier element, you mainly want to know about the star's mass.
<u>Explanation:</u>
Stellar evolution is procedure by which star experiences a succession of radical change during its lifetime. Depending upon the mass of stars, this lifetimes range from just two or three million years for most big to the trillions of years for a least massive, which is significantly longer or more than the age of universe.
All stars are conceived from falling billows of gas and residue, frequently called nebulae or sub-atomic mists. Throughout a large number of years, the protostar settles down into a condition of balance, turning out to be what is also known as the main- sequence - star.
Answer:
N = 23.4 N
Explanation:
After reading that long sentence, let's solve the question
The contact force is the so-called normal in this case we can find it by writing the translational equilibrium equation for the y axis
N - w₁ -w₂ =
N = m₁ g + m₂ g
N = g (m₁ + m₂)
let's calculate
N = 9.8 (0.760 + 1.630)
N = 23.4 N
This is the force of the support of the two blocks on the surface.
Answer:
p2 = 9.8×10^4 Pa
Explanation:
Total pressure is constant and PT = P = 1/2×ρ×v^2
So p1 + 1/2×ρ×(v1)^2 = p2 + 1/2×ρ×(v2)^2
from continuity we have ρ×A1×v1 = ρ×A2×v2
v2 = v1×A1/A2
and
r2 = 2×r1
then:
A2 = 4×A1
so,
v2 = (v1)/4
then:
p2 = p1 + 1/2×ρ×(v1)^2 - 1/2×ρ×(v2)^2 = p1 + 1/2×ρ×(v1)^2 - 1/2×ρ×(v1/4)^2
p2 = 3.0×10^4 Pa + 1/2×(1000 kg/m^3)×(12m/s)^2 - 1/2×(1000kg/m^3)×(12^2/16)
= 9.75×10^4 Pa
= 9.8×10^4 Pa
Therefore, the pressure in the wider section is 9.8×10^4 Pa
