Answer:
d. -y direction
Explanation:
Given that
m₁=0.145 kg, u₁=2.1 j m/s
m₂= 0.0570 kg , u₂ = -8.8 j m/s
We know that linear momentum is the product of mass and velocity
P= Mass x velocity
Therefore the total linear momentum P
P= m₁u₁+ m₂u₂
Now by putting the values
P = 0.145 x ( 2.1 j) + 0.057 ( - 8.8 j) kg.m/s
P =0.3045 j - 0.5016 j kg.m/s
P= - 0.1971 j kg.m/s
P= 0.1971 (- j) kg.m/s
So we can say that the direction of the liner momentum of the system will be in the negative y -direction.
d. -y direction
The speed by which the spaceprobe travels when all the rocket fuel is used up will be 29.262 m/sec.
<h3>What is the law of conservation of linear momentum?</h3>
According to the law of conservation of linear momentum before the collision is equal to the momentum after the collision. These laws state how momentum gets conserved.
Unit conversion;
1 km/sec = 1000 m/sec
Given data;
Spaceprobe speed = 1.795 km/s = 1795 m /sec
Probe mass = 635.0 kg
Fuel mass = 4092.0 kg
Expelled propellent velocity = 4.161 km/s = 41461 m/sec
From the momentum conservation principle;
Hence, the speed by which the spaceprobe travels when all the rocket fuel is used up will be 29.262 m/sec.
To learn more about the law of conservation of momentum refer to:
brainly.com/question/1113396
#SPJ1
The solution would be like this for this specific problem:
Given:
Oil spill radius constant rate increase = 1 m/s
r = 34m
Let the area of the spill
be A and then let its radius be r. <span>
Then A = π r². </span>
Differentiating with
respect to t: <span>
dA / dt = 2π r dr / dt. </span>
Substituting r = 34 and dr
/ dt = 1:
dA/dt = 2 π * r * dr / dt
dA/dt = 2π * 34 * 1 = 68π
= 214 m²/s, to 3 significant figures.
<span>So, given that the radius
is 34m, then the area of the spill is increasing at 214 m²/s.</span>
Young people
Explanation:
cuz old people can't do sports
Answer: 5.5km
Explanation:
Atmospheric pressure will be 500 mb (that is half of the total 1000mb air pressure).
Pressure decreases with increasing altitude. This is because at At higher altitudes, there are fewer air molecules above a the known or given surface than a similar surface at lower levels.
Pressure decreasing with higher altitudes also means that air pressure decreases rapidly at lowerevels but more slowly at higher levels.
It is also known that more than half of the atmospheric molecules are located below 5.5 km(that is atmospheric pressure decreases within the lowest 5.5 km to about fifty(50) percent( that is 500 millibar).
