<span>change in velocity = final velocity - initial velocity = v - u
for comet:
uc = initial velocity of comet (before impact)
vc = final velocity of comet
mc= mass of comet
uc = 40000 kmph
vc = ?
mc= 10 x 10^14 kg
for probe:
up = initial velocity of probe (before impact)
vp = final velocity of probe
mp= mass of probe
up= 37000 kmph
vp= ?
mp= 372 kg
Now,
by principle of conservation of momentum
(mc x uc) - (mp x up) = (mc x vc) + (mp x vp)
Since probe is in comet after collision, vp= vc = V
then,
(mc x uc) - (mp x up) = V (mc + mp )
V = [(mc x uc) - (mp x up)] / (mc + mp )
= ((10 × 10^14 × 40000) - (372 × 37000)) ÷ ((10 × 10^14) + 372)
= ???
then,
change in velocity of the comet = ??? - (40000) =
</span>
Answer:
50J
Explanation:
At the top you have(A)
KE_a = O
PE_a = 100J
KE + PE = 100J
At the bottom you have (C)
KE_c= 100J
PE_c=0J
KE+PE = 100J
At point C:
You are at half the height.
We know that at H, PE =100J
PE_c = mgH
At C,
PE_c= mg (H/2) *at half the height
*m and g stay the same
Intuitively, the higher you are, the more potential energy you have.
If you decrease the height by a half, your PE will also decrease
At A:
PE_a / (mg) = H
At B:
PE_b / (mg) = H/2
to also get H on the right hand side, multiply by 2
2 (PE_b/ (mg))= H
2PE_b / (mg) = H
Ok, now that we have set up 2 equations (where H is isolated), find PE at B
AT A = AT B *This way you are saying that H = H (you compare both equations)
PE_a / (mg) = 2x PE_b / (mg)
*mg are the same for both cancel them (you can do that because of the = sign)
PE_a = 2PE_b
We know that PE_a = 100J
100J/2 = PE_b
PE at b = 50J
**FIND KE at b
We know that
KE_b + PE_b is always 100J
100J = 50J + KE_b
KE_b = 50J
Answer:
d. interaction atmosphere and biosphere interaction
Explanation:
hydrosphere, lithosphere, and biosphere interaction.
Answer:
Explanation:
Due to change in the position of 3 kg mass , the moment of inertia of the system changes , due to which angular speed changes . We shall apply conservation of angular momentum , because no external torque is acting .
Initial moment of inertia I₁ = M R² = 3 x 1 ² = 3 kg m²
Final moment of inertia I₂ = M R² = 3 x .3 ² = 0.27 kg m²
Applying law of conservation of angular momentum
I₁ ω₁ = I₂ ω₂
Putting the values ,
3 x .75 = .27 x ω₂
ω₂ = 8.33 rad / s
New angular speed = 8.33 rad /s .
Answer:
This is because it has ns1 electron configuration like the alkali metals.
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