Answer:
Pressure is equal to the ratio of thrust to the area in contact. Upthrust is a force exerted by the fluids on an object placed in the fluid . Upthrust acts in upward direction.
Answer:
2nd one if you ask me
almost every baby stats to talk after their a year old
I believe the answer is free electrons
Answer with Explanation:
We are given that
Initial velocity,u=4.5 m/s
Time=t =0.5 s
Final velocity=v=0m/s
We have to find the deceleration and estimate the force exerted by wall on you.
We know that
Acceleration=
Using the formula
Acceleration=
deceleration=a=
We know that
Force =ma
Using the formula and suppose mass of my body=m=40 kg
The force exerted by wall on you
Force=
Answer:
Explanation:
Given that,
At one instant,
Center of mass is at 2m
Xcm = 2m
And velocity =5•i m/s
One of the particle is at the origin
M1=? X1 =0
The other has a mass M2=0.1kg
And it is at rest at position X2= 8m
a. Center of mass is given as
Xcm = (M1•X1 + M2•X2) / (M1+M2)
2 = (M1×0 + 0.1×8) /(M1 + 0.1)
2 = (0+ 0.8) /(M1 + 0.1)
Cross multiply
2(M1+0.1) = 0.8
2M1 + 0.2 =0.8
2M1 = 0.8-0.2
2M1 = 0.6
M1 = 0.6/2
M1 = 0.3kg
b. Total momentum, this is an inelastic collision and it momentum after collision is given as
P= (M1+M2)V
P = (0.3+0.1)×5•i
P = 0.4 × 5•i
P = 2 •i kgm/s
c. Velocity of particle at origin
Using conversation of momentum
Momentum before collision is equal to momentum after collision
P(before) = M1 • V1 + M2 • V2
We are told that M2 is initially at rest, then, V2=0
So, P(before) = 0.3V1
We already got P(after) = 2 •i kgm/s in part b of the question
Then,
P(before) = P(after)
0.3V1 = 2 •i
V1 = 2/0.3 •i
V1 = 6 ⅔ •i m/s
V1 = 6.667 •i m/s
