Answer:
a). 1.218 m/s
b). R=2.8![^{-3}](https://tex.z-dn.net/?f=%5E%7B-3%7D)
Explanation:
![m_{bullet}=6.99g*\frac{1kg}{1000g}=6.99x10^{-3}kg](https://tex.z-dn.net/?f=m_%7Bbullet%7D%3D6.99g%2A%5Cfrac%7B1kg%7D%7B1000g%7D%3D6.99x10%5E%7B-3%7Dkg)
![v_{bullet}=341\frac{m}{s}](https://tex.z-dn.net/?f=v_%7Bbullet%7D%3D341%5Cfrac%7Bm%7D%7Bs%7D)
Momentum of the motion the first part of the motion have a momentum that is:
![P_{1}=m_{bullet}*v_{bullet}](https://tex.z-dn.net/?f=P_%7B1%7D%3Dm_%7Bbullet%7D%2Av_%7Bbullet%7D)
![P_{1}=6.99x10^{-3}kg*341\frac{m}{s} \\P_{1}=2.3529](https://tex.z-dn.net/?f=P_%7B1%7D%3D6.99x10%5E%7B-3%7Dkg%2A341%5Cfrac%7Bm%7D%7Bs%7D%20%5C%5CP_%7B1%7D%3D2.3529)
The final momentum is the motion before the action so:
a).
![P_{2}=m_{b1}*v_{fbullet}+(m_{b2}+m_{bullet})*v_{f}}](https://tex.z-dn.net/?f=P_%7B2%7D%3Dm_%7Bb1%7D%2Av_%7Bfbullet%7D%2B%28m_%7Bb2%7D%2Bm_%7Bbullet%7D%29%2Av_%7Bf%7D%7D)
![P_{2}=1.202 kg*0.554\frac{m}{s}+(1.523kg+6.99x10^{-3}kg)*v_{f}](https://tex.z-dn.net/?f=P_%7B2%7D%3D1.202%20kg%2A0.554%5Cfrac%7Bm%7D%7Bs%7D%2B%281.523kg%2B6.99x10%5E%7B-3%7Dkg%29%2Av_%7Bf%7D)
![P_{1}=P_{2}](https://tex.z-dn.net/?f=P_%7B1%7D%3DP_%7B2%7D)
![2.529=0.665+(1.5299)*v_{f}\\v_{f}=\frac{1.864}{1.5299}\\v_{f}=1.218 \frac{m}{s}](https://tex.z-dn.net/?f=2.529%3D0.665%2B%281.5299%29%2Av_%7Bf%7D%5C%5Cv_%7Bf%7D%3D%5Cfrac%7B1.864%7D%7B1.5299%7D%5C%5Cv_%7Bf%7D%3D1.218%20%5Cfrac%7Bm%7D%7Bs%7D)
b).
kinetic energy
![K=\frac{1}{2}*m*(v)^{2}](https://tex.z-dn.net/?f=K%3D%5Cfrac%7B1%7D%7B2%7D%2Am%2A%28v%29%5E%7B2%7D)
Kinetic energy after
![Ka=\frac{1}{2}*1.202*(0.554)^{2}+\frac{1}{2}*1.523*(1.218)^{2}\\Ka=1.142 J](https://tex.z-dn.net/?f=Ka%3D%5Cfrac%7B1%7D%7B2%7D%2A1.202%2A%280.554%29%5E%7B2%7D%2B%5Cfrac%7B1%7D%7B2%7D%2A1.523%2A%281.218%29%5E%7B2%7D%5C%5CKa%3D1.142%20J)
Kinetic energy before
![Kb=\frac{1}{2}*mb*(vf)^{2}\\Kb=\frac{1}{2}*6.99x10^{-3}kg*(341)^{2}\\Kb=406.4J](https://tex.z-dn.net/?f=Kb%3D%5Cfrac%7B1%7D%7B2%7D%2Amb%2A%28vf%29%5E%7B2%7D%5C%5CKb%3D%5Cfrac%7B1%7D%7B2%7D%2A6.99x10%5E%7B-3%7Dkg%2A%28341%29%5E%7B2%7D%5C%5CKb%3D406.4J)
Ratio =![\frac{Ka}{Kb}](https://tex.z-dn.net/?f=%5Cfrac%7BKa%7D%7BKb%7D)
![R=\frac{1.14}{406.4}\\R=2.8x10^{-3}](https://tex.z-dn.net/?f=R%3D%5Cfrac%7B1.14%7D%7B406.4%7D%5C%5CR%3D2.8x10%5E%7B-3%7D)
Answer:
B
Explanation:
OOf we are doing this stuff atm
So if its faster at the front and slow at the back you can tell that its not slowing down because less of a force is there however at the front there is more of a force. Friction is low which means that its not makimg much contact so no sudden change of forces thats also why its B
Answer:
x(t) = - 6 cos 2t
Explanation:
Force of spring = - kx
k= spring constant
x= distance traveled by compressing
But force = mass × acceleration
==> Force = m × d²x/dt²
===> md²x/dt² = -kx
==> md²x/dt² + kx=0 ------------------------(1)
Now Again, by Hook's law
Force = -kx
==> 960=-k × 400
==> -k =960 /4 =240 N/m
ignoring -ve sign k= 240 N/m
Put given data in eq (1)
We get
60d²x/dt² + 240x=0
==> d²x/dt² + 4x=0
General solution for this differential eq is;
x(t) = A cos 2t + B sin 2t ------------------------(2)
Now initially
position of mass spring
at time = 0 sec
x (0) = 0 m
initial velocity v= = dx/dt= 6m/s
from (2) we have;
dx/dt= -2Asin 2t +2B cost 2t = v(t) --- (3)
put t =0 and dx/dt = v(0) = -6 we get;
-2A sin 2(0)+2Bcos(0) =-6
==> 2B = -6
B= -3
Putting B = 3 in eq (2) and ignoring first term (because it is not possible to find value of A with given initial conditions) - we get
x(t) = - 6 cos 2t
==>
Answer:
When waves overlap in-phase (crest meets crest or trough meets trough) the waves energy is additive and the amplitude increases.
Explanation:
When waves overlap out-of-phase (crest meets trough) the waves cancel and the amplitude (energy) decreases. When two interfering waves cancel each other out.
Answer: ![M = 5.98\times 10^{24} kg](https://tex.z-dn.net/?f=M%20%3D%205.98%5Ctimes%2010%5E%7B24%7D%20kg)
Explanation:
We know that force acting on an object due to Earth's gravity on the surface is given by:
![mg = G\frac{Mm}{r^2}\\ \Rightarrow g = \frac{GM}{r^2}](https://tex.z-dn.net/?f=mg%20%3D%20G%5Cfrac%7BMm%7D%7Br%5E2%7D%5C%5C%20%5CRightarrow%20g%20%3D%20%5Cfrac%7BGM%7D%7Br%5E2%7D)
where g is the acceleration due to gravity, r would be radius of Earth, M is the mass of Earth and G is the gravitational constant.
It is given that at pole, g = 9.830 m/s² and r = 6371 km = 6371 × 10³ m
![\Rightarrow M = \frac{g\times r^2}{G}](https://tex.z-dn.net/?f=%5CRightarrow%20M%20%3D%20%5Cfrac%7Bg%5Ctimes%20r%5E2%7D%7BG%7D)
![M = \frac{9.830 m/s^2 \times (6371 \times 10^3 m )^2}{6.67 \times 10^{-11} m^3 kg^{-1} s^{-2}}](https://tex.z-dn.net/?f=M%20%3D%20%5Cfrac%7B9.830%20m%2Fs%5E2%20%5Ctimes%20%286371%20%5Ctimes%2010%5E3%20m%20%29%5E2%7D%7B6.67%20%5Ctimes%2010%5E%7B-11%7D%20m%5E3%20kg%5E%7B-1%7D%20s%5E%7B-2%7D%7D)
![\Rightarrow M = 5.98\times 10^{24} kg](https://tex.z-dn.net/?f=%5CRightarrow%20M%20%3D%205.98%5Ctimes%2010%5E%7B24%7D%20kg)
Hence, Earth's mass is ![5.98\times 10^{24} kg](https://tex.z-dn.net/?f=%205.98%5Ctimes%2010%5E%7B24%7D%20kg)