Answer:
North of west
Explanation:
Given
40,000-ton luxury line traveling 20 knots towards west and
60,000 ton freighter traveling towards North with 10 knots
suppose v is the common velocity after collision
conserving momentum in west direction
![20\times 40,000=(20,000+60,000)v\cos \theta](https://tex.z-dn.net/?f=20%5Ctimes%2040%2C000%3D%2820%2C000%2B60%2C000%29v%5Ccos%20%5Ctheta)
suppose the final velocity makes \theta angle with x axis
![v\cos \theta =10----1](https://tex.z-dn.net/?f=v%5Ccos%20%5Ctheta%20%3D10----1)
Conserving Momentum in North direction
![10\times 60,000=80,000\times v\sin\theta](https://tex.z-dn.net/?f=10%5Ctimes%2060%2C000%3D80%2C000%5Ctimes%20v%5Csin%5Ctheta)
![v\sin \theta =\frac{15}{2}----2](https://tex.z-dn.net/?f=v%5Csin%20%5Ctheta%20%3D%5Cfrac%7B15%7D%7B2%7D----2)
divide 1 and 2
![\tan \theta =\frac{3}{4}](https://tex.z-dn.net/?f=%5Ctan%20%5Ctheta%20%3D%5Cfrac%7B3%7D%7B4%7D)
![\theta =37^{\circ}](https://tex.z-dn.net/?f=%5Ctheta%20%3D37%5E%7B%5Ccirc%7D)
so search in the area
North of west to find the ship
Given:
The given value is
.
To find:
The value of the given expression by using the Binomial approximation.
Explanation:
We have,
![(1.0004)^{\frac{1}{2}}](https://tex.z-dn.net/?f=%281.0004%29%5E%7B%5Cfrac%7B1%7D%7B2%7D%7D)
It can be written as:
![(1.0004)^{\frac{1}{2}}=(1+0.0004)^{\frac{1}{2}}](https://tex.z-dn.net/?f=%281.0004%29%5E%7B%5Cfrac%7B1%7D%7B2%7D%7D%3D%281%2B0.0004%29%5E%7B%5Cfrac%7B1%7D%7B2%7D%7D)
![[\because (1+x)^n=1+nx]](https://tex.z-dn.net/?f=%5B%5Cbecause%20%281%2Bx%29%5En%3D1%2Bnx%5D)
![(1.0004)^{\frac{1}{2}}=1+0.0002](https://tex.z-dn.net/?f=%281.0004%29%5E%7B%5Cfrac%7B1%7D%7B2%7D%7D%3D1%2B0.0002)
![(1.0004)^{\frac{1}{2}}=1.0002](https://tex.z-dn.net/?f=%281.0004%29%5E%7B%5Cfrac%7B1%7D%7B2%7D%7D%3D1.0002)
Therefore, the approximate value of the given expression is 1.0002.
Answer:
12.7m/s
Explanation:
Given parameters:
Mass of the diver = 77kg
Height = 8.18m
Unknown:
Final velocity = ?
Solution:
To solve this problem, we use one of the motion equations.
v² = u² + 2gh
v is the final velocity
u is the initial velocity
g is the acceleration due to gravity
h is the height
v² = 0² + (2 x 9.8 x 8.18)
v² = 160.3
v = 12.7m/s
Answer:
To write a number in scientific notation. First write a decimal point in the numbers so that there's only one digit to the left of the decimal point.
Answer:
No the given statement is not necessarily true.
Explanation:
We know that the kinetic energy of a particle of mass 'm' moving with velocity 'v' is given by
![K.E=\frac{1}{2}mv^{2}](https://tex.z-dn.net/?f=K.E%3D%5Cfrac%7B1%7D%7B2%7Dmv%5E%7B2%7D)
Similarly the momentum is given by ![m\times v](https://tex.z-dn.net/?f=m%5Ctimes%20v)
For 2 particles with masses
and moving with velocities
respectively the respective kinetic energies is given by
![K.E_{1}=\frac{1}{2}m_{1}v_{1}^{2}](https://tex.z-dn.net/?f=K.E_%7B1%7D%3D%5Cfrac%7B1%7D%7B2%7Dm_%7B1%7Dv_%7B1%7D%5E%7B2%7D)
![K.E_{2}=\frac{1}{2}m_{2}v_{2}^{2}](https://tex.z-dn.net/?f=K.E_%7B2%7D%3D%5Cfrac%7B1%7D%7B2%7Dm_%7B2%7Dv_%7B2%7D%5E%7B2%7D)
Similarly For 2 particles with masses
and moving with velocities
respectively the respective momenta are given by
![p_{1}=m_{1}\times v_{1}](https://tex.z-dn.net/?f=p_%7B1%7D%3Dm_%7B1%7D%5Ctimes%20v_%7B1%7D)
![p_{2}=m_{2}\times v_{2}](https://tex.z-dn.net/?f=p_%7B2%7D%3Dm_%7B2%7D%5Ctimes%20v_%7B2%7D)
Now since it is given that the two kinetic energies are equal thus we have
![\frac{1}{2}m_{1}v_{1}^{2}=\frac{1}{2}m_{2}v_{2}^{2}\\\\(m_{1}v_{1})\times v_{1}=(m_{2}v_{2})\times v_{2}\\\\p_{1}\times v_{1}=p_{2}\times v_{2}\\\\\therefore \frac{p_{1}}{p_{2}}=\frac{v_{2}}{v_{1}}............(i)](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7Dm_%7B1%7Dv_%7B1%7D%5E%7B2%7D%3D%5Cfrac%7B1%7D%7B2%7Dm_%7B2%7Dv_%7B2%7D%5E%7B2%7D%5C%5C%5C%5C%28m_%7B1%7Dv_%7B1%7D%29%5Ctimes%20v_%7B1%7D%3D%28m_%7B2%7Dv_%7B2%7D%29%5Ctimes%20v_%7B2%7D%5C%5C%5C%5Cp_%7B1%7D%5Ctimes%20v_%7B1%7D%3Dp_%7B2%7D%5Ctimes%20v_%7B2%7D%5C%5C%5C%5C%5Ctherefore%20%5Cfrac%7Bp_%7B1%7D%7D%7Bp_%7B2%7D%7D%3D%5Cfrac%7Bv_%7B2%7D%7D%7Bv_%7B1%7D%7D............%28i%29)
Thus we infer that the moumenta are not equal since the ratio on right of 'i' is not 1 , and can be 1 only if the velocities of the 2 particles are equal which becomes a special case and not a general case.