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vampirchik [111]
3 years ago
13

(PLS HELP 20 POINTS, IM TOO DUMB FOR THIS) What is the total momentum of the system after the collision?​

Physics
1 answer:
Dmitry_Shevchenko [17]3 years ago
6 0
Using the law of conservation of momentum
m1u1+m2u2=m1v1+m2v2
Where m1 is mass of first object
m2 is mass of second object
u1 and u2 are initial velocities of object 1 and 2 respectively
v1 and v2 are final velocities of object 1 and 2 respectively
Here, they are moving as a system after collision. Thus they will posses same final velocity
m1u1 +m2u2=v(m1+m2)
Substituting values
600*4+0=v(600+400)
2400=v*1000
v=2.4 m/s

Now momentum of system
p=Mv
p=(600+400)*2.4
p=1000*2.4
Therefore p=2400 kg m/s
Hope this helps :)
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Answer:

2.92682 m

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250000000000 Hz

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Explanation:

c = Speed of light = 3\times 10^8\ m/s

\lambda = Wavelength

f = Frequency

Wavelength is given by

\lambda=\dfrac{c}{f}\\\Rightarrow \lambda=\dfrac{3\times 10^8}{102.5\times 10^6}\\\Rightarrow \lambda=2.92682\ m

The wavelength is 2.92682 m

Frequency is given by

f=\dfrac{c}{\lambda}\\\Rightarrow f=\dfrac{3\times 10^8}{0.2\times 10^{-9}}\\\Rightarrow f=1.5\times 10^{18}\ Hz

The frequency is 1.5\times 10^{18}\ Hz

f=\dfrac{c}{\lambda}\\\Rightarrow f=\dfrac{3\times 10^8}{1.2\times 10^{-3}}\\\Rightarrow f=250000000000\ Hz

The frequency is 250000000000 Hz

\lambda=\dfrac{c}{f}\\\Rightarrow \lambda=\dfrac{3\times 10^8}{1.04\times 10^{16}}\\\Rightarrow \lambda=2.88462\times 10^{-8}\ m

The wavelength is 2.88462\times 10^{-8}\ m

\lambda=\dfrac{c}{f}\\\Rightarrow \lambda=\dfrac{3\times 10^8}{1.42\times 10^{9}}\\\Rightarrow \lambda=0.21126\ m

The wavelength is 0.21126 m

\lambda=\dfrac{c}{f}\\\Rightarrow \lambda=\dfrac{3\times 10^8}{2.45\times 10^9}\\\Rightarrow \lambda=0.12244\ m

The wavelength is 0.12244 m

8 0
3 years ago
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