M1U1 + M2V2 = (M1+M2)V, where M1 is the mass of the moving car, M2 is the mass of the stationary car, U1 is the initial velocity, and V is the common velocity after collision.
therefore;
(1060× 16) + (1830 ×0) = (1060 +1830) V
16960 = 2890 V
V = 5.869 m/s
The velocity of the cars after collision will be 5.689 m/s
Answer:
Option B, Some of the cars' kinetic energy was converted to sound and heat energy.
Explanation:
In an elastic collision, no energy is lost during and after collision. Thus, it can be said that in an elastic collision both momentum and kinetic energy remains conserved.
While in non-elastic collision, kinetic energy of the system is lost. However, the momentum of the system is conserved. Generally, during and after collision some of the kinetic energy is lost as thermal energy, sound energy etc.
Hence, option B is correct
A because of the resistors are four in this options first option is multiplied by 4
35 protons are present in an element whose atomic number is 35.
<u>Explanation:
</u>
In an atom of an element, the number of protons = atomic number
Number of protons = number of electrons
Mass Number = number of protons + number of neutrons
Hence, an atom with atomic number 35 will have 35 protons.
Answer:
![Q=7.9\times 10^{-10}\ C](https://tex.z-dn.net/?f=Q%3D7.9%5Ctimes%2010%5E%7B-10%7D%5C%20C)
Explanation:
Given that
V= 12 V
K=3
d= 2 mm
Area=5.00 $ 10#3 m2
Assume that
$ = Multiple sign
# = Negative sign
![A=5\times 10^{-3}\ m^2](https://tex.z-dn.net/?f=A%3D5%5Ctimes%2010%5E%7B-3%7D%5C%20m%5E2)
We Capacitance given as
For air
![C_1=\dfrac{\varepsilon _oA}{d}](https://tex.z-dn.net/?f=C_1%3D%5Cdfrac%7B%5Cvarepsilon%20_oA%7D%7Bd%7D)
![C_1=\dfrac{\varepsilon _oA}{d}](https://tex.z-dn.net/?f=C_1%3D%5Cdfrac%7B%5Cvarepsilon%20_oA%7D%7Bd%7D)
![C_1=\dfrac{8.85\times 10^{-12}\times 5\times 10^{-3}}{2\times 10^{-3}}](https://tex.z-dn.net/?f=C_1%3D%5Cdfrac%7B8.85%5Ctimes%2010%5E%7B-12%7D%5Ctimes%205%5Ctimes%2010%5E%7B-3%7D%7D%7B2%5Ctimes%2010%5E%7B-3%7D%7D)
![C_1=2.2\times 10^{-11}\ F](https://tex.z-dn.net/?f=C_1%3D2.2%5Ctimes%2010%5E%7B-11%7D%5C%20F)
![C_2=\dfrac{K\varepsilon _oA}{d}](https://tex.z-dn.net/?f=C_2%3D%5Cdfrac%7BK%5Cvarepsilon%20_oA%7D%7Bd%7D)
![C_2=\dfrac{3\times 8.85\times 10^{-12}\times 5\times 10^{-3}}{2\times 10^{-3}}\ F](https://tex.z-dn.net/?f=C_2%3D%5Cdfrac%7B3%5Ctimes%208.85%5Ctimes%2010%5E%7B-12%7D%5Ctimes%205%5Ctimes%2010%5E%7B-3%7D%7D%7B2%5Ctimes%2010%5E%7B-3%7D%7D%5C%20F)
![C_2=6.6\times 10^{-11}\ F](https://tex.z-dn.net/?f=C_2%3D6.6%5Ctimes%2010%5E%7B-11%7D%5C%20F)
Net capacitance
C=C₁+C₂
![C=8.8\times 10^{-11}\ F](https://tex.z-dn.net/?f=C%3D8.8%5Ctimes%2010%5E%7B-11%7D%5C%20F)
We know that charge Q given as
Q= C V
![Q=12\times 6.6\times 10^{-11}\ C](https://tex.z-dn.net/?f=Q%3D12%5Ctimes%206.6%5Ctimes%2010%5E%7B-11%7D%5C%20C)
![Q=7.9\times 10^{-10}\ C](https://tex.z-dn.net/?f=Q%3D7.9%5Ctimes%2010%5E%7B-10%7D%5C%20C)