<span>D. A burning candle. (chemical energy into energy of heat and light, i.e. thermal and wave)</span>
There is not enough information given to answer with. The force of gravity at the planet's surface depends on the planet's radius as well as its mass. The planet could have exactly the same mass as Earth has. But if it's radius is only 71% of Earth's radius, then gravity on its surface will be twice as strong as gravity on Earth.
Answer:
(a)0.531m/s
(b)0.00169
Explanation:
We are given that
Mass of bullet, m=4.67 g=![4.67\times 10^{-3} kg](https://tex.z-dn.net/?f=4.67%5Ctimes%2010%5E%7B-3%7D%20kg)
1 kg =1000 g
Speed of bullet, v=357m/s
Mass of block 1,![m_1=1177g=1.177kg](https://tex.z-dn.net/?f=m_1%3D1177g%3D1.177kg)
Mass of block 2,![m_2=1626 g=1.626 kg](https://tex.z-dn.net/?f=m_2%3D1626%20g%3D1.626%20kg)
Velocity of block 1,![v_1=0.681m/s](https://tex.z-dn.net/?f=v_1%3D0.681m%2Fs)
(a)
Let velocity of the second block after the bullet imbeds itself=v2
Using conservation of momentum
Initial momentum=Final momentum
![mv=m_1v_1+(m+m_2)v_2](https://tex.z-dn.net/?f=mv%3Dm_1v_1%2B%28m%2Bm_2%29v_2)
![4.67\times 10^{-3}\times 357+1.177(0)+1.626(0)=1.177\times 0.681+(4.67\times 10^{-3}+1.626)v_2](https://tex.z-dn.net/?f=4.67%5Ctimes%2010%5E%7B-3%7D%5Ctimes%20357%2B1.177%280%29%2B1.626%280%29%3D1.177%5Ctimes%200.681%2B%284.67%5Ctimes%2010%5E%7B-3%7D%2B1.626%29v_2)
![1.66719=0.801537+1.63067v_2](https://tex.z-dn.net/?f=1.66719%3D0.801537%2B1.63067v_2)
![1.66719-0.801537=1.63067v_2](https://tex.z-dn.net/?f=1.66719-0.801537%3D1.63067v_2)
![0.865653=1.63067v_2](https://tex.z-dn.net/?f=0.865653%3D1.63067v_2)
![v_2=\frac{0.865653}{1.63067}](https://tex.z-dn.net/?f=v_2%3D%5Cfrac%7B0.865653%7D%7B1.63067%7D)
![v_2=0.531m/s](https://tex.z-dn.net/?f=v_2%3D0.531m%2Fs)
Hence, the velocity of the second block after the bullet imbeds itself=0.531m/s
(b)Initial kinetic energy before collision
![K_i=\frac{1}{2}mv^2](https://tex.z-dn.net/?f=K_i%3D%5Cfrac%7B1%7D%7B2%7Dmv%5E2)
![k_i=\frac{1}{2}(4.67\times 10^{-3}\times (357)^2)](https://tex.z-dn.net/?f=k_i%3D%5Cfrac%7B1%7D%7B2%7D%284.67%5Ctimes%2010%5E%7B-3%7D%5Ctimes%20%28357%29%5E2%29)
![k_i=297.59 J](https://tex.z-dn.net/?f=k_i%3D297.59%20J)
Final kinetic energy after collision
![K_f=\frac{1}{2}m_1v^2_1+\frac{1}{2}(m+m_2)v^2_2](https://tex.z-dn.net/?f=K_f%3D%5Cfrac%7B1%7D%7B2%7Dm_1v%5E2_1%2B%5Cfrac%7B1%7D%7B2%7D%28m%2Bm_2%29v%5E2_2)
![K_f=\frac{1}{2}(1.177)(0.681)^2+\frac{1}{2}(4.67\times 10^{-3}+1.626)(0.531)^2](https://tex.z-dn.net/?f=K_f%3D%5Cfrac%7B1%7D%7B2%7D%281.177%29%280.681%29%5E2%2B%5Cfrac%7B1%7D%7B2%7D%284.67%5Ctimes%2010%5E%7B-3%7D%2B1.626%29%280.531%29%5E2)
![K_f=0.5028 J](https://tex.z-dn.net/?f=K_f%3D0.5028%20J)
Now, he ratio of the total kinetic energy after the collision to that before the collision
=![\frac{k_f}{k_i}=\frac{0.5028}{297.59}](https://tex.z-dn.net/?f=%5Cfrac%7Bk_f%7D%7Bk_i%7D%3D%5Cfrac%7B0.5028%7D%7B297.59%7D)
=0.00169
Answer:
The image height is 3.0 cm
Explanation:
Given;
object distance,
= 15.0 cm
image distance,
= 5.0 cm
height of the object,
= 9.0 cm
height of the image,
= ?
Apply lens equation;
![\frac{h_i}{h_o} = -\frac{d_i}{d_o}\\\\ h_i = h_o(-\frac{d_i}{d_o})\\\\h_i = -9(\frac{5}{15} )\\\\h_i = -3 \ cm](https://tex.z-dn.net/?f=%5Cfrac%7Bh_i%7D%7Bh_o%7D%20%3D%20-%5Cfrac%7Bd_i%7D%7Bd_o%7D%5C%5C%5C%5C%20h_i%20%3D%20h_o%28-%5Cfrac%7Bd_i%7D%7Bd_o%7D%29%5C%5C%5C%5Ch_i%20%3D%20-9%28%5Cfrac%7B5%7D%7B15%7D%20%29%5C%5C%5C%5Ch_i%20%3D%20-3%20%5C%20cm)
Therefore, the image height is 3.0 cm. The negative values for image height indicate that the image is an inverted image.
Answer:
Electric force, ![F=-3.59\times 10^6\ N](https://tex.z-dn.net/?f=F%3D-3.59%5Ctimes%2010%5E6%5C%20N)
Explanation:
Given that,
Electric charge 1, ![q_1=+40\ C](https://tex.z-dn.net/?f=q_1%3D%2B40%5C%20C)
Electric charge 2, ![q_2=-40\ C](https://tex.z-dn.net/?f=q_2%3D-40%5C%20C)
Distance, ![d=2\ km=2\times 10^3\ m](https://tex.z-dn.net/?f=d%3D2%5C%20km%3D2%5Ctimes%2010%5E3%5C%20m)
To find,
The electric force between these two sets of charges.
Solution,
There exists a force between two electric charges and this force is called electrostatic force. It is equal to the product of electric charged divided by square of distance between them.
![F=k\dfrac{q_1q_2}{d^2}](https://tex.z-dn.net/?f=F%3Dk%5Cdfrac%7Bq_1q_2%7D%7Bd%5E2%7D)
k is the electrostatic constant
![F=8.99\times 10^9\times \dfrac{40\times (-40)}{(2\times 10^3)^2}](https://tex.z-dn.net/?f=F%3D8.99%5Ctimes%2010%5E9%5Ctimes%20%5Cdfrac%7B40%5Ctimes%20%28-40%29%7D%7B%282%5Ctimes%2010%5E3%29%5E2%7D)
![F=-3.59\times 10^6\ N](https://tex.z-dn.net/?f=F%3D-3.59%5Ctimes%2010%5E6%5C%20N)
So, the electric force between these two sets of charges is
.