Hmm doesnt soujd familiar
Answer:
Showing results for Two point charge q, separated by 1.5cm have change value of +2.0 and -4.0AND/C respectively what is the magnitude of the Electric force midway between them?
Search instead for Two point charge q, seperated by 1.5cm have change value of +2.0 and -4.0N/C respectively what is the magnitude of the Electric force midway between them?
Answer:
a) 0.138J
b) 3.58m/S
c) (1.52J)(I)
Explanation:
a) to find the increase in the translational kinetic energy you can use the relation
![\Delta E_k=W=W_g-W_p](https://tex.z-dn.net/?f=%5CDelta%20E_k%3DW%3DW_g-W_p)
where Wp is the work done by the person and Wg is the work done by the gravitational force
By replacing Wp=Fh1 and Wg=mgh2, being h1 the distance of the motion of the hand and h2 the distance of the yo-yo, m is the mass of the yo-yo, then you obtain:
![Wp=(0.35N)(0.16m)=0.056J\\\\Wg=(0.062kg)(9.8\frac{m}{s^2})(0.32m)=0.19J\\\\\Delta E_k=W=0.19J-0.056J=0.138J](https://tex.z-dn.net/?f=Wp%3D%280.35N%29%280.16m%29%3D0.056J%5C%5C%5C%5CWg%3D%280.062kg%29%289.8%5Cfrac%7Bm%7D%7Bs%5E2%7D%29%280.32m%29%3D0.19J%5C%5C%5C%5C%5CDelta%20E_k%3DW%3D0.19J-0.056J%3D0.138J)
the change in the translational kinetic energy is 0.138J
b) the new speed of the yo-yo is obtained by using the previous result and the formula for the kinetic energy of an object:
![\Delta E_k=\frac{1}{2}mv_f^2-\frac{1}{2}mv_o^2](https://tex.z-dn.net/?f=%5CDelta%20E_k%3D%5Cfrac%7B1%7D%7B2%7Dmv_f%5E2-%5Cfrac%7B1%7D%7B2%7Dmv_o%5E2)
where vf is the final speed, vo is the initial speed. By doing vf the subject of the formula and replacing you get:
![v_f=\sqrt{\frac{2}{m}}\sqrt{\Delta E_k+(1/2)mv_o^2}\\\\v_f=\sqrt{\frac{2}{0.062kg}}\sqrt{0.138J+1/2(0.062kg)(2.9m/s)^2}=3.58\frac{m}{s}](https://tex.z-dn.net/?f=v_f%3D%5Csqrt%7B%5Cfrac%7B2%7D%7Bm%7D%7D%5Csqrt%7B%5CDelta%20E_k%2B%281%2F2%29mv_o%5E2%7D%5C%5C%5C%5Cv_f%3D%5Csqrt%7B%5Cfrac%7B2%7D%7B0.062kg%7D%7D%5Csqrt%7B0.138J%2B1%2F2%280.062kg%29%282.9m%2Fs%29%5E2%7D%3D3.58%5Cfrac%7Bm%7D%7Bs%7D)
the new speed is 3.58m/s
c) in this case what you can compute is the quotient between the initial rotational energy and the final rotational energy
![\frac{E_{fr}}{E_{fr}}=\frac{1/2I\omega_f^2}{1/2I\omega_o^2}=\frac{\omega_f^2}{\omega_o^2}\\\\\omega_f=\frac{v_f}{r}\\\\\omega_o=\frac{v_o}{r}\\\\\frac{E_{fr}}{E_{fr}}=\frac{v_f^2}{v_o^2}=\frac{(3.58m/s)}{(2.9m/s)^2}=1.52J](https://tex.z-dn.net/?f=%5Cfrac%7BE_%7Bfr%7D%7D%7BE_%7Bfr%7D%7D%3D%5Cfrac%7B1%2F2I%5Comega_f%5E2%7D%7B1%2F2I%5Comega_o%5E2%7D%3D%5Cfrac%7B%5Comega_f%5E2%7D%7B%5Comega_o%5E2%7D%5C%5C%5C%5C%5Comega_f%3D%5Cfrac%7Bv_f%7D%7Br%7D%5C%5C%5C%5C%5Comega_o%3D%5Cfrac%7Bv_o%7D%7Br%7D%5C%5C%5C%5C%5Cfrac%7BE_%7Bfr%7D%7D%7BE_%7Bfr%7D%7D%3D%5Cfrac%7Bv_f%5E2%7D%7Bv_o%5E2%7D%3D%5Cfrac%7B%283.58m%2Fs%29%7D%7B%282.9m%2Fs%29%5E2%7D%3D1.52J)
hence, the change in Er is about 1.52J times the initial rotational energy
Answer:
To derive the fourth equation of motion, first we have to consider the equation for acceleration and then to rearrange it. or v2 = u2 + 2as and this equation of motion can be used to find the final velocity or the distance travelled if the other values are given.
Explanation:
v= u + at
s =( u + v ) t /2
s = ut + at2/2
v2 = u2 + 2as
Answer:
The <u>nucleus</u> is the core of an atom.
Step-by-step explanation:
That's just how it is.