it allows only a reduced number of electrons to flow through it.
Find the horizontal components vcos30 ...one goes right and one goes left so they cancel each other.
Find vertical components vsin30.....there are two of them.... so 2vcos30....hey presto... resultant velocity = 2vCos30
Answer:
trigonometry (guessing)
Explanation:
ellipse: is the shape of an orbit : looks like an oval
periapsis : shortest distance between something like the moon and the planet its orbiting around like the earth
parallax is triangulation. like how gps works. looking at a star one day and then looking at it again 6 months later, an astronomer can see a difference in the viewing angle for the star. With trigonometry, the different angles yield a distance. This technique works for stars within about 400 light years of earth
https://science.howstuffworks.com/question224.htm
By comparing the intrinsic brightness to the star's apparent brightness we can calculate the distance of stars
1/r^2 rule states that the apparent brightness of a light source is proportional to the square of its distance.Jan 11, 2022
https://www.space.com/30417-parallax.html
alternative distance measurement for stars used by most astronomers is the parsec. A star with a parallax angle of 1 arcsecond has a distance of 1 parsec, or 1 parsec per arcsecond of parallax, which is about 3.26 light years
blossoms.mit.edu
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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:
1. 2.5s
Explanation:
1. For time, divide Distance / speed
25m / 10
=2.5s