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timama [110]
3 years ago
7

Which field(s) are created by an electron when it moves?

Physics
2 answers:
Jet001 [13]3 years ago
7 0
Positive charge i think im right pls let me know   
amid [387]3 years ago
7 0
Dude thats a hard one give me a sec
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When light in vacuum is incident at the polarizing angle (Brewster's angle) on a certain glass slab, the angle of refraction is
marysya [2.9K]

Answer:

Explanation:

θ( p ) + θ( r ) = 90  

θ (r) = angle of refraction and θ ( p ) is polarising angle.

given θ ( r ) = 31.8

θ ( p ) = 90 - 31.8 = 58.2 degree

ii ) Tanθ ( p ) = n ( refractive index ) = Tan 58.2 = 1.6

3 0
3 years ago
Streak is a reliable identifier of a mineral. true or false
inysia [295]

Answer:

true?

Explanation:

Im positive but not 100% sure wait for someone else to answer and see if they say the same.

8 0
3 years ago
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The type of exercise that you choose will determine what _____will be worked
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Does muscle go in the blank?
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3 years ago
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You pull straight up on the string of a yo-yo with a force 0.35 N, and while your hand is moving up a distance 0.16 m, the yo-yo
jarptica [38.1K]

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

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

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

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}

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

hence, the change in Er is about 1.52J times the initial rotational energy

5 0
2 years ago
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The small ball of mass m and its supporting wire become a simple pendulum when the horizontal cord is severed. Determine the rat
natali 33 [55]

Answer:

See the attached image and the explanation below

Explanation:

We must draw a schematic of the described problem, after the sketch it is necessary to make a free body diagram, at the time before and after cutting the cord.

These free body diagrams can be seen in the attached image.

First we perform a sum of forces on the x & y axes before cutting the cord, to be able to find the T tension of the wire. (This analysis can be seen in the attached image).

In this way we get the T-wire tension equation, before cutting.

Now we make another free body diagram, for the moment when the wire is cut (see in the attached diagram).

It is important to clarify that when the cord is cut, the system will no longer be in statically, therefore newton's second law will be used for summation of forces which will be equal to the product of mass by acceleration.

Finally with equations 1 and 2 we can find the K ratio.

5 0
3 years ago
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