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3 years ago

9

(I speak portuguese, but I will make a question) Se eu estou num carro, com velocidade constante de 100km/h e há uma abelha dent

ro do carro, ela está voando a 100km/h também?

2 answers:

denis23 [38]3 years ago

8 0

Yes the bee is flying with the car at 100km/h

dusya [7]3 years ago

3 0

Sim a abelha esta movendo na mesma velocidade do carro

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Is the refraction different entering medium that has a higher index of refraction compared to entering amedium that has a lower

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Answer:

Yes

Explanation:

The speed of light when it travels through glass, diamond, etc, the light travels at different speed from the speed of light. Speed of the light in material is related to the index of refraction.

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<u>When the light goes into a medium with the higher index of the refraction, light bends towards normal. Conversely, if the light traveling goes from higher refractive index to lower refractive index, it will bend away from the normal. </u>

<u>Hence, the refraction is different in both the scenario.</u>

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How is a conducter different from an insulater

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Conductors have magnetic fields; insulators do not have magnetic fields. Conductors do not have magnetic fields; insulators do have magnetic fields. ... In a conductor, electric current cannot flow freely; in an insulator, it can flow freely.

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2 years ago

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Calculate the mass of an object with a density of 102.5 g/mL and volume of 375 mL.

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Answer:

38,437.5

Explanation:

Density(d)= 102.5g/ml

Volume (v)=375ml

Mass(m) = ?

D =m/v

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3 years ago

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

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Answer:

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about 1250 km

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Explanation:

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