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Sveta_85 [38]
3 years ago
8

I need help to know what resemble means for since I’m really having trouble so pls help ASAP

Physics
1 answer:
Marizza181 [45]3 years ago
8 0

Answer:

<em>Resemble means when someone or something has a similar appearance to or qualities in common with a certain person or </em><em>thing.</em>

<em>Hope </em><em>this </em><em>helps </em><em>you,</em><em> </em><em>please</em><em> </em><em>mark</em><em> me</em><em> as</em><em> brainliest</em><em>!</em>

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Sophie [7]

Answer:

2.00m

Explanation:

All we need to do to find the wavelength of the first intereference maximum is subtract both values we are given.

8.00 - 6.00 = 2.00m

Best of Luck!

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3 years ago
What is the speed of a wave with a wavelength of 0.60 m and frequency of 240 Hz
kakasveta [241]

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Explanation:

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3 years ago
CAN SOMEONE PLEASE HELP ME!!?<br> I NEED HELP PLEASE!!
Dmitry [639]
The first one is decreases because there are less molecules in the air.
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3 years ago
There is given an ideal capacitor with two plates at a distance of 3 mm. The capacitor is connected to a voltage source with 12
Licemer1 [7]

The kinematic energy of the positive charge is 2 10⁻⁸ J

This electrostatics exercise must be done in parts, the first part: let's start by finding the charge of the capacitor, the capacitance is defined by

        C = \frac{Q}{\Delta V}

        C = ε₀ \frac{A}{d}

we solve for the charge (Q)

        \frac{Q}{\Delta V} = \epsilon_o \frac{A}{d}

indicates that for the initial point d₁ = 3 mm = 0.003 m and the voltage is DV₁ = 12

         Q = \epsilon_o \  \frac{A \ \Delta V_1 }{d_1}

Now the voltage source is disconnected so the charge remains constant across the ideal capacitor.

For the second part, the condenser is separated at d₂ = 5mm = 0.005 m

         Q = \epsilon_o \  \frac{A \ \Delta V_2 }{d_2}

we match the expressions of the charge and look for the voltage

          \frac{\Delta V_1}{d_1} = \frac{\Delta V_2}{d_2}

          ΔV₂ = \frac{d_2}{d_1 } \ \Delta V_1

The third part we use the concepts of conservation of energy

starting point. With the test load (q = 1 nC = 1 10⁻⁹ C) next to the left plate

          Em₀ = U = q DV₂

          Em₀ = q  \frac{d_2}{d_1 } \ \Delta V_1

           

final point. Proof load on the right plate

         Em_f = K

energy is conserved

         Em₀ = em_f

         q  \frac{d_2}{d_1 } \ \Delta V_1 = K

   

we calculate

         K = 1 10⁻⁹  12  \frac{0.005}{0.003}  

         K = 20 10⁻⁹ J

In this exercise, as the conditions at two different points of separation give, the area of ​​the condenser is not necessary and with conservation of energy we find the final kinetic energy of 2 10⁻⁸ J

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3 years ago
Winner of five swimming gold medals at the 1988 olympics
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Matt Biondi..?
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3 years ago
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