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3 years ago

13

The negative will be to the positive. Therefore, they are pulling each other, and creating and electrical

force.

1 answer:

Ierofanga [76]3 years ago

3 0

Answer:

attracted... (not sure) close to...

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Which is a correct step in a scientific experiment involving acids and bases?

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Answer:

b) using an indicator to measure the hydrogen ion concentration of a solution

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4 years ago

A physicist is calibrating a spectrometer that uses a diffraction grating to separate light in order of increasing wavelength (λ

cluponka [151]

Answer:

Explanation:

Given

$N=3680 cm^{-1}$

therefore slit spacing $d=\frac{1}{N}=\frac{1}{3680}=2.717\times 10^{-4}\ cm$

since $d\sin \theta =n\lambda$

for $n=1$

$d\sin \theta =\lambda$

Now,at $\theta _1=12.9^{\circ},\Rightarrow \lambda _1=6.0657\times 10^{-7}\ m=606.57\ nm$

at $\theta _2=14.2^{\circ}\Rightarrow \lambda_2=666.5\ nm$

at $\theta _3=15^{\circ}\Rightarrow \lambda_3=703.21\ nm$

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3 years ago

If a hockey puck is made to slide along a straight line on a completely frictionless surface, which statement about the motion o

77julia77 [94]

C. the hockey puck will accelerate and continue moving in a straight line

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3 years ago

Read 2 more answers

A 0.5 μF and a 11 μF capacitors are connected in series. Then the pair are connected in parallel with a 1.5 μF capacitor. What i

NISA [10]

Answer:

$C_{eq}=1.97\ \mu F$

Explanation:

Given that,

Capacitance 1, $C_1=0.5\ \mu F$

Capacitance 2, $C_2=11\ \mu F$

Capacitance 3, $C_3=1.5\ \mu F$

C₁ and C₂ are connected in series. Their equivalent is given by :

$\dfrac{1}{C'}=\dfrac{1}{C_1}+\dfrac{1}{C_2}$

$\dfrac{1}{C'}=\dfrac{1}{0.5}+\dfrac{1}{11}$

$C'=0.47\ \mu F$

Now C' and C₃ are connected in parallel. So, the final equivalent capacitance is given by :

$C_{eq}=C'+C_3$

$C_{eq}=0.47+1.5$

$C_{eq}=1.97\ \mu F$

So, the equivalent capacitance of the combination is 1.97 micro farad. Hence, this is the required solution.

3 0

3 years ago

Two parallel conducting plates are separated by 10.0 cm, and one of them is taken to be at zero volts. (a) what is the electric

Alex_Xolod [135]

The expression for the magnitude of the electric field between two uniform conducting plates is

$E=\frac{V}{d}$

Here, V is potential difference between plates and d is separation between plates.

As the potential 6.00 cm from the zero volt plate (and 4.00 cm from the other) is 420 V.

Therefore,

$E = \frac{420 \ V}{6 \times 10^{-2} m } = 70 \times 10^2 \ V/m$

Thus, the electric field strength between the plates is 7000 V/ m

8 0

3 years ago