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3 years ago

When a dielectric slab is inserted between the plates of one of the two identical capacitors

Physics

2 answers:

tatiyna3 years ago

4 0

Answer:(A-P,S;B-P,S;C-Q,S;D-P,S)

Solution

(A)→P,S,(B)→P,S,(C)→Q,S,(D)→P,S.

Explanation:

Akimi4 [234]3 years ago

3 0

This could be like 711 u know they be selling hot dogs they good try them

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In which surface does the toy car move easily or fast?why?

statuscvo [17]

sandpaper and gravel surface

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3 years ago

Standing waves are set up on two strings fixed at each end, as shown in the drawing. The two strings have the same tension and m

uysha [10]

The beat frequency produced by the two standing waves is 13 Hz.

<h3>The wavelength of the shorter string</h3>

The wavelength of the shorter string is calculated as follows;

$L = \frac{\lambda}{2} \\\\\lambda = 2L\\\\\lambda = \frac{v}{f} \\\\\lambda = \frac{41.9}{225} \\\\\lambda = 0.186 \ m\\\\\lambda = 18.6 \ cm\\\\L= \frac{\lambda }{2} \\\\L = \frac{18.6 \ cm}{2} = 9.3\ cm$

<h3>The length of the longer string</h3>

$L_2 = 0.58 \ cm \ + 9.3 \ cm\\\\L_2 = 9.88 \ cm \\\\\lambda _2 = 2L_2\\\\\lambda _2 = 2(9.88 \ cm)\\\\\lambda_2 = 19.76 \ cm = 0.1976 \ m$

The frequency of the longer string is calculated as follows;

$v_1 = v_2\\\\f_2 = \frac{v_2}{\lambda_2} \\\\f_2 = \frac{41.9}{0.1976} \\\\f_2 = 212 \ Hz$

<h3>Beat frequency</h3>

The beat frequency produced by the two standing waves is calculated as follows;

$F_b = 225 \ Hz \ - \ 212 \ Hz\\\\F_b = 13 \ Hz$

Learn more about beat frequency here: brainly.com/question/3086912

8 0

3 years ago

Before a collision, a 50.0-kg object is moving at +5.0 m/s. Find the impulse that acted on the object if, after the collision, i

Virty [35]

Answer:

Explanation:

<u>Impulse And Momentum </u>

Suppose a particle is moving at a certain speed $v_1$ and changes it to $v_2$ . The impulse J is equivalent to the change of linear momentum. The momentum can be computed by

$p=mv$

The initial and final momentums are given, respectively, by:

$p_1=mv_1,\ p_2=mv_2$

Thus, the change of momentum is

$\Delta p=p_2-p_1=m(v_2-v_1)$

It's equal to the Impulse J

$J=\Delta p$

$J=m(v_2-v_1)$

Our data is

$m=50\ kg,\ v_1=5\ m/s,\ v_2=17\ m/s$

$J=50\ (17-5)$

$J=600\ kg\ m/s$

7 0

3 years ago

An object accelerates to a velocity of 230 m/s over a time of 2. 5 s. The acceleration it experienced was 42 m/s2. What was its

11Alexandr11 [23.1K]

An object accelerates to a velocity of 230 m/s over a time of 2. 5 s. The acceleration it experienced was 42 m/s2. Its initial velocity will be 125 m/s

final velocity = 230 m/s

time = 2.5 second

acceleration = 42 $m/s^{2}$

initial velocity = ?

acceleration = final velocity - initial velocity / time

42 = (230 - u) / 2.5

u = 125 m/s

Its initial velocity will be 125 m/s

To learn more about acceleration here :

brainly.com/question/12550364

#SPJ4

8 0

1 year ago

A 30 μFμF capacitor initially charged to 30 μCμC is discharged through a 1.9 kΩkΩ resistor. How long does it take to reduce the

Neko [114]

Given Information:

Resistance = R = 1.9 k Ω

Capacitance = C = 30 uF

Initial charge = 30 uC

Final charge = 5 uC

Required Information:

Time taken to reduce the capacitor's charge to 5.0 μC = ?

Answer:

t = 0.101 seconds

Explanation:

The voltage across the capacitor is given by

V = V₀e^(–t/τ)

Where τ is time constant, V₀ is the initial voltage, V is the voltage after some time t

τ = RC

τ = 1900*30x10⁻⁶

τ = 0.057 sec

The initial voltage across the capacitor was

V₀ = Q/C

V₀ = 30/30

V₀ = 1 V

Voltage to reduce the charge to 5 uF

V = 5/30

V = 0.167 V

V = V₀e^(–t/τ)

0.167 = 1*e^(–t/0.057)

take ln on both sides

ln(0.167) = ln(e^(–t/0.057))

-1.789 = -t/0.057

t = 1.789*0.057

t = 0.101 seconds

3 0

3 years ago