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2 years ago

11

In the circuit shown in the figure above, suppose that the value of R1 is 100 k ohms and the value of R2 is 470 k ohms. At which

of the following locations in the circuit would you measure the highest voltage with your meter?
A. Between points A and B
B. Between points A and C
C. Between points B and E
D. Between points B and C

1 answer:

Nesterboy [21]2 years ago

7 0

Answer:

A

Explanation:

voltage between A and C is equal battery's voltage.

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An airplane flying at 115 m/s due east makes a gradual turn following a circular path to fly south. The turn takes 15 seconds to

ankoles [38]

Answer:

The magnitude of the centripetal acceleration during the turn is $a=12.04\ m/s^2.$

Explanation:

Given :

Speed to the airplane in circular path , v = 115 m/s.

Time taken by plane to turn , t= 15 s.

Also , the plane turns from east to south i.e. quarter of a circle .

Therefore, time taken to complete whole circle is , $T=t\times 4=60\ s.$

Now , Velocity ,

$v=\dfrac{2\pi r}{T}\\\\115=\dfrac{2\times 3.14\times r}{60}\\\\r=1098.73\ m.$

Also , we know :

Centripetal acceleration ,

$a=\dfrac{v^2}{r}$

Putting all values we get :

$a=12.04\ m/s^2.$

Hence , this is the required solution .

5 0

1 year ago

A variable-length air column is placed just below a vibrating wire that is fixed at both ends. The length of the air column, ope

pogonyaev

Answer:

The speed of transverse waves in the wire is 234.26 m/s.

Explanation:

Given that,

Length of wire = 129 cm

Speed of sound in air = 340 m/s

First position of resonance = 31.2 cm

We need to calculate the wavelength

For pipe open at one end and closed at other, there is node at closed end and an anti node at open end for 1st resonance.

At 1st resonance,

$L=\dfrac{\lambda}{4}$

$\lambda=4\times L$

Put the value into the formula

$\lambda=4\times31.2\times10^{-2}$

$\lambda=1.248\ m$

We need to calculate the frequency of sound in pipe

Using formula of frequency

$f=\dfrac{v}{\lambda}$

Put the value into the formula

$f=\dfrac{340}{1.248}$

$f=272.4\ Hz$

We need to calculate the node distance

For the wire, there are 3 segments, so 4 nodes

Node-node distance ,

$L=\dfrac{l}{3}$

$L = \dfrac{129}{3}$

$L=43\ cm$

We need to calculate the wavelength of the wire

Using formula of length

$L=\dfrac{\lambda}{2}$

$\lambda=L\times 2$

Put the value into the formula

$\lambda=43\times2$

$\lambda=86\ cm$

We need to calculate the speed of the wave

Using formula of frequency

$f=\dfrac{v}{\lambda}$

$v=f\times\lambda$

Put the value into the formula

$v=272.4\times86\times10^{-2}$

$v=234.26\ m/s$

Hence, The speed of transverse waves in the wire is 234.26 m/s.

4 0

2 years ago

The short vertical parts adjacent to it also reach into the magnetic field and should experience forces. why can we neglect them

hammer [34]

It is not only the horizontal part of the loop that dips into the magnetic field. We can neglect or disregard the horizontal parts of the loop that hollows into the magnetic fields since only the parts perpendicular to the magnetic field complement to it.

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4 0

1 year ago

A computer that is 87% efficient consumes 375 kWh of energy. How much useful energy does it provide?

tekilochka [14]

Answer:

326.25 kWh

Explanation:

Efficiency of a machine is defined as the ratio of useful energy to that of the energy consumed by the machine.

Here, efficiency is given as 87% and the energy consumed by the computer is 375 kWh.

Efficiency, $\eta=\frac{\textrm{Useful energy}}{\textrm{Energy consumed}}$

Plug in the values of $\eta=0.87$ and 375 kWh for energy consumed. Solve for useful energy. This gives,

Efficiency, $\eta=\frac{\textrm{Useful energy}}{\textrm{Energy consumed}}\\ 0.87=\frac{\textrm{Useful energy}}{375}\\ \textrm{Useful energy}=0.87\times 375=326.25 \textrm{ kWh}$

Therefore, the useful energy provided by the computer is 326.25 kWh.

3 0

1 year ago

A resistor with resistance R and an air-gap capacitor of capacitance C are connected in series to a battery (whose strength is "

blsea [12.9K]

Answer:

a) Q = C*emf

b) Reduction in electric field strength and electric potential

c) Initial current through the resistor = emf/R

d) The final charge = K*C*emf

Explanation:

a) The resistors and capacitors are connected in series with the battery

Using Kirchoff's voltage law, sum of all voltages in the circuit is zero

Let $V_{R}$ = Voltage dropped across the Resistor

$V_{c}$ = Voltage dropped across the capacitor

Applying KVL;

$emf - V_{R} - V_{c} = 0\\$ .........................(1)

Since the connection is in series, the same current flow through the circuit

$V_{R} = IR\\Q = CV_{c} \\V_{c} = Q/C$

Putting $V_{c}$ and $V_{R}$ into equation (1)

$emf - IR - Q/C = 0$

At the final charge, the capacitor in fully charged, and current drops to zero due to equilibrium

$I = 0A\\emf = Q/C\\Q = C* emf$

b) Current starts running through the plate because as the sheet of plastic is inserted between the plates both the electric field intensity and the electric potential reduces. The charge also reduces, then current flows

c) The current through the resistor is the current through the entire circuit ( series connection)

$I = I_{o} \exp(\frac{-t}{RC} )\\At time the initial time, t\\t = 0\\ I_{o} = \frac{emf}{R} \\$

Putting the values of t and I₀ into the formula for I written above

$I = \frac{emf}{R} \exp(0)\\I = \frac{emf}{R}$

d) NB: The initial charge on the capacitor = C * emf

The final charge will be:

$Q = K* Q_{initial} \\Q_{initial} = C *emf\\Q_{final} = KCemf$

4 0

2 years ago