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iren2701 [21]
2 years ago
15

The electrical voltage between the ends of a resistor is equal to 8 V. A student has two voltmeters available that measure up to

6 V. The voltages of the voltmeters are equal to 500 ohms and 600 ohms, respectively. What voltages will the connected voltmeters indicate? (Fig. 3.45)​

Physics
1 answer:
Rufina [12.5K]2 years ago
6 0

Answer:

I = V / R

I1 = 6 / 500 = .012   amps for full scale deflection

I2 = 6 / 600 = .01   amps for full scale deflection

I = 8 / 1100 = .00727   amps thru load (actual current)

D1 = .00727 / .012 = .606 of full scale deflection

D2 = .00727 / .01 = .727    of full scale deflection

V1 = .606 * 6 = 3.64 V  (of 6 V)

V2 = .727 * 6 = 4.36 V   (of 6 V)

Check: V1 + V2 = 3.64 + 4.36 = 8 V

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What will be the pressure exerterd by the 0bject if 4000n is acting on an area of 50msqure
Ivahew [28]
<h3>Given, </h3>

Force,F = 4000 N

Area,a = 50 m²

<h3>We know that, </h3>

Pressure = Force/Area

★ Putting the values in the above formula,we get:

\sf \rightarrow \: pressure =  \dfrac{4000}{50}

\sf \rightarrow pressure = 80 \: N {m}^{ - 2}

7 0
3 years ago
An infinite sheet of charge, oriented perpendicular to the x-axis, passes through x = 0. It has a surface charge density σ1 = -2
pishuonlain [190]

Answer:

E_{total}=4.82*10^6N/C

vector with direction equal to the axis X.

Explanation:

We use the Gauss Law and the superposition law in order to solve this problem.

<u>Superposition Law:</u> the Total Electric field is the sum of the electric field of the first infinite sheet and the Electric field of the second infinite sheet:

E_{total}=E_1+E_2

<u>Thanks Gauss Law</u> we know that the electric field of a infinite sheet with density of charge σ is:

E=\sigma/(2\epsilon_o)

Then:

E_{total}=(\sigma_1+\sigma_2)/(2\epsilon_o)=(-2.7*10^{-6}+88*10^{-6})/(2*8.85*10^{-12})=4.82*10^6N/C

This electric field has a direction in the axis perpendicular to the sheets, that means it has the same direction as the axis X.

7 0
3 years ago
Read 2 more answers
the gravitational force between two objects is 1600 and what will be the gravitational force between the objects if the distance
Xelga [282]

I believe this is what you have to do:

The force between a mass M and a point mass m is represented by

F = G\frac{Mm}{r^{2} }

So lets compare it to the original force before it doubles, it would just be the exact formula so lets call that F₁

So F₁ = G(Mm/r^2)

Now the distance has doubled so lets account for this in F₂:

F₂ = G(Mm/(2r)^2)

Now square the 2 that gives you four and we can pull that out in front to give

F₂ = \frac{1}{4} G(Mm/r^2)

Now we can replace G(Mm/r^2) with F₁ as that is the value of the force before alterations

now we see that:

F₂ = \frac{1}{4} F₁

So the second force will be 0.25 (1/4) x 1600 or 400 N.



6 0
3 years ago
Two loudspeakers are 1.60 m apart. A person stands 3.00 m from one speaker and 3.50 m from the other. (a) What is the lowest fre
VMariaS [17]

Answer:

Explanation:

Given

Distance between two loud speakers d=1.6\ m

Distance of person from one speaker x_1=3\ m

Distance of person from second speaker x_2=3.5\ m

Path difference between the waves is given by

x_2-x_1=(2m+1)\cdot \frac{\lambda }{2}

for destructive interference m=0 I.e.

x_2-x_1=\frac{\lambda }{2}

3.5-3=\frac{\lambda }{2}

\lambda =0.5\times 2

\lambda =1\ m

frequency is given by

f=\frac{v}{\lambda }

where v=velocity\ of\ sound\ (v=343\ m/s)

f=\frac{343}{1}=343\ Hz

For next frequency which will cause destructive interference is

i.e. m=1 and m=2

3.5-3=\frac{2\cdot 1+1}{2}\cdot \lambda

\lambda =\frac{1}{3}\ m

frequency corresponding to this is

f_2=\frac{343}{\frac{1}{3}}=1029\ Hz

for m=2

3.5-3=\frac{5}{2}\cdot \lambda

\lambda =\frac{1}{5}\ m

Frequency corresponding to this wavelength

f_3=\frac{343}{\frac{1}{5}}

f_3=1715\ Hz                        

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