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Phoenix [80]
3 years ago
15

One of the checks that you could do for problem 1) would be to check the output resistance of the Wheatstone bridge to make sure

the bridge is not damaged. With no power connected to the bridge, you connect anOhmmeter to the outputpins. If each arm(branch)of the Wheatstone Bridge is 350Ω, what should the resistance read and why? (Draw the circuit and show your calculations.)
Physics
1 answer:
Ray Of Light [21]3 years ago
6 0

Answer:

Explanation:

Let the four resistances of th wheat stone bridge is

P, Q, R and S and the value of each is 350 ohm.

Here, P and Q are in series.

R' = P + Q = 350 + 350 = 700 ohm

Then R and S are in series

R' = R + S = 350 + 350 = 700 ohm

Now R' and R'' are in parallel.

So, the equivalent resistance is

Req = R' x R'' / ( R' + R'')

Req = 700 / 2 = 350 ohm

Thus, the reading of ohmmeter is 350 ohm.

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A car with mass m traveling at speed v has kinetic energy k. what is the kinetic energy of a second car that has the same mass m
vampirchik [111]
Kinetic energy, KE, is modeled by the formula KE =  \frac{1}{2}mv^2, where m is the mass in kg and v is the velocity in m/s.

In this scenario, mass and one-half are constant but the velocity changes. 

You can see that by squaring twice the velocity, that is equal to four times the original KE. Therefore, the answer is 4k.
7 0
3 years ago
the total positive charge is QQQ = 1.62×10−6 CC , what is the magnitude of the electric field caused by this charge at point P,
balu736 [363]

Answer:

6.1 × 10^9 Nm-1

Explanation:

The electric field is given by

E= Kq/d^2

Where;

K= Coulombs constant = 9.0 × 10^9 C

q = magnitude of charge = 1.62×10−6 C

d = distance of separation = 1.53 mm = 1.55 × 10^-3 m

E= 9.0 × 10^9 × 1.62×10−6/(1.55 × 10^-3 )^2

E= 14.58 × 10^3/2.4 × 10^-6

E= 6.1 × 10^9 Nm-1

8 0
3 years ago
the maximum intensity levels of a trumpet, trombone, and a bass drum, each at a distance of 3m are 94 dB, 107dB, and 113dB respe
Gwar [14]

Answer:

β = 114 db

Explanation:

The intensity of sound in decibles is

          β = 10 log \frac{I}{I_{o}}

in most cases Io is the hearing threshold 1 10-12 W / cm²

let's calculate the intensity of each instrument

            I / I₀ = 10 (β / 10)

            I = I₀ 10 (β / 10)

trumpet

            I1 = 1 10⁻¹² 10 (94/10)

            I1 = 2.51 10⁻³ / cm²

Thrombus

           I2 = 1 10⁻¹² 10 (107/10)

           I2 = 5.01 10-2 W / cm²

low

           I3 =1 1-12    (113/10) W/cm²

            I3 = 1,995 10-1 W / cm²

when we place the three instruments together their sounds reinforce

           I_total = I₁ + I₂ + I₃

           I_ttoal = 2.51 10-3 + 5.01 10-2 + ​​1.995 10-1

           I_total = 0.00251 + 0.0501 + 0.1995

           I_total = 0.25211 W / cm²

let's bring this amount to the SI system

         β = 10 log (0.25211 / 1 10⁻¹²)

           β = 114 db

7 0
3 years ago
The magnetic field in a solenoid is . A circular wire of radius 8 cm is concentric with a solenoid of radius 2 cm and length d =
defon

Answer:

6.03 mV

Explanation:

length of solenoid, L = 2 m, N = 12000, di/dt = 40 A/s,

Magnetic field due to solenoid

B = μ0 n i = μ0 N i / L

dB/dt = μ0 N / L x di / dt

dB /dt = (4 x 3.14 x 10^-7 x 12000 x 40) / 2 = 0.3 T/s

Induced emf, e = rate of change of magnetic flux

e = dΦ / dt = A x dB / dt

e = 3.14 x 0.08 x 0.08 x 0.3 = 6.03 x 10^-3 V = 6.03 mV

7 0
3 years ago
A wire loop with 70 turns is formed into a square with sides of length ???? . The loop is in the presence of a 1.20 T uniform ma
Oksi-84 [34.3K]

Answer:

l= 3.002 cm

Explanation:

Given that

n= 70 turns

B= 1.2 T

θ= 15°

I= 1.5 A

τ = 0.0294 N⋅m

Lets take length of sides is l.

We know that

τ  = n I A B sin θ

Area of square ,A= l²

Now by putting the value

τ  = n I A B sin θ

0.0294  = 70 x  1.5 x  l² x  1.2 x  sin 15°

l² = 0.000901 m²

l² = 9.01 cm²

l= 3.002 cm

3 0
4 years ago
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