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2 years ago

An ac generator has a maximum output emf of 215 V. What is the rms potential difference?

1 answer:

7 0

The rms potential difference of an ac generator is given by:
$V_{rms} = \frac{V_0}{ \sqrt{2} }$
where $V_0$ is the maximum value of the voltage.

For the ac generator in our problem, $V_0=215 V$ , therefore the rms value of the potential difference is
$V_{rms} = \frac{V_0}{ \sqrt{2} } = \frac{215 V}{ \sqrt{2} }=152 V$

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A rotating beacon is located 2 miles out in the water. Let A be the point on the shore that is closest to the beacon. As the bea

Mice21 [21]

Answer:

$v = 3369.2 m/s$

Explanation:

As we know that Beacon is rotating with angular speed

$f = 10 rev/min$

so we have

$\omega = 2\pi f$

$\omega = 2\pi(\frac{10}{60})$

$\omega = 1.047 rad/s$

now we know that

$v = r \omega$

here we will have

$r = 2 miles = 2(1609 m)$

$r = 3218 m$

so we have

$v = 3218(1.047)$

$v = 3369.2 m/s$

6 0

1 year ago

A toy car is given an initial velocity of 5.0 m/s and experiences a constant acceleration of 2.0 m/s656-03-02-00-00_files/i02900

Whitepunk [10]

It would be 17 m/s

If we use

V2 = V1 + a*t
Sub in 5 for v1
2m/s*2 for a
And
6 for t
That should give you the answer.

5 0

1 year ago

Read 2 more answers

Steam at 0.6 MPa, 200 oC, enters an insulated nozzle with a velocity of 50 m/s. It leaves at a pressure of 0.15 MPa and a veloci

Rudiy27

Answer:

x2 = 0.99

Explanation:

from superheated water table

at pressure p1 = 0.6MPa and temperature 200 degree celcius

h1 = 2850.6 kJ/kg

From energy equation we have following relation

$\dot m( h1+\frac{v1^2}{2}+ gz1 )+ Q = \dot m( h2+\frac{v2^2}{2}+ gz1) + W$

$\dot m( h1+\frac{v1^2}{2}) = \dot m( h2+\frac{v2^2}{2})$

$h1+\frac{v1^2}{2} = h2+\frac{v2^2}{2}$

$2850.6 + [\frac{50^2}{2} * \frac{1 kJ/kg}{1000 m^2/S^2}] = h2 +[ \frac{600^2}{2} * \frac{1 kJ/kg}{1000 m^2/S^2}]$

h2 = 2671.85 kJ/kg

from superheated water table

at pressure p2 = 0.15MPa

specific enthalpy of fluid hf = 467.13 kJ/kg

enthalpy change hfg = 2226.0 kJ/kg

specific enthalpy of the saturated gas hg = 2693.1 kJ/kg

as it can be seen from above value hf>h2>hg, so phase 2 is two phase region. so we have

quality of steam x2

h2 = hf + x2(hfg)

2671.85 = 467.13 +x2*2226.0

x2 = 0.99

6 0

2 years ago

The length of a wire 2.00 m is measured as 2.02m. What is the percentage error in the measurement?

n200080 [17]

Answer:

Explanation:

Percent error can be found by dividing the absolute error (difference between measure and actual value) by the actual value, then multiplying by 100.

$Percent Error=\frac{V_{measured}- V_{true} } {V_{true}} *100$

The measured value is 2.02 meters and the actual value is 2.00 meters.

$V_{measured}=2.02\\\\V_{true}=2.00$

$Percent Error=\frac{2.02-2.00}{2.00} *100$

First, evaluate the fraction. Subtract 2.00 from 2.02

$Percent Error=\frac{0.02}{2.00}*100$

Next, divide 0.02 by 2.00

$PercentError=0.01 *100$

Finally, multiply 0.01 and 100.

$Percent Error=1\\Percent Error= 1 \%$

The percent error is 1%.

6 0

2 years ago

Dylan has two cubes of iron. The larger cube has twice the mass of the smaller cube. He measures the smaller cube. Its mass is 2

liubo4ka [24]

Answer:

The volume of the larger cube is 5.08 g/cm³.

Explanation:

Given that,

Mass of smaller cube = 20 g

Density of smaller cube $\rho= 7.87 g/cm^2$

Dylan has two cubes of iron.

The larger cube has twice the mass of the smaller cube.

$M_{l}=2m_{s}$

Density is same for both cubes because both cubes are same material.

The density is equal to the mass divided by the volume.

$\rho=\dfrac{m}{V}$

$V=\dfrac{m}{\rho}$

Where, V = volume

m = mass

$\rho=density$

We need to calculate the volume of smaller mass

The volume of smaller mass

$V_{s}=\dfrac{m_{s}}{\rho_{s}}$

$V_{s}=\dfrac{20}{7.87}$

$V_{s}=2.54\ cm^3$

Now, We need to calculate the volume of large cube

$V_{l}=\dfrac{m_{l}}{\rho_{l}}$

$V_{l}=\dfrac{2\times20}{7.87}$

$V_{l}=5.08\ g/cm^3$

Hence, The volume of the larger cube is 5.08 g/cm³.

8 0

2 years ago