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3 years ago

6

Which expression should appear in the blue box to compute percent error? Accepted value – Experimental value Accepted value × Ex

perimental value Experimental value + Accepted value Experimental value – Accepted value

1 answer:

Alexxx [7]3 years ago

5 0

The last choice ... Experimental value -- <span>Accepted value ... is the place
to start, but it doesn't quite get you there.

To find the FRACTIONAL error, you'd have to divide that difference
by the Accepted Value.

And then . . .

To find the PERCENTAGE error, you'd have to multiply
the fractional error by 100 .
</span>

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Calculate the phase angle (in radians) for a circuit with a maximum voltage of 12 V and w-50 Hz. The voltage source is connected

Vinvika [58]

Answer:

The phase angle is 0.0180 rad.

(c) is correct option.

Explanation:

Given that,

Voltage = 12 V

Angular velocity = 50 Hz

Capacitance $C= 20\times10^{-2}\ F$

Inductance $L=20\times10^{-3}\ H$

Resistance $R= 50\ Omega$

We need to calculate the impedance

Using formula of impedance

$z=\sqrt{R^2+(\omega L-\dfrac{1}{\omega C})^2}$

$z=\sqrt{50^2+(50\times20\times10^{-3}-\dfrac{1}{50\times20\times10^{-2}})^2}$

$z=50.00$

We need to calculate the phase angle

Using formula of phase angle

$\theta=\cos^{-1}(\dfrac{R}{z})$

$\theta=\cos^{-1}(\dfrac{50}{50.00})$

$\theta=0.0180\ rad$

Hence, The phase angle is 0.0180 rad.

3 0

3 years ago

Learning Goal: To understand the behavior ofthe electric field at the surface of a conductor, and itsrelationship to surface cha

Ivan

Complete Question

The complete question is shown on the first uploaded image

Answer:

a it is always zero

b 0

c $\eta = -\epsilon _o E$

Explanation:ss

Here the net charge is on the outer surface of the conductor thus this means that the net charge inside the conductor is zero

Generally the charge density of a conductor is dependent on the charge per unit area which implies that the charge density is dependent on the net charge so this means that the charge density inside the conductor is zero

Generally the direction of electric field this from the positive charge to the negative charge so from the question we can deduce that the negative charge is located on the surface of the conductor

So We can mathematically define the charge density on the surface of the electric field as

∮ $E \cdot dA = \frac{-Q}{\epsilon _o}$

Where E is the electric field

$dA$ change in unit area

$-Q$ is the negative charge

$\epsilon _o$ is the permittivity of free space

So

$EA = \frac{-Q}{\epsilon _o }$

$\frac{Q}{A} = -\epsilon _o E$

$\eta = -\epsilon _o E$

Where $\eta$ is the charge density

8 0

3 years ago

Nitrogen at 100 kPa and 25oC in a rigid vessel is heated until its pressure is 300 kPa. Calculate (a) the work done and (b) the

nignag [31]

Answer:

A. The work done during the process is W = 0

B. The value of heat transfer during the process Q = 442.83 $\frac{KJ}{kg}$

Explanation:

Given Data

Initial pressure $P_{1}$ = 100 k pa

Initial temperature $T_{1}$ = 25 degree Celsius = 298 Kelvin

Final pressure $P_{2}$ = 300 k pa

Vessel is rigid so change in volume of the gas is zero. so that initial volume is equal to final volume.

⇒ $V_{1}$ = $V_{2}$ ------------- (1)

Since volume of the gas is constant so pressure of the gas is directly proportional to the temperature of the gas.

⇒ P ∝ T

⇒ $\frac{P_{2} }{P_{1}}$ = $\frac{T_{2} }{T_{1}}$

⇒ Put all the values in the above formula we get the final temperature

⇒ $T_{2}$ = $\frac{300}{100}$ × 298

⇒ $T_{2}$ = 894 Kelvin

(A). Work done during the process is given by W = P × ( $V_{2} -V _{1}$ )

From equation (1), $V_{1}$ = $V_{2}$ so work done W = P × 0 = 0

⇒ W = 0

Therefore the work done during the process is zero.

Heat transfer during the process is given by the formula Q = m $C_{v}$ ( $T_{2}$ - $T_{1}$ )

Where m = mass of the gas = 1 kg

$C_{v}$ = specific heat at constant volume of nitrogen = 0.743 $\frac{KJ}{kg k}$

Thus the heat transfer Q = 1 × 0.743 × ( 894- 298 )

⇒ Q = 442.83 $\frac{KJ}{kg}$

Therefore the value of heat transfer during the process Q = 442.83 $\frac{KJ}{kg}$

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SpyIntel [72]

Simple the answer is A and only 2 ones

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What does the area under a speed-time graph represent

Tom [10]

Answer: It represents the whole distance traveled. Hope this helps!

Explanation:

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