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4 years ago

10

Nicholas works as a teacher. In his first year of teaching, Nicholas claims two allowances on his W-4. The second year, he claim

s one allowance. The third year, he does not claim any allowances. Each year Nicholas receives a federal tax refund. Assuming nothing else has changed for Nicholas in those three years, he will get the _____.
largest refund in the first year
smallest refund in the second year
largest refund in the third year
same amount refunded each year

1 answer:

frez [133]4 years ago

7 0

Small-lest refund in the third year .

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A small steel wire of diameter 1.0 mm is connected to an oscillator and is under a tension of 7.5 N. The frequency of the oscill

avanturin [10]

Answer:

a

The output power is $P= 0.764Watt$

b

The Amplitude would decrease by $\frac{1}{2}$

Explanation:

From the question we are told that

The diameter of the steel wire is = 1.0mm $= \frac{1}{1000} = 1.0*10^{-3}m$

The raduis of this steel wire is $r = \frac{1*10^{-3}}{2} = 0.5*10^{-3}m$

Now from the question we can deduce that the power output is equal to the power being transmitted by wave on the wire this is mathematically represented as

$P = \frac{1}{2} \mu w^2 A^2 v ----(1)$

Where $\mu$ is the mass per unit length of the wire

This is mathematically evaluated as

$\mu = a* \rho$

Where a is the area of the the wire = $\pi r^2 = (3.142 * 0.5*10^{-3})^2 =7.855*10^{-7}m^2$

$\rho$ is the density of steel with a generally value of $7850 kg/m^3$

So

$\mu = 7.855*10^{-7} *7850$

$= 6.162*10^{-3}kg/m$

$w$ is velocity of the wave

This is mathematically evaluated as

$w=2 \pi f$

substituting 60Hz for f

We have

$w = 2 *3.142 * 60$

$=377.04 \ rad/s$

$A$ is the amplitude with a given value of 0.50 cm $= \frac{0.50}{100} = 0.50 *10^{-2}m$

v is the linear velocity of the wave

This is mathematically evaluated as

$v = \sqrt{\frac{T}{\mu} }$

Where T is the tension with a given value of $7.5N$

$v = \sqrt{\frac{7.5}{6.162*10^{-3}} }$

$=34.89 m/s$

Substituting values into equation 1

$P = 6.162*10^{-3}* 377.04^2 * (0.5*10^{-2})^2 * 34.89$

$P= 0.764Watt$

Since the doubling of the frequency does not affect the amplitude and from equation one the output power is $\ \frac{1}{2}$ of the Amplitude, Then the Amplitude would decrease by $\frac{1}{2}$

4 0

3 years ago

The force required to maintain an object at a constant speed in free space is equal to __________

rusak2 [61]

The correct answer is: zero.

8 0

3 years ago

Use the graph to answer the questions.

UkoKoshka [18]

Answer:

- decreases

- inverse proportionality

- 1 A

Explanation:

Just take my word for it lol:)

3 0

3 years ago

Read 2 more answers

A charged particle moving along the +x-axis enters a uniform magnetic field pointing along the +z-axis. A uniform electric field

Brums [2.3K]

Answer:

the electric field direction should be in positive y axis

Explanation:

Lets assume that charge on particle is positive and it isequal to +q

First calculate the magnetic force on it

$F_B = q V\times B =qVBsin\theta$

for direction

use Right Hand Rule which will give the direction and by using his the direction will come towards negative y axis.

As given in the question that charge particle does not change their velocity so we need to apply electric field in such a way that electric force direction should be opposite to the magnectic field.

and magnitude should be same as magnectic force and also direcion of electric force depend on the direction of elecric field when charge is positive because electric force $F_E = qE$

Hence the electric field direction should be in positive y axis

3 0

4 years ago

Help me plz do it fastt

PSYCHO15rus [73]

Go to quiz let it helps

5 0

3 years ago