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2 years ago

11

Transfer payments are used to redistribute money to various segments of society. please select the best answer from the choices

provided t f

1 answer:

mote1985 [20]2 years ago

7 0

Answer:

True

Explanation:

You pay back what you owe for what youve purchased

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Q 2 Two mirrors meet at right angles. A ray of light is incident on one at an angle of 30°

serg [7]

a ray of light is incident towards a plane mirror at an angles of 30degrees with the mirror surface. what will be the angles of reflection is 60degree.

3 0

3 years ago

kristine speeds past a parked police car at 32 m/s. The police car starts from rest with a uniform acceleration of 2.5 m/s^2. Ho

Digiron [165]

$32 = 0 \times t + \frac{1}{2} \times 2.5 \times t^{2} \\ 32 = 0 + 1.25 \times t {}^{2} \\ 32 = 1.25t {}^{2} \\ \frac{32}{1.25} = \frac{1.25t {}^{2} }{1.25} \\ t {}^{2} = 25.6 \\ \sqrt{t {}^{2} } = \sqrt{25.6} \\ t = 5.1seconds \\$

7 0

3 years ago

One of the main factors driving improvements in the cost and complexity of integrated circuits (ICs) is improvements in photolit

nika2105 [10]

Answer:

0.000003782 m

0.000001891 m

0.000001197125 m

Explanation:

$\lambda$ = Wavelength = 248 nm

D = Diameter of beam = 1 cm

f = Focal length = 0.625 cm

The angle is given by

$\theta=\dfrac{1.22\lambda}{D}$

The width is given by

$d=2\theta f\\\Rightarrow d=2\dfrac{1.22\lambda f}{D}\\\Rightarrow d=2\dfrac{1.22\times 248\times 10^{-9}\times 6.25\times 10^{-2}}{1\times 10^{-2}}\\\Rightarrow d=0.000003782\ m$

The required width is 0.000003782 m

Minimum resolvable line separation is given by

$\dfrac{0.000003782}{2}=0.000001891\ m$

The minimum resolvable line separation between adjacent lines is 0.000001891 m

when $\lambda=157\ nm$

$d=2\dfrac{1.22\times 157\times 10^{-9}\times 6.25\times 10^{-2}}{1\times 10^{-2}}\\\Rightarrow d=0.00000239425\ m$

The new minimum resolvable line separation between adjacent lines is

$\dfrac{0.00000239425}{2}=0.000001197125\ m$

6 0

3 years ago

It's not important to monitor your heart rate during moderate intensity activities. True False

Mama L [17]

I think it is False

hope this helps :3

5 0

3 years ago

Read 2 more answers

A wheel is rotating freely at angular speed 420 rev/min on a shaft whose rotational inertia is negligible. A second wheel, initi

Mashcka [7]

Answer:60 rev/min

Explanation:

Given

angular speed of first shaft $\omega _1=420\ rev/min$

Moment of inertia of second shaft is seven times times the rotational speed of the first i.e. If I is the moment of inertia of first wheel so moment of inertia of second is 7 I

As there is no external torque therefore angular momentum is conserved

$L_1=L_2$

$I_1\omega _1=I_2\omega _2$

$I\times (420)=7 I\times (\omega _2)$

$\omega _2=\frac{420}{7}$

$\omega _2=60\ rev/min$

8 0

3 years ago