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3 years ago

9

Jack and jill are married. this year jack earned $72,000 and jill earned $80,000 and they received $4,000 of interest income fro

m a joint savings account. how much gross income would jack report if he files married-filing-separate from jill?

1 answer:

MaRussiya [10]3 years ago

4 0

<span>If Jack is filing married-filing-separate he would report $76,000 gross income as head of household.</span>

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Suppose that the Mars orbiter was to have established orbit at 155 km and that one group of engineers specified this distance as

MAXImum [283]

Answer:

108 km

Explanation:

The conversion factor between meters and feet is

1 m = 3.28 ft

So the second altitude, written in feet, can be rewritten in meters as

$h_2 = 1.55 \cdot 10^5 ft \cdot \frac{1}{3.28 ft/m}=4.7\cdot 10^4 m$

or in kilometers,

$h_2 = 47 km$

the first altitude in kilometers is

$h_1 = 155 km$

so the difference between the two altitudes is

$\Delta h = 155 km - 47 km = 108 km$

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3 years ago

what is the magnitude of the electric force between charges of 0.25 C and 0.11 C at a separation of 0.88 m? if the separation be

leonid [27]

$F = k \cdot \frac{q_1 q_2}{d} = 9 \cdot 10^{9} \cdot \frac{0.25 \cdot 0.11}{0.88} = 144 \cdot 5^{9} \ N.$

If the separation between the charges is increased then the magnitude of the force will increase in fact how the distance is being used in that formula.

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3 years ago

A 16-slug mass is raised by 10 ft. the PE of the mass increased by?

Eva8 [605]

Answer:

<em>The PE of the mass increased by 6,972.95 J</em>

Explanation:

<u>Gravitational Potential Energy</u>

It's the energy stored in an object because of its vertical position or height in a gravitational field.

It can be calculated with the equation:

U=m.g.h

Where m is the mass of the object, h is the height with respect to a fixed reference, and g is the acceleration of gravity or 9.8 m/s^2.

We are given the mass of m=16 slug raised by a height h=10 ft. Both units will be converted to SI standard:

1 slug = 14.59 Kg, thus

16 slug = 16*14.59 Kg=233.44 Kg

1 ft = 0.3048 m, thus:

10 ft = 10*0.3048 m = 3.048 m

Thus, the PE of the mass increased by:

U = 233.44 * 9.8 * 3.048 = 6,972.95 J

the PE of the mass increased by 6,972.95 J

4 0

3 years ago

A crane motor exerts 3,500 W to lift a load up 5 floors in 25 seconds. How much work did the crane do?

jok3333 [9.3K]

Answer: 87500J

Explanation:

Given that,

Power exerted by crane = 3,500 W

Time taken = 25 seconds

work done by crane = ?

Since power is the rate of work done per unit time, then power is workdone by the crane divided by the time taken.

i.e power = work / time

3,500 W = work / 25 seconds

Work = 3500W x 25 seconds

Work = 87500J

Thus, 87500 joules of work was done by the crane.

4 0

3 years ago

The figure shows the arrangement of master cylinder and slave cylinder of a part of braking system. The master cylinder piston,

Neko [114]

Answer:

a) 35 kPa

d) 140 N

c) 1) Increasing the brake fluid pressure

2) Increasing the slave piston surface area.

Explanation:

The parameters given are;

a) Force applied to the master cylinder piston = 28 N

Cross sectional area of the master cylinder piston = 8 cm² = 8 × 10⁻⁴ m²

The pressure P on the brake fluid is given by the formula for pressure as follows;

$P= \dfrac{Applied \ force}{Area \over \ which \ force \ is \ applied} = \dfrac{28 \, N}{8 \times 10^{-4} \, m^2} = 35,000 \, N/m^2 = 35,000 \, Pa$

The pressure on the brake fluid, P, produced by the master cylinder piston = 35,000 Pa = 35 kPa

b) Given that the area of the slave piston = 40 cm² = 0.004 m², we have from the formula for pressure, P;

$P= \dfrac{Applied \ force}{Area \over \ which \ force \ is \ applied} = \dfrac{Applied \ force}{4 \times 10^{-3} \, m^2} = 35,000 \, N/m^2$

Therefore;

Applied force on the slave piston = 4 × 10⁻³ m² × 35,000 N/m² = 140 N

c) The force, F produced by the slave cylinder piston is given by the relation;

F = Pressure × Area

Therefore, the two ways of increasing the force produced by the slave cylinder piston is as follows;

1) Increasing the pressure in the brake fluid by increasing the force exerted by the master cylinder piston

2) Increasing the surface area of the slave piston.

4 0

3 years ago