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2 years ago

Suppose a consumption function is given as c = $500 0.75yd. the marginal propensity to consume is:______

1 answer:

7 0

In the the case above, if the consumption function is: C = $500 + 0.75 × YD, from the equation, the The marginal propensity to consume is option C) 0.75.

The marginal propensity to consume which is MPC is the aspect of each extra money that is dollar of a household's income that is said to be used, consumed or spent.

Note that Consumer behavior in regards to saving or spending has a very vital impact on the economy of a country.

From the formula given, the breakdown is:

YD = disposable income.

Autonomous consumption = $500.

The marginal propensity to consume = 0.75.

Therefore, In the the case above, if the consumption function is: C = $500 + 0.75 × YD, from the equation, the The marginal propensity to consume is option C) 0.75.

See full question below

Suppose that the consumption function is: C = $500 + 0.75 × YD

(Scenario: Consumption Spending) The marginal propensity to consume is:

A) $500.

B) 0.

C) 0.75

D) 0.2.

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Which of the following statements about electromagnetic waves in free space are true? (There could be more than one correct choi

Vanyuwa [196]

Answer:

The electric field carries the same mount of energy as the magnetic field

The frequency of the magnetic field is the same as the frequency of the electric field.

Explanation:

Let's analyze each option in detail:

The electric and magnetic fields have equal amplitudes. --> FALSE. In fact, the amplitudes of the electric and magnetic field are related by the equation:

$E=cB$

where

E is the amplitude of the electric field

B is the amplitude of the magnetic field

c is the speed of light

Therefore, from the equation we see that the amplitude of the electric field is much larger than that of the magnetic field.

The electric field carries the same mount of energy as the magnetic field. --> TRUE.

The energy carried by the electric field is:

$u_E = \frac{1}{2}\epsilon_0 E^2$

where $\epsilon_0$ is the vacuum permittivity.

The energy carried by the magnetic field is:

$u_B = \frac{1}{2\mu_0}B^2$

where $\mu_0$ is the vacuum permeability.

Given the following relationship:

$c=\frac{1}{\sqrt{\epsilon_0 \mu_0}}$

We can write

$E=cB=\frac{1}{\sqrt{\epsilon_0 \mu_0}}B$

$u_E = \frac{1}{2}\epsilon_0 (\frac{1}{\sqrt{\epsilon_0 \mu_0}}B)^2=\frac{1}{2\mu_0}B^2$

which means $u_E=u_B$ .

The electric field carries more energy than the magnetic field. --> FALSE, because in disagreement with the calculations above.

The frequency of the magnetic field is the same as the frequency of the electric field. --> TRUE. In an electromagnetic waves, electric field and magnetic field oscillate at the same frequency.

The frequency of the electric field is higher than the frequency of the magnetic field. --> FALSE, because in disagreement with the previous statement.

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3 years ago

What is the mass of the sun plus all the planets in solar system?

Savatey [412]

It’s a very huge mass

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2 years ago

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Name and state the rule to find out the direction of magnetic field inside and around the loop.

goldfiish [28.3K]

Answer:

The right hand rule states that: to determine the direction of the magnetic force on a positive moving charge, ƒ, point the thumb of the right hand in the direction of v, the fingers in the direction of B, and a perpendicular to the palm points in the direction of F.

Explanation:

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3 years ago

a certain hydraulic jack is used for lifting load. a force of 250N is applied on a small piston with a diameter of 80cm while th

anzhelika [568]

The maximum mass of a load that can be lifted by the jack and the distance covered are:

m = 160.2 Kg

h = 25 cm

Given that a certain hydraulic jack is used for lifting load. a force of 250N is applied on a small piston with a diameter of 80cm while the diameter of the larger piston is 2.0m.

The parameters given are

$F_{1}$ = 250

$A_{1}$ = Area of the small piston = π $r^{2}$

$A_{1}$ = 22/7 x $0.4^{2}$

$A_{1}$ = 0.5 $m^{2}$

$F_{2}$ = ?

$A_{2}$ = Area of the large piston = π $r^{2}$

$A_{2}$ = π x 1

$A_{2}$ = 3.14 $m^{2}$

To calculate the force on the large piston, we will use the below formula

$F_{1}$ / $A_{1}$ = $F_{2}$ / $A_{2}$

Substitute all the parameters into the equation

250/0.5 = $F_{2}$ /3.14

$F_{2}$ = 1570 N

To calculate the maximum mass of a load that can be lifted by the jack, let us apply Newton second law

F = mg

1570 = 9.8m

m = 1570/9.8

m = 160.2 Kg

.(take g=9.81ms^-2)

If the applied force moves through a distance of 25cm, the distance through which the load is lifted will be

$F_{1}$ / 0.25 $A_{1}$ = $F_{2}$ / $A_{2}$ h

250/0.125 = 1570/3.14h

make h the subject of the formula

6280h = 1570

h = 1570/6280

h = 0.25 m

Therefore, the distance through which the load is lifted is 25 cm

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2 years ago

How do you think that changing the mass of the pendulum bob will affect the period of the pendulum swing?

myrzilka [38]

(Mass does not affect the pendulum's swing. The longer the length of string, the farther the pendulum falls; and therefore, the longer the period, or back and forth swing of the pendulum. The greater the amplitude, or angle, the farther the pendulum falls; and therefore, the longer the period.)

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3 years ago