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3 years ago

Incentives influence people's economic decisions by:

Physics

2 answers:

Serggg [28]3 years ago

7 0

Answer:

making it easier to perform a cost-benefit analysis of a decision.reducing the marginal benefits for a particular choice.

Explanation:

brilliants [131]3 years ago

6 0

Answer: Making certain choices more beneficial than others.

Explanation: Just took test - Apx

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On planet X, the absolute pressure at a depth of 2.00 m below the surface of a liquid nitrogen lake is 5.00 × 105 N/m2 . At a de

eimsori [14]

Answer:

300000.01008 Pa

123.76237 m/s²

Explanation:

$\rho$ = Density of liquid nitrogen = 808 kg/m³

h = Depth

g = Acceleration due to gravity

P = Atmospheric pressure

Absolute Pressure is given by

Below 2 m from surface

$P_a=P+\rho gh\\\Rightarrow 5\times 10^5=P+808g\times 2$

Below 5 m from surface

$P_a=P+\rho gh\\\Rightarrow 8\times 10^5=P+808g\times 5$

Subtracting the above equations we get

$-3\times 10^5=-808g3\\\Rightarrow g=\frac{3\times 10^5}{808\times 3}\\\Rightarrow g=123.76237\ m/s^2$

The acceleration due to gravity on the planet is 123.76237 m/s²

Equating the value of g in the first equation

$5\times 10^5=P+808\times 123.76237\times 2\\\Rightarrow P=5\times 10^5-808\times 123.76237\times 2\\\Rightarrow P=300000.01008\ Pa$

The atmospheric pressure on the planet is 300000.01008 Pa

7 0

4 years ago

A 1200 kg car starts from rest and travels 100m in a time of 10 seconds . A) what is the acceleration of the car? B) what force

SVEN [57.7K]

Given:
Mass (m) = 1200 kg
Distance (s) = 100 m
Time (t) = 10 seconds
Now,
$velocity (v) = \frac{distance}{time}$

= $\frac{100~ m}{10~seconds}$

= 10 m/s
Note that this one is the final velocity.
We also know that,
initial velocity (u) = 0 m/s ....... because the car starts from rest.
Now,
$acceleration (a)= \frac{change~in~velocity}{time}$

= $\frac{v-u}{t}$

= $\frac{10-0}{10}$

= 1 m/s²
Now,
Force (F) = mass (m) * acceleration (a)
= 1200 kg * 1 m/s²
= 1200 kg.m/s²
= 1200 N
Now,
Work Done (W) = Force (F) * displacement (s) ....note that displacement is same as distance.
 = 1200 N * 100 m
 = 120000 N.m
= 120000 J
Now,
Power (P) = $\frac{Work~done(W)}{time(t)}$

= $\frac{120000~J}{10s}$

= 12000 J/s
= 12000 watt
SO,
A) The acceleration of the car is 1 m/s².
B) 1200 Newton (N) force must have acted on the car.
C) The velocity of the car after 10 seconds is 10 m/s.
D) 120000 Joule (J) work was done on the car.
E) The engine produced a minimum power of 12000 watt.

5 0

4 years ago

What is the work of the force F (6.0N)(4.0N)j(-2.0N)k when the object moves from an initial point with coordinates (1.5 m, 3.0m,

Alina [70]

Answer:

option (c) - 10 j

Explanation:

F = (6 i + 4 j - 2 k) N

r1 = (1.5, 3, -4.5) m = (1.5 i + 3j - 4.5 k) m

r2 = (4, -2.5, - 3) m = (4 i - 2.5 j - 3 k) m

displacement, r = r2 - r1 = ( 2.5 i - 5.5 j + 1.5 k) m

Work done is defined as the dot product of force vector and teh displacement vector.

$W = \overrightarrow{F}.\overrightarrow{r}$

W = (6 i + 4 j - 2 k) . ( 2.5 i - 5.5 j + 1.5 k)

W = 15 - 22 - 3 = - 10 J

4 0

3 years ago

Which type of graph is best for analyzing quantitative dependent and independent variables?

34kurt

Answer:

line graph

Explanation:

on edge

8 0

3 years ago

Compare the gravitational force on a 4 kg

Eddi Din [679]

Answer:

The gravitational force by the Earth on the object, and by the object on the Earth is

F = GMm/R²

= 6.674×10−11 m3⋅kg−1⋅s−2 × 6 × 10^24 kg × 44.5 kg/(6.4 × 10^6 m)²

= 435 N

Please note that the ration between the gravitation force 435 and the mass 44.5

should be gravitational acceleration

435/44.5 = 9.78

I attribute the discrepancy between 9.78 and the usual 9.81 to rounding off in the

Earth's weight and radius.

The mass of the Moon is M / 81.3.

The radius of the Moon is R × 0.27.

The gravitational force on the moon would be

G(M/81.3)m/(R×0.27)² = 0.17×GMm/R²

The gravitational force on the moon is smaller by the factor of about 0.17.

6 0

3 years ago