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Viktor [21]
3 years ago
12

A code of ethics for psychologists helps to do all of the following except __________.

Physics
2 answers:
s2008m [1.1K]3 years ago
6 0
I'm pretty sure the answer is B: <span>establish public doubt
Hope this helps!</span>
natulia [17]3 years ago
4 0
Answer: B


I hope this helps!
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A large bottle contains 150 L of water, and is open to the atmosphere. If the bottle has a flat bottom with an area of 2 ft, cal
Yuri [45]

Answer:

Total pressure exerted at bottom =  119785.71 N/m^2

Explanation:

given data:

volume of water in bottle = 150 L = 0.35 m^3

Area of bottle = 2 ft^2

density of water = 1000 kg/m

Absolute pressure on bottom of bottle will be sum of atmospheric pressure and pressure due to water

Pressure due to water P = F/A

F, force exerted by water = mg

m, mass of water = density * volume

                             =  1000*0.350 = 350 kg

F  = 350*9.8 = 3430 N

A = 2 ft^2 = 0.1858 m^2  

so, pressure P = 3430/ 0.1858 = 18460.71 N/m^2

Atmospheric pressure

At sea level atmospheric pressure is 101325 Pa

Total pressure exerted at bottom  = 18460.71 + 101325 = 119785.71 N/m^2

Total pressure exerted at bottom =  119785.71 N/m^2

6 0
4 years ago
"KATZPSEF1 7.P.053.MI. MY NOTES ASK YOUR TEACHER PRACTICE ANOTHER Two black holes (the remains of exploded stars), separated by
eduard

Answer:

There are two possible solutions.

M1 = 4.68*10^30kg,  M2 = 5.53*10^30kg

M1 = 5.53*10^30kg,  M2 = 4.7*10^29kg

Explanation:

In order to find the mass of each black hole, you take into account the gravitational force between them and the sum of their masses.

You use the formula for the gravitational force between two masses:

F_g=G\frac{M_1M_2}{r^2}              (1)

G: Cavendish's constant = 6.674*10^-11 m^3/kg.s^2

M1, M2: mass of each black hole = ?

r: distance between the black holes = 10.0 AU = 10.0(1.50*10^11m) = 1.5*10^12m

Fg: gravitational force between the black holes = 7.70*10^25N

Furthermore, you take into account that the sum of the masses M1 and M2 is:

M1 + M2 = 6.00*10^30 kg        (2)

You solve the equation (2) for M2.

M_2=6.00*10^{30}-M_1

Next, you replace the obtained expression for M2 into the equation (1) and solve for M1, as follow (for simplicity, you do not add the units):

F_g=G\frac{M_1(6.00*10^{30}-M_1)}{r^2}\\\\\frac{r^2F_g}{G}=6.00*10^{30}M_1-M_1^2\\\\\frac{(1.5*10^{12})^2(7.70*10^{25})}{(6.674*10^{-11}}=6.00*10^{30}M_1-M_1^2\\\\2.59*10^{60}=6.00*10^{30}M_1-M_1^2\\\\M_1^2-6.00*10^{30}M_1+2.59*10^{60}=0

Then, you have obtained a quadratic polynomial. You solve it with the quadratic formula:

M_1=\frac{-(-6.00*10^{30})\pm \sqrt{(-6.00*10^{30})^2-4(1)(2.59*10^{60}))}}{2(1)}\\\\M_1=\frac{6.00*10^{30}\pm 5.06*10^{30}}{2}\\\\M_1=4.68*10^{29}\\\\M_1=5.53*10^{30}

Both results are consistent, then the mass of one black hole can be 4.68*10^30kg and also 5.53*10^30kg.

The other black hole has a mass of:

M_2=6.00*10^{30}kg-4.68*10^{29}kg=5.53*10^{30}kg\\\\M_2=6.00*10^{30}kg-5.53*10^{30}kg=4.7*10^{29}kg

Hence, you have a pair of solutions:

M1 = 4.68*10^30kg,  M2 = 5.53*10^30kg

M1 = 5.53*10^30kg,  M2 = 4.7*10^29kg

3 0
3 years ago
Which is this the correct variable for the equation "____= displacement divided by time"?
zheka24 [161]
The answer is velocity
3 0
3 years ago
the base of a rectangular vessel measure 10m by 18cm. water is poured into a depth of 4cm. (a) what is the pressure on the base?
Alex787 [66]

Answer:

a) P =392.4[Pa]; b) F = 706.32[N]

Explanation:

With the input data of the problem we can calculate the area of the tank base

L = length = 10[m]

W = width = 18[cm] = 0.18[m]

A = W * L = 0.18*10

A = 1.8[m^2]

a)

Pressure can be calculated by knowing the density of the water and the height of the water column within the tank which is equal to h:

P = density * g *h

where:

density = 1000[kg/m^3]

g = gravity = 9.81[m/s^2]

h = heigth = 4[cm] = 0.04[m]

P = 1000*9.81*0.04

P = 392.4[Pa]

The force can be easily calculated knowing the relationship between pressure and force:

P = F/A

F = P*A

F = 392.4*1.8

F = 706.32[N]

4 0
3 years ago
Which component is used to change alternating current to direct current
Ad libitum [116K]
A rectifier is used.
3 0
3 years ago
Read 2 more answers
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