This is my exam question be serious

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

s2008m [1.1K]

I'm pretty sure the answer is B: <span>establish public doubt
Hope this helps!</span>

6 0

3 years ago

How much work would it take to push two protons very slowly from a separation of 2.00×10−10m (a typical atomic distance) to 3.00

laiz [17]

Answer:

Work= -7.68×10⁻¹⁴J

Explanation:

Given data

q₁=q₂=1.6×10⁻¹⁹C

r₁=2.00×10⁻¹⁰m

r₂=3.00×10⁻¹⁵m

To find

Work

Solution

The work done on the charge is equal to difference in potential energy

W=ΔU

$Work=U_{1}-U_{2}\\ Work=-kq_{1}q_{2}[\frac{1}{r_{2}}-\frac{1}{r_{1}} ]\\Work=(-9*10^{9})*(1.6*10^{-19} )^{2}[\frac{1}{3.0*10^{-15} }-\frac{1}{2*10^{-10} } ]\\ Work=-7.68*10^{-14}J$

4 0

4 years ago

The air is less dense at higher elevations, so skydivers reach a high terminal speed. The highest recorded speed for a skydiver

swat32

Answer:

the terminal velocity v_t= 202.96 m/s≅203 m/s

Explanation:

The expression for the terminal velocity

$v_t= \sqrt{\frac{2mg}{\rho AC_d} }$

here, C_d is the drag coefficient for the cylinder is 1.15

The surface density of the air at 20°C is

ρ_surface = 1.2041 kg/m^3

the density of air at an altitude of 39000 m

ρ= 4.3/100×39000 = 0.05177 kg/m^3

now substitute these values in equation above

we get

$v_t= \sqrt{\frac{2\times90\times9.81}{0.05177\times0.72\times1.15} }$

v_t= 202.96 m/s≅203 m/s

the terminal velocity v_t= 202.96 m/s≅203 m/s

5 0

4 years ago

Assume that a clay model of a lion has a mass of 0.225 kg and travels on the ice at a speed of 0.85 m/s. It hits another clay mo

Charra [1.4K]

Answer:

Final velocity will be equal to 0.321 m/sec

Explanation:

We have given mass of clay model of lion $m_1=0.225kg$

Its speed is 0.85 m/sec, so $v_1=0.85m/sec$

Mass of another clay model $m_2=0.37kg$

It is given that second clay is motionless

So its velocity $v_2=0m/sec$

Now according to conservation of momentum

Momentum before collision will be equal to momentum after collision

So $m_1v_!+m_2v_2=(m_1+m_2)v$ , here v is velocity after collision

So $0.225\times 0.85+0.37\times 0+(0.225+0.37)v$

$0.595v=0.191$

v = 0.321 m/sec

So final velocity will be equal to 0.321 m/sec

5 0

3 years ago

What gravitational force does the earth exert on a person

lara [203]

The gravitational force exerted by the earth on a person standing on the earth's surface is 602.74 N.

<h3>What is the gravitational force of the earth on the person?</h3>

The gravitational force exerted by the earth on a person standing on the earth's surface is given below as follows:

$F = \frac{Gm^{1}m^{2}}{r^{2}}$

where

G = 6.67 * 10⁻¹¹

m¹ = 62 kg

m² = 5.97 * 10²⁷ kg

r = 6.4 * 10⁶ m

$F = \frac{5.97*10^{24}*62*6.67*10^{-11}}{(6.4*10^{6}){2}} = 602.74\:N$

Therefore, the gravitational force exerted by the earth on a person standing on the earth's surface is 602.74 N.

Learn more about gravitational force at: brainly.com/question/940770

#SPJ1

7 0

3 years ago